Tag: CONTINUITY AND DIFFERENTIABILITY

Exercise-Miscellaneous, Class 12th, Maths, Chapter 5, NCERT (11-22)
Question 11.

$y = x^{(x^{2} - 3)} + (x - 3)^{x^{2}}, x > 3$

Formula Used

To differentiate $a^{u (x)}$ , where base is $a (x)$ or exponent is a function:

When both base and exponent are functions of x:

$\frac{d}{d x} (f (x)^{g (x)}) = f (x)^{g (x)} [g^{'} (x) \ln f (x) + g (x) \frac{f^{'} (x)}{f (x)}]$

We use logarithmic differentiation.

Solution

Term 1: $x^{x^{2} - 3}$

Let $u = x$ , $v = x^{2} - 3$

$\frac{d}{d x} (x^{x^{2} - 3}) = x^{x^{2} - 3} [(2 x) \ln x + \frac{x^{2} - 3}{x}]$ Term 2: $(x - 3)^{x^{2}}$

Let $u = x - 3$ , $v = x^{2}$

$\frac{d}{d x} ((x - 3)^{x^{2}}) = (x - 3)^{x^{2}} [(2 x) \ln (x - 3) + \frac{x^{2}}{x - 3}]$

Final Answer

$\frac{d y}{d x} = x^{x^{2} - 3} (2 x \ln x + \frac{x^{2} - 3}{x}) + (x - 3)^{x^{2}} (2 x \ln (x - 3) + \frac{x^{2}}{x - 3})$

Question 12

Find $\frac{d y}{d x}$ , if $y = 12 (1 - \cos t), x = 10 (t - \sin t), - \frac{π}{2} < t < \frac{π}{2}$

Solution:

For parametric equations: $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Step 1: Differentiate y w.r.t. t

$y = 12 (1 - \cos t)$

$\frac{d y}{d t} = 12 (0 + \sin t) = 12 \sin t$

Step 2: Differentiate x w.r.t. t

$x = 10 (t - \sin t)$

$\frac{d x}{d t} = 10 (1 - \cos t) = 10 (1 - \cos t)$

Step 3: Substitute in formula

$\frac{d y}{d x} = \frac{12 \sin t}{10 (1 - \cos t)}$

We have: $\frac{d y}{d x} = \frac{6 \sin t}{5 (1 - \cos t)}$

Use trigonometric identities

$1 - \cos t = 2 \sin^{2} \frac{t}{2}$

Substitute these in: $\frac{d y}{d x} = \frac{6 (2 \sin \frac{t}{2} \cos \frac{t}{2})}{5 \cdot 2 \sin^{2} \frac{t}{2}}$

Cancel 2 from numerator and denominator:

$\frac{d y}{d x} = \frac{6 (\sin \frac{t}{2} \cos \frac{t}{2})}{5 \sin^{2} \frac{t}{2}}$

Simplify:

$\frac{d y}{d x} = \frac{6}{5} \cdot \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}$

Final Answer

$\frac{d y}{d x} = \frac{6}{5} \cot \frac{t}{2}$

Question 13

Find $\frac{d y}{d x}$ , if

$y = \sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}, 0 < x < 1$

Solution

Differentiate term by term.

Term 1: $\sin^{- 1} x$

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

Term 2: $\sin^{- 1} \sqrt{1 - x^{2}}$

Let $u = \sqrt{1 - x^{2}} = (1 - x^{2})^{1 / 2}$

$\frac{d u}{d x} = \frac{1}{2} (1 - x^{2})^{- 1 / 2} (- 2 x) = \frac{- x}{\sqrt{1 - x^{2}}}$

Now,

$\frac{d}{d x} (\sin^{- 1} u) = \frac{1}{\sqrt{1 - u^{2}}} \cdot \frac{d u}{d x}$

But,

$u^{2} = (1 - x^{2}) \Rightarrow 1 - u^{2} = 1 - (1 - x^{2}) = x^{2}$

So,

$\frac{d}{d x} (\sin^{- 1} \sqrt{1 - x^{2}}) = \frac{1}{\sqrt{x^{2}}} \cdot \frac{- x}{\sqrt{1 - x^{2}}}$

For $0 < x < 1$ , $\sqrt{x^{2}} = x$

$\frac{d}{d x} (\sin^{- 1} \sqrt{1 - x^{2}}) = \frac{- x}{x \sqrt{1 - x^{2}}} = \frac{- 1}{\sqrt{1 - x^{2}}}$

Combine both derivatives

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} = 0$

Final Answer

$\frac{d y}{d x} = 0$

Question 14

If

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0, - 1 < x < 1$

prove that $\frac{d y}{d x} = - \frac{1}{(1 + x)^{2}}$

Solution

Differentiate both sides w.r.t. $x$ :

$\frac{d}{d x} (x \sqrt{1 + y}) + \frac{d}{d x} (y \sqrt{1 + x}) = 0$

Use product rule for each term

First term:

$\sqrt{1 + y} + x \cdot \frac{1}{2 \sqrt{1 + y}} \cdot \frac{d y}{d x}$

Second term:

$\frac{d y}{d x} \sqrt{1 + x} + y \cdot \frac{1}{2 \sqrt{1 + x}}$

Put them together:

$\sqrt{1 + y} + \frac{x}{2 \sqrt{1 + y}} \frac{d y}{d x} + \sqrt{1 + x} \frac{d y}{d x} + \frac{y}{2 \sqrt{1 + x}} = 0$

Now group $\frac{d y}{d x}$ terms:

$\frac{d y}{d x} (\frac{x}{2 \sqrt{1 + y}} + \sqrt{1 + x}) = - \sqrt{1 + y} - \frac{y}{2 \sqrt{1 + x}}$

Use original equation to simplify

Given:

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ $y \sqrt{1 + x} = - x \sqrt{1 + y}$

Divide both sides by $2 \sqrt{1 + x}$ :

$\frac{y}{2 \sqrt{1 + x}} = - \frac{x \sqrt{1 + y}}{2 (1 + x)}$

Substitute this into RHS:

$- \sqrt{1 + y} + \frac{x \sqrt{1 + y}}{2 (1 + x)} = - \sqrt{1 + y} (1 - \frac{x}{2 (1 + x)})$ $= - \sqrt{1 + y} \frac{2 + x}{2 (1 + x)}$

Simplify LHS expression

$\frac{x}{2 \sqrt{1 + y}} + \sqrt{1 + x} = \frac{x + 2 \sqrt{(1 + x) (1 + y)}}{2 \sqrt{1 + y}}$

But from original equation,

$\sqrt{(1 + x) (1 + y)} = - \frac{x \sqrt{1 + y}}{y}$

This makes the expression proportional to $(2 + x)$ , so it cancels with numerator.

So: $\frac{d y}{d x} = \frac{- \sqrt{1} \frac{2 + x}{2 (1 + x)}}{\frac{2 + x}{2 (1 + x)}}$ Final Answer

$\frac{d y}{d x} = - \frac{1}{(1 + x)^{2}}$

Question 15

For the curve

$(x - a)^{2} + (y - b)^{2} = c^{2}, c > 0$

prove that

$\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}}$

is a constant independent of $a$ and $b$ .

Solution

The given equation represents a circle with center $(a, b)$ and radius $c$ .

$(x - a)^{2} + (y - b)^{2} = c^{2}$

Differentiate w.r.t. $x$ :

First derivative

$2 (x - a) + 2 (y - b) \frac{d y}{d x} = 0$

$(x - a) + (y - b) \frac{d y}{d x} = 0$

$\frac{d y}{d x} = - \frac{x - a}{y - b}$

Second derivative

Differentiate again w.r.t. $x$ using quotient rule:

$\frac{d}{d x} (\frac{d y}{d x}) = - \frac{(y - b) - (x - a) \frac{d y}{d x}}{(y - b)^{2}}$

Substitute $\frac{d y}{d x} = - \frac{x - a}{y - b}$ :

$\frac{d^{2} y}{d x^{2}} = - \frac{(y - b) + \frac{(x - a)^{2}}{y - b}}{(y - b)^{2}}$

Take LCM in numerator:

$\frac{d^{2} y}{d x^{2}} = - \frac{(y - b)^{2} + (x - a)^{2}}{(y - b)^{3}}$

But from the original equation:

$(x - a)^{2} + (y - b)^{2} = c^{2}$

So: $\frac{d^{2} y}{d x^{2}} = - \frac{c^{2}}{(y - b)^{3}}$

Compute $1 + {(\frac{d y}{d x})}^{2}$

$1 + {(\frac{d y}{d x})}^{2} = 1 + {(\frac{x - a}{y - b})}^{2} = \frac{(y - b)^{2} + (x - a)^{2}}{(y - b)^{2}} = \frac{c^{2}}{(y - b)^{2}}$

Raise to power $3 / 2$ :

${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = {(\frac{c^{2}}{(y - b)^{2}})}^{3 / 2} = \frac{c^{3}}{∣ y - b ∣^{3}}$

Now evaluate the required expression

$\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = \frac{\frac{c^{3}}{∣ y - b ∣^{3}}}{\frac{- c^{2}}{(y - b)^{3}}} = \frac{c^{3}}{c^{2}} \cdot - 1 = - c$

Final Proven Result

$\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = - c$

Conclusion
- The expression is a constant.
- It is independent of $a$ and $b$ (center of the circle).
- The constant equals the radius $c$ (up to sign).
Question 16

If

$\cos y = x \cos (a + y), \cos a \neq \pm 1$

prove that

$\frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\sin a}$

Solution

Given:

$\cos y = x \cos (a + y)$ Differentiate both sides with respect to $x$ :

LHS

$\frac{d}{d x} (\cos y) = - \sin y \cdot \frac{d y}{d x}$

RHS

$\frac{d}{d x} (x \cos (a + y)) = 1 \cdot \cos (a + y) + x \cdot \frac{d}{d x} (\cos (a + y))$

Now,

$\frac{d}{d x} [\cos (a + y)] = - \sin (a + y) \cdot \frac{d y}{d x}$

So RHS becomes:

$\cos (a + y) - x \sin (a + y) \frac{d y}{d x}$

Now equate derivatives

$- \sin y \frac{d y}{d x} = \cos (a + y) - x \sin (a + y) \frac{d y}{d x}$

Group $\frac{d y}{d x}$ terms:

$- x \sin (a + y) \frac{d y}{d x} + \sin y \frac{d y}{d x} = \cos (a + y)$

$\frac{d y}{d x} (\sin y - x \sin (a + y)) = \cos (a + y)$

Use original equation for substitution

Original:

$\cos y = x \cos (a + y)$

So: $x = \frac{\cos y}{\cos (a + y)}$

Substitute in the factor:

$\sin y - x \sin (a + y) = \sin y - \frac{\cos y}{\cos (a + y)} \sin (a + y)$

$= \frac{\sin y \cos (a + y) - \cos y \sin (a + y)}{\cos (a + y)}$ Use identity:

$\sin y \cos (a + y) - \cos y \sin (a + y) = \sin (y - (a + y)) = \sin (- a) = - \sin a$

So: $\sin y - x \sin (a + y) = \frac{- \sin a}{\cos (a + y)}$

Final step

$\frac{d y}{d x} = \frac{\cos (a + y)}{- \sin a / \cos (a + y)} = \frac{\cos^{2} (a + y)}{\sin a}$

Question 17

If

$x = a (\cos t + t \sin t), y = a (\sin t - t \cos t)$

find $\frac{d^{2} y}{d x^{2}}$

Solution

Step 1: First derivatives w.r.t. $t$

$x = a (\cos t + t \sin t)$

$\frac{d x}{d t} = a (- \sin t + \sin t + t \cos t) = a t \cos t$

$y = a (\sin t - t \cos t)$

$\frac{d y}{d t} = a (\cos t - (\cos t - t \sin t)) = a t \sin t$

Step 2: First derivative $\frac{d y}{d x}$

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a t \sin t}{a t \cos t} = \tan t (t \neq 0)$

Step 3: Second derivative $\frac{d^{2} y}{d x^{2}}$

Formula:

$\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$

$\frac{d}{d t} (\tan t) = \sec^{2} t$ $\frac{d x}{d t} = a t \cos t$

So: $\frac{d^{2} y}{d x^{2}} = \frac{\sec^{2} t}{a t \cos t}$

Final Answer

$\frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} t}{a t}$ or equivalently:

$\frac{d^{2} y}{d x^{2}} = \frac{1}{a t \cos^{3} t}$

since $\sec^{2} t = \frac{1}{\cos^{2} t}$

Question 18

If

$f (x) = ∣ x ∣^{3}$

show that $f^{''} (x)$ exists for all real $x$ and find it.

Solution

First, rewrite the function without modulus

$∣ x ∣ = {\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}$

So: $∣ x ∣^{3} = {\begin{cases} x^{3}, & x \geq 0 \\ (- x)^{3} = - x^{3}, & x < 0 \end{cases}$

Thus $f (x) = {\begin{cases} x^{3}, & x \geq 0 \\ - x^{3}, & x < 0 \end{cases}$

First derivative

$f^{'} (x) = {\begin{cases} 3 x^{2}, & x > 0 \\ - 3 x^{2}, & x < 0 \end{cases}$

Check at $x = 0$ :

$f^{'} (0) = \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = \lim_{x \to 0} \frac{∣ x ∣^{3}}{x} = \lim_{x \to 0} ∣ x ∣^{2} = 0$ So: $f^{'} (x) = {\begin{cases} 3 x^{2}, & x > 0 \\ 0, & x = 0 \\ - 3 x^{2}, & x < 0 \end{cases}$

Second derivative

$f^{''} (x) = {\begin{cases} 6 x, & x > 0 \\ - 6 x, & x < 0 \end{cases}$

Check at $x = 0$ :

$f^{''} (0) = \lim_{x \to 0} \frac{f^{'} (x) - f^{'} (0)}{x - 0} = \lim_{x \to 0} \frac{f^{'} (x)}{x}$

Right-hand limit:

$\lim_{x \to 0^{+}} \frac{3 x^{2}}{x} = \lim 3 x = 0$

Left-hand limit: $\lim_{x \to 0^{-}} \frac{- 3 x^{2}}{x} = \lim - 3 x = 0$

Both limits exist and equal → 0.

So $f^{''} (0) = 0$

Question 19

Using the fact that

$\sin (A + B) = \sin A \cos B + \cos A \sin B$

and differentiation, obtain the sum formula for cosines.

Solution

Differentiate both sides with respect to $B$ :

Given identity:

$\sin (A + B) = \sin A \cos B + \cos A \sin B$

Differentiate LHS

$\frac{d}{d B} [\sin (A + B)] = \cos (A + B)$

Differentiate RHS
- $\sin A$ is constant w.r.t $B$
- $\cos B$ differentiates to $- \sin B$
- $\cos A$ is constant w.r.t $B$
- $\sin B$ differentiates to $\cos B$
So:

$\frac{d}{d B} [\sin A \cos B + \cos A \sin B] = \sin A (- \sin B) + \cos A (\cos B)$ $= \cos A \cos B - \sin A \sin B$

Equate derivatives of both sides

$\cos (A + B) = \cos A \cos B - \sin A \sin B$

Final Answer

$\cos (A + B) = \cos A \cos B - \sin A \sin B$

Question 20

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Answer

Yes, such a function does exist.

Example

$f (x) = ∣ x ∣ + ∣ x - 1 ∣$

Check Continuity
- Both $∣ x ∣$ and $∣ x - 1 ∣$ are continuous everywhere.
- Sum of continuous functions is also continuous.
$\Rightarrow f (x) is continuous for all real x .$

Check Differentiability

A function involving $∣ x ∣$ is not differentiable where the inside part becomes zero.

For $∣ x ∣$ :

Not differentiable at $x = 0$

For $∣ x - 1 ∣$ :

Not differentiable at $x = 1$

So $f (x)$ is not differentiable exactly at two points: $x = 0$ and $x = 1$

Everywhere else, the derivative exists.

Conclusion

$Yes, such a function exists. One example is f (x) = ∣ x ∣ + ∣ x - 1 ∣$

Question 21

If $y = ∣ \begin{array}{ccc} f (x) & g (x) & h (x) \\ l & m & n \\ a & b & c \end{array} ∣$

prove that

$\frac{d y}{d x} = ∣ \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{array} ∣$

Simple and Direct Proof

Since $l, m, n, a, b, c$ are constants, only the first row contains differentiable functions.

Expand determinant along the first row:

$y = f (x) ∣ \begin{array}{cc} m & n \\ b & c \end{array} ∣ - g (x) ∣ \begin{array}{cc} l & n \\ a & c \end{array} ∣ + h (x) ∣ \begin{array}{cc} l & m \\ a & b \end{array} ∣$

Let the minors be constants:

$A = ∣ \begin{array}{cc} m & n \\ b & c \end{array} ∣, B = ∣ \begin{array}{cc} l & n \\ a & c \end{array} ∣, C = ∣ \begin{array}{cc} l & m \\ a & b \end{array} ∣$

So rewrite:

$y = f (x) A - g (x) B + h (x) C$

Differentiate both sides:

$\frac{d y}{d x} = f^{'} (x) A - g^{'} (x) B + h^{'} (x) C$

Substitute determinants back:

$\frac{d y}{d x} = f^{'} (x) ∣ \begin{array}{cc} m & n \\ b & c \end{array} ∣ - g^{'} (x) ∣ \begin{array}{cc} l & n \\ a & c \end{array} ∣ + h^{'} (x) ∣ \begin{array}{cc} l & m \\ a & b \end{array} ∣$

This is exactly:

$\frac{d y}{d x} = ∣ \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{array} ∣$

Question 22

If

$y = e^{a \cos^{- 1} x}, - 1 \leq x \leq 1$

show that

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0$

Solution

Let

$y = e^{a \cos^{- 1} x}$

Let

$u = \cos^{- 1} x$

so that

$y = e^{a u}$

Then:

$\frac{d u}{d x} = - \frac{1}{\sqrt{1 - x^{2}}}$

First derivative

$\frac{d y}{d x} = e^{a u} \cdot a \frac{d u}{d x} = y \cdot a (- \frac{1}{\sqrt{1 - x^{2}}})$ $\frac{d y}{d x} = - \frac{a y}{\sqrt{1 - x^{2}}}$

Second derivative

Differentiate again w.r.t $x$ :

$\frac{d^{2} y}{d x^{2}} = - \frac{a}{\sqrt{1 - x^{2}}} \cdot \frac{d y}{d x} - a y \cdot \frac{d}{d x} (1 - x^{2})^{- 1 / 2}$

Compute derivative of $(1 - x^{2})^{- 1 / 2}$ :

$\frac{d}{d x} (1 - x^{2})^{- 1 / 2} = - \frac{1}{2} (1 - x^{2})^{- 3 / 2} (- 2 x) = \frac{x}{(1 - x^{2})^{3 / 2}}$

So:

$\frac{d^{2} y}{d x^{2}} = - \frac{a}{\sqrt{1 - x^{2}}} \cdot \frac{d y}{d x} - a y \cdot \frac{x}{(1 - x^{2})^{3 / 2}}$

Substitute $\frac{d y}{d x} = - \frac{a y}{\sqrt{1 - x^{2}}}$ :

$\frac{d^{2} y}{d x^{2}} = - \frac{a}{\sqrt{1 - x^{2}}} (- \frac{a y}{\sqrt{1 - x^{2}}}) - \frac{a x y}{(1 - x^{2})^{3 / 2}}$ $= \frac{a^{2} y}{1 - x^{2}} - \frac{a x y}{(1 - x^{2})^{3 / 2}}$

Multiply both sides by $1 - x^{2}$

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y - \frac{a x y}{\sqrt{1 - x^{2}}}$

Now substitute again

$\frac{d y}{d x} = - \frac{a y}{\sqrt{1 - x^{2}}}$

So:

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y + x \frac{d y}{d x}$

Rearrange:

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0$
November 28, 2025
Exercise-Miscellaneous, Class 12th, Maths, Chapter 5, NCERT(1-10)

Question 1

Differentiate w.r.t. x:

$y = (3 x^{2} - 9 x + 5)^{9}$

Solution

Using the chain rule, differentiate the outer function first and then multiply by the derivative of the inner function:

$\frac{d y}{d x} = 9 (3 x^{2} - 9 x + 5)^{8} \times \frac{d}{d x} (3 x^{2} - 9 x + 5)$

Now differentiate the bracket:

$\frac{d}{d x} (3 x^{2} - 9 x + 5) = 6 x - 9$

So, $\frac{d y}{d x} = 9 (3 x^{2} - 9 x + 5)^{8} (6 x - 9)$

$\frac{d y}{d x} =27 (2 x - 3) (3 x^{2} - 9 x + 5)^{8}$

Question 2

Differentiate w.r.t. $x$ :

$y = \sin^{3} x + \cos^{6} x$

Solution

Differentiate each term separately. Using the chain rule:

$\frac{d}{d x} (\sin^{3} x) = 3 \sin^{2} x \cdot \frac{d}{d x} (\sin x) = 3 \sin^{2} x \cos x$ $\frac{d}{d x} (\cos^{6} x) = 6 \cos^{5} x \cdot \frac{d}{d x} (\cos x) = 6 \cos^{5} x (- \sin x)$

So, $\frac{d y}{d x} = 3 \sin^{2} x \cos x - 6 \cos^{5} x \sin x$

Now factor common terms: $3 \sin x \cos x$

$\frac{d y}{d x} = 3 \sin x \cos x (\sin x - 2 \cos^{4} x)$

Question 3

Differentiate w.r.t. $x$ :

$y = (5 x)^{3 \cos 2 x}$

Solution

Take log on both sides:

$\ln y = 3 \cos 2 x \cdot \ln (5 x)$

Differentiate w.r.t. $x$ (using product rule):

$\frac{1}{y} \frac{d y}{d x} = 3 \ln (5 x) \frac{d}{d x} (\cos 2 x) + 3 \cos 2 x \frac{d}{d x} (\ln (5 x))$

$= 3 \ln (5 x) (- 2 \sin 2 x) + 3 \cos 2 x \cdot \frac{1}{x}$

$= - 6 \ln (5 x) \sin 2 x + \frac{3 \cos 2 x}{x}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = (5 x)^{3 \cos 2 x} (\frac{3 \cos 2 x}{x} - 6 \ln (5 x) \sin 2 x)$

Question 4

Differentiate w.r.t. $x$ :

$y = \sin^{- 1} (x \sqrt{x}), 0 \leq x \leq 1$

Solution

First rewrite the function inside:

$x \sqrt{x} = x^{3 / 2}$

$y = \sin^{- 1} (x^{3 / 2})$

Now differentiate:

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - (x^{3 / 2})^{2}}} \cdot \frac{d}{d x} (x^{3 / 2})$

$\frac{d}{d x} (x^{3 / 2}) = \frac{3}{2} x^{1 / 2}$

So, $\frac{d y}{d x} = \frac{\frac{3}{2} x^{1 / 2}}{\sqrt{1 - x^{3}}}$

$\frac{d y}{d x} = \frac{3 \sqrt{x}}{2 \sqrt{1 - x^{3}}}, 0 \leq x \leq 1$

Question 5 $y = \frac{\cos^{- 1} (\frac{x}{2})}{\sqrt{2 x + 7}}, - 2 < x < 2$ Answer :

To Find: $\frac{d y}{d x}$

Let: $y = \cos^{- 1} (\frac{x}{2}) (2 x + 7)^{- 1 / 2}$

Use Product Rule:

$\frac{d y}{d x} = \frac{d u}{d x} \cdot v + u \cdot \frac{d v}{d x}$

Where: $u = \cos^{- 1} (\frac{x}{2}), v = (2 x + 7)^{- 1 / 2}$

Step 1: Differentiate $u = \cos^{- 1} (x / 2)$

$\frac{d u}{d x} = - \frac{1}{\sqrt{1 - {(\frac{x}{2})}^{2}}} \cdot \frac{d}{d x} (\frac{x}{2})$ $\frac{d u}{d x} = - \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} \cdot \frac{1}{2}$ $\frac{d u}{d x} = - \frac{1}{2 \sqrt{1 - \frac{x^{2}}{4}}}$

Now,

$\sqrt{1 - \frac{x^{2}}{4}} = \frac{\sqrt{4 - x^{2}}}{2}$

So,

$\frac{d u}{d x} = - \frac{1}{2 \cdot \frac{\sqrt{4 - x^{2}}}{2}} = - \frac{1}{\sqrt{4 - x^{2}}}$

Step 2: Differentiate $v = (2 x + 7)^{- 1 / 2}$

$\frac{d v}{d x} = - \frac{1}{2} (2 x + 7)^{- 3 / 2} \cdot 2$ $\frac{d v}{d x} = - (2 x + 7)^{- 3 / 2}$

Apply Product Rule

$\frac{d y}{d x} = (- \frac{1}{\sqrt{4 - x^{2}}}) (2 x + 7)^{- 1 / 2} + (\cos^{- 1} (\frac{x}{2})) (- (2 x + 7)^{- 3 / 2})$ $\frac{d y}{d x} = - \frac{1}{\sqrt{4 - x^{2}} \sqrt{2 x + 7}} - \frac{\cos^{- 1} (x / 2)}{(2 x + 7)^{3 / 2}}$

Question 6

$y = \cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}), 0 < x < \frac{π}{2}$

Solution

Let

$A = \sqrt{1 + \sin x}, B = \sqrt{1 - \sin x}$

So the expression inside $\cot^{- 1}$ becomes:

$\frac{A + B}{A - B}$

Simplify the expression

Multiply numerator and denominator by $(A + B)$ :

$\frac{A + B}{A - B} \cdot \frac{A + B}{A + B} = \frac{(A + B)^{2}}{A^{2} - B^{2}}$ Expand numerator:

$(A + B)^{2} = A^{2} + 2 A B + B^{2}$ And denominator:

$A^{2} - B^{2} = (1 + \sin x) - (1 - \sin x) = 2 \sin x$

Now calculate numerator:

$A^{2} + B^{2} = (1 + \sin x) + (1 - \sin x) = 2$ $A B = \sqrt{(1 + \sin x) (1 - \sin x)} = \sqrt{1 - \sin^{2} x} = \cos x$

So: $(A + B)^{2} = 2 + 2 \cos x$

Thus: $\frac{A + B}{A - B} = \frac{2 + 2 \cos x}{2 \sin x} = \frac{1 + \cos x}{\sin x}$

Use Trigonometric Identity

$\frac{1 + \cos x}{\sin x} = \cot \frac{x}{2}$

So: $y = \cot^{- 1} (\cot \frac{x}{2})$

Given domain $0 < x < \frac{π}{2}$ , we have:

$0 < \frac{x}{2} < \frac{π}{4}$

In this domain, $\cot^{- 1} (\cot θ) = θ$
Thus: $y = \frac{x}{2}$

Differentiate

$\frac{d y}{d x} = \frac{1}{2}$

Question 7

$y = (\log x)^{\log x}, x > 1$

We need to find:

$\frac{d y}{d x}$ Solution

Take logarithm on both sides

$y = (\log x)^{\log x}$ $\ln y = \log x \cdot \ln (\log x)$

Differentiate both sides w.r.t. $x$

Use implicit differentiation:

Left side:

$\frac{1}{y} \frac{d y}{d x}$

Right side — Product rule:

$\frac{d}{d x} [\log x \cdot \ln (\log x)]$

Let $u = \log x$ and $v = \ln (\log x)$

$u^{'} = \frac{1}{x}, v^{'} = \frac{1}{\log x} \cdot \frac{1}{x}$

Apply product rule:

$\frac{d}{d x} [u v] = u^{'} v + u v^{'}$ $= \frac{1}{x} \ln (\log x) + \log x \cdot \frac{1}{x \log x}$

$= \frac{\ln (\log x)}{x} + \frac{1}{x}$

Equate both sides

$\frac{1}{y} \frac{d y}{d x} = \frac{\ln (\log x)}{x} + \frac{1}{x}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = y (\frac{\ln (\log x)}{x} + \frac{1}{x})$

Substitute $y = (\log x)^{\log x}$

$\frac{d y}{d x} = (\log x)^{\log x} (\frac{\ln (\log x)}{x} + \frac{1}{x})$

Final Answer

$\frac{d y}{d x} = (\log x)^{\log x} (\frac{\ln (\log x)}{x} + \frac{1}{x})$

Question 8

$y = \cos (a \cos x + b \sin x)$

where $a$ and $b$ are constants.

We have to find:

$\frac{d y}{d x}$ Solution

Let:

$u = a \cos x + b \sin x$

So the function becomes:

$y = \cos u$

Differentiate using Chain Rule

$\frac{d y}{d x} = - \sin u \cdot \frac{d u}{d x}$

Now differentiate $u$ :

$\frac{d u}{d x} = a \frac{d}{d x} (\cos x) + b \frac{d}{d x} (\sin x)$

$\frac{d u}{d x} = a (- \sin x) + b (\cos x)$

$\frac{d u}{d x} = - a \sin x + b \cos x$

Substitute back into derivative

$\frac{d y}{d x} = - \sin (a \cos x + b \sin x) \cdot (- a \sin x + b \cos x)$

Final Answer

$\frac{d y}{d x} = \sin (a \cos x + b \sin x) (a \sin x - b \cos x)$

Question 9

$y = (\sin x - \cos x)^{(\sin x - \cos x)}, \frac{π}{4} < x < \frac{3 π}{4}$

We must find:

$\frac{d y}{d x}$

Solution

Let:

$y = (\sin x - \cos x)^{(\sin x - \cos x)}$

Take natural log on both sides (logarithmic differentiation):

$\ln y = (\sin x - \cos x) \ln (\sin x - \cos x)$

Differentiate both sides w.r.t $x$ :

Left side:

$\frac{1}{y} \frac{d y}{d x}$

Right side (product rule):

Let $u = \sin x - \cos x$ and $v = \ln (\sin x - \cos x)$

$u^{'} = \cos x + \sin x$ $v^{'} = \frac{1}{\sin x - \cos x} (\cos x + \sin x)$

Apply product rule:

$\frac{d}{d x} [u v] = u^{'} v + u v^{'}$

$= (\cos x + \sin x) \ln (\sin x - \cos x) + (\sin x - \cos x) \frac{\cos x + \sin x}{\sin x - \cos x}$

Simplify the second term:

$= (\cos x + \sin x) \ln (\sin x - \cos x) + (\cos x + \sin x)$

Factor out $(\cos x + \sin x)$

$= (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Now equate both sides

$\frac{1}{y} \frac{d y}{d x} = (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Multiply both sides by $y$ :

$\frac{d y}{d x} = y (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Substitute $y = (\sin x - \cos x)^{(\sin x - \cos x)}$

$\frac{d y}{d x} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Final Answer

$\frac{d y}{d x} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Question 10

$y = x^{x} + x^{a} + a^{x} + a^{a}, a > 0, x > 0$

We need to find:

$\frac{d y}{d x}$

Differentiate term by term

1. $x^{x}$

Use logarithmic differentiation:

$\frac{d}{d x} (x^{x}) = x^{x} (\ln x + 1)$

2. $x^{a}$

Here $a$ is constant:

$\frac{d}{d x} (x^{a}) = a x^{a - 1}$

3. $a^{x}$

Exponential with constant base:

$\frac{d}{d x} (a^{x}) = a^{x} \ln a$

4. $a^{a}$

Constant term:

$\frac{d}{d x} (a^{a}) = 0$

Combine all results

$\frac{d y}{d x} = x^{x} (\ln x + 1) + a x^{a - 1} + a^{x} \ln a + 0$

Final Answer

$\frac{d y}{d x} = x^{x} (\ln x + 1) + a x^{a - 1} + a^{x} \ln a$

November 27, 2025
Exercise-5.7, Class 12th, Maths, Chapter 5, NCERT

$Find the second order derivatives of the functions given in Exercises 1 to 10.$

Q1. $y = x^{2} + 3 x + 2$

Solution

$\frac{d y}{d x} = 2 x + 3$ $\frac{d^{2} y}{d x^{2}} = 2$

Q2. $y = x^{20}$

Solution

$\frac{d y}{d x} = 20 x^{19}$ $\frac{d^{2} y}{d x^{2}} = 380 x^{18}$

Q3. $y = x \cos x$

Solution

Using product rule:

$\frac{d y}{d x} = \cos x - x \sin x$

Now differentiate again:

$\frac{d^{2} y}{d x^{2}} = - \sin x - (\sin x + x \cos x)$ $\frac{d^{2} y}{d x^{2}} = - 2 \sin x - x \cos x$

Q4. $y = \log x$

Solution

$\frac{d y}{d x} = \frac{1}{x}$ $\frac{d^{2} y}{d x^{2}} = - \frac{1}{x^{2}}$

Q5. $y = x^{3} \log x$

Solution

$\frac{d y}{d x} = 3 x^{2} \log x + x^{2}$ $\frac{d^{2} y}{d x^{2}} = 6 x \log x + 5 x$

Q6. $y = e^{x} \sin 5 x$

Solution

$\frac{d y}{d x} = e^{x} (\sin 5 x + 5 \cos 5 x)$ $\frac{d^{2} y}{d x^{2}} = e^{x} (- 24 \sin 5 x + 10 \cos 5 x)$

Q7. $y = e^{6 x} \cos 3 x$

Solution $\frac{d y}{d x} = e^{6 x} (6 \cos 3 x - 3 \sin 3 x)$ $\frac{d^{2} y}{d x^{2}} = e^{6 x} (27 \cos 3 x - 36 \sin 3 x)$

Q8. $y = \tan^{- 1} x$

Solution $\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ $\frac{d^{2} y}{d x^{2}} = - \frac{2 x}{(1 + x^{2})^{2}}$

Q9. $y = \log (\log x)$

Solution $\frac{d y}{d x} = \frac{1}{x \log x}$ $\frac{d^{2} y}{d x^{2}} = - \frac{\log x + 1}{x^{2} (\log x)^{2}}$

Q10. $y = \sin (\log x)$

Solution $\frac{d y}{d x} = \frac{\cos (\log x)}{x}$ $\frac{d^{2} y}{d x^{2}} = \frac{- \cos (\log x) - \sin (\log x)}{x^{2}}$

Q11. If $y = 5 \cos x - 3 \sin x$ , show that $\frac{d^{2} y}{d x^{2}} + y = 0$

Solution

$\frac{d y}{d x} = - 5 \sin x - 3 \cos x$ $\frac{d^{2} y}{d x^{2}} = - 5 \cos x + 3 \sin x$

Now, $\frac{d^{2} y}{d x^{2}} + y = (- 5 \cos x + 3 \sin x) + (5 \cos x - 3 \sin x)$

$\frac{d^{2} y}{d x^{2}} + y = 0$

$Hence proved$

November 27, 2025
Exercise-5.6, Class 12th, Maths, Chapter 5, NCERT

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx.

Question 1

If $x$ and $y$ are connected parametrically by the equations:

$x = 2 a t^{2}, y = a t^{4}$

Find $\frac{d y}{d x}$ without eliminating the parameter.

Solution

Differentiate both equations with respect to parameter $t$ :

$\frac{d x}{d t} = \frac{d}{d t} (2 a t^{2}) = 4 a t$ $\frac{d y}{d t} = \frac{d}{d t} (a t^{4}) = 4 a t^{3}$

Now apply the formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{4 a t^{3}}{4 a t} = t^{2}$

Final Answer

$\frac{d y}{d x} = t^{2}$

Question 2

If $x = a \cos θ, y = b \cos θ$

find $\frac{d y}{d x}$ .

Solution

Differentiate w.r.t. $θ$ :

$\frac{d x}{d θ} = - a \sin θ$ $\frac{d y}{d θ} = - b \sin θ$

Apply:

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{- b \sin θ}{- a \sin θ}$

$\frac{d y}{d x} = \frac{b}{a}$

Final Answer

$\frac{d y}{d x} = \frac{b}{a}$

Question 3

If $x = \sin t, y = \cos 2 t$

find $\frac{d y}{d x}$ .

Solution

Differentiate both equations with respect to parameter $t$ :

$\frac{d x}{d t} = \cos t$

$\frac{d y}{d t} = \frac{d}{d t} (\cos 2 t) = - 2 \sin 2 t$

Apply parametric derivative formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{- 2 \sin 2 t}{\cos t}$

Now use identity:

$\sin 2 t = 2 \sin t \cos t$

So: $\frac{d y}{d x} = \frac{- 2 (2 \sin t \cos t)}{\cos t}$ $\frac{d y}{d x} = - 4 \sin t$

Question 4

If $x = 4 t, y = \frac{4}{t}$

find $\frac{d y}{d x}$ .

Solution

Differentiate both equations with respect to parameter $t$ :

$\frac{d x}{d t} = 4$ $\frac{d y}{d t} = \frac{d}{d t} (\frac{4}{t}) = - \frac{4}{t^{2}}$

Now apply:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{- \frac{4}{t^{2}}}{4}$ $\frac{d y}{d x} = - \frac{1}{t^{2}}$

Question 5

If $x = \cos θ - \cos 2 θ, y = \sin θ - \sin 2 θ$

find $\frac{d y}{d x}$ .

Solution

Differentiate both equations with respect to $θ$ :

Differentiate $x$

$\frac{d x}{d θ} = \frac{d}{d θ} (\cos θ) - \frac{d}{d θ} (\cos 2 θ)$ $= - \sin θ - (- \sin 2 θ) (2)$ $= - \sin θ + 2 \sin 2 θ$

Differentiate $y$

$\frac{d y}{d θ} = \frac{d}{d θ} (\sin θ) - \frac{d}{d θ} (\sin 2 θ)$ $= \cos θ - (2 \cos 2 θ)$

Apply parametric derivative formula

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{\cos θ - 2 \cos 2 θ}{- \sin θ + 2 \sin 2 θ}$

Question 6

If $x = a (θ - \sin θ), y = a (1 + \cos θ)$

find $\frac{d y}{d x}$ .

Solution

Differentiate both $x$ and $y$ with respect to $θ$ :

Differentiate $x$

$\frac{d x}{d θ} = a (\frac{d}{d θ} (θ) - \frac{d}{d θ} (\sin θ))$ $= a (1 - \cos θ)$

Differentiate $y$

$\frac{d y}{d θ} = a \frac{d}{d θ} (1 + \cos θ)$ $= a (0 - \sin θ)$ $= - a \sin θ$

Apply the formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{- a \sin θ}{a (1 - \cos θ)}$ $\frac{d y}{d x} = \frac{- \sin θ}{1 - \cos θ}$

Use identity:

$1 - \cos θ = 2 \sin^{2} \frac{θ}{2}, \sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}$ $\frac{d y}{d x} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}}$ $= - \frac{\cos \frac{θ}{2}}{\sin \frac{θ}{2}}$ $= - \cot \frac{θ}{2}$

Question 7

If

$x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}}$

Solution

Rewrite $x$ and $y$ as:

$x = \sin^{3} t (\cos 2 t)^{- 1 / 2}, y = \cos^{3} t (\cos 2 t)^{- 1 / 2}$

Divide $y$ by $x$ :

$\frac{y}{x} = \frac{\cos^{3} t}{\sin^{3} t} = \cot^{3} t$

Take cube root:

${(\frac{y}{x})}^{1 / 3} = \cot t$ Let:

$u = \tan t = {(\frac{x}{y})}^{1 / 3}$ Differentiate implicitly:

$\sec^{2} t \frac{d t}{d x} = \frac{1}{3} {(\frac{x}{y})}^{- 2 / 3} \cdot \frac{y - x \frac{d y}{d x}}{y^{2}}$

$\frac{d y}{d x} = - \cot 3 t$

Question 8

If $x = a (\cos t + \log \tan \frac{t}{2}), y = a \sin t$

find $\frac{d y}{d x}$ .

Solution

Differentiate both $x$ and $y$ with respect to $t$ .

Differentiate $y$

$y = a \sin t$

$\frac{d y}{d t} = a \cos t$

Differentiate $x$

$x = a (\cos t + \log \tan \frac{t}{2})$

$\frac{d x}{d t} = a (- \sin t + \frac{d}{d t} \log \tan \frac{t}{2})$

Now recall the identity:

$\frac{d}{d t} (\log \tan \frac{t}{2}) = \frac{1}{\sin t}$

So:

$\frac{d x}{d t} = a (- \sin t + \frac{1}{\sin t})$ $\frac{d x}{d t} = a (\frac{- \sin^{2} t + 1}{\sin t})$ $\frac{d x}{d t} = a (\frac{1 - \sin^{2} t}{\sin t})$ $\frac{d x}{d t} = a (\frac{\cos^{2} t}{\sin t})$ $\frac{d x}{d t} = a \cos^{2} t \csc t$

Now apply parametric formula

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{a \cos t}{a \cos^{2} t \csc t}$ Cancel $a$ :

$\frac{d y}{d x} = \frac{\cos t \sin t}{\cos^{2} t}$

$\frac{d y}{d x} = \sin t \cdot \sec t$

$\frac{d y}{d x} = \tan t$

Question 9

If

$x = a \sec θ, y = b \tan θ$

find $\frac{d y}{d x}$ .

Solution

Differentiate both with respect to $θ$ :

Differentiate $x$

$\frac{d x}{d θ} = a \sec θ \tan θ$

Differentiate $y$

$\frac{d y}{d θ} = b \sec^{2} θ$

Apply parametric derivative formula

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{b \sec^{2} θ}{a \sec θ \tan θ}$ $\frac{d y}{d x} = \frac{b \sec θ}{a \tan θ}$ $\frac{d y}{d x} = \frac{b}{a} \cdot \frac{\sec θ}{\tan θ}$ Use identity:

$\frac{\sec θ}{\tan θ} = \csc θ$ Thus:

$\frac{d y}{d x} = \frac{b}{a} \csc θ$

Question 10

If

$x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ)$

find $\frac{d y}{d x}$ .

Solution

Differentiate $x$ and $y$ with respect to $θ$ .

Differentiate $x$

$x = a (\cos θ + θ \sin θ)$

Apply product rule to $θ \sin θ$ :

$\frac{d x}{d θ} = a (- \sin θ + \sin θ + θ \cos θ)$

$\frac{d x}{d θ} = a (θ \cos θ)$

Differentiate $y$

$y = a (\sin θ - θ \cos θ)$ $\frac{d y}{d θ} = a (\cos θ - (\cos θ - θ \sin θ))$

$\frac{d y}{d θ} = a (θ \sin θ)$

Apply parametric derivative formula

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{a θ \sin θ}{a θ \cos θ}$

Cancel $a θ$ :

$\frac{d y}{d x} = \frac{\sin θ}{\cos θ}$ $\frac{d y}{d x} = \tan θ$

Question

If $x = \sqrt{a^{\sin^{- 1} t}}, y = \sqrt{a^{\cos^{- 1} t}}$

show that

$\frac{d y}{d x} = - \frac{y}{x}$

Solution

First rewrite expressions using exponent rules.

$x = {(a^{\sin^{- 1} t})}^{1 / 2} = a^{\frac{1}{2} \sin^{- 1} t}$ $y = {(a^{\cos^{- 1} t})}^{1 / 2} = a^{\frac{1}{2} \cos^{- 1} t}$

Differentiate w.r.t. $t$ using log differentiation:

Differentiate $x$

$\log x = \frac{1}{2} (\sin^{- 1} t) \log a$

Differentiate:

$\frac{1}{x} \frac{d x}{d t} = \frac{1}{2} \log a \cdot \frac{1}{\sqrt{1 - t^{2}}}$

$\frac{d x}{d t} = x \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}$

Differentiate $y$

$\log y = \frac{1}{2} (\cos^{- 1} t) \log a$

Differentiate:

$\frac{1}{y} \frac{d y}{d t} = \frac{1}{2} \log a \cdot \frac{- 1}{\sqrt{1 - t^{2}}}$

$\frac{d y}{d t} = - y \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}$

Now apply parametric derivative formula

$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ $\frac{d y}{d x} = \frac{- y \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}}{x \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}}$

Cancel common factors:

$\frac{d y}{d x} = \frac{- y}{x}$

November 27, 2025
Exercise-5.5, Class 12th, Maths, Chapter 5, NCERT

Question 1

Differentiate the following function w.r.t. $x$ :

$y = \cos x \cdot \cos 2 x \cdot \cos 3 x$

Solution

Given:

$y = \cos x \cdot \cos 2 x \cdot \cos 3 x$

This is a product of three functions.
Use product rule:

$(u v w)^{'} = u^{'} v w + u v^{'} w + u v w^{'}$

Let:

$u = \cos x, v = \cos 2 x, w = \cos 3 x$

Then:

$u^{'} = - \sin x, v^{'} = - 2 \sin 2 x, w^{'} = - 3 \sin 3 x$

Applying product rule:

$y^{'} = u^{'} v w + u v^{'} w + u v w^{'}$ $y^{'} = (- \sin x) (\cos 2 x) (\cos 3 x) + (\cos x) (- 2 \sin 2 x) (\cos 3 x)$

$+ (\cos x) (\cos 2 x) (- 3 \sin 3 x)$ $y^{'} = - \sin x \cos 2 x \cos 3 x - 2 \cos x \sin 2 x \cos 3 x - 3 \cos x \cos 2 x \sin 3 x$

Question 2

Differentiate the following w.r.t. $x$ :

$y = \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}}$

Solution

Rewrite using exponent form:

$y = {(\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)})}^{1 / 2}$

Taking log on both sides (log differentiation method):

$\log y = \frac{1}{2} [\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)]$

Differentiate w.r.t. $x$ :

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{2} [\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}]$

Multiply both sides by $y$ :

$\frac{d y}{d x} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} [\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}]$

Final Answer

$\frac{d y}{d x} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} [\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}]$

Question 3

Differentiate w.r.t. $x$ :

$y = (\log x)^{\cos x}$

Solution :

Take natural logarithm on both sides:

$\log y = \cos x \cdot \log (\log x)$

Differentiate both sides w.r.t. $x$ :

Left side:

$\frac{1}{y} \cdot \frac{d y}{d x}$

Right side (product rule):

$\frac{d}{d x} (\cos x \cdot \log (\log x)) = - \sin x \cdot \log (\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$

So:

$\frac{1}{y} \frac{d y}{d x} = - \sin x \log (\log x) + \frac{\cos x}{x \log x}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = y [- \sin x \log (\log x) + \frac{\cos x}{x \log x}]$

Substitute $y = (\log x)^{\cos x}$ :

$\frac{d y}{d x} = (\log x)^{\cos x} [\frac{\cos x}{x \log x} - \sin x \log (\log x)]$

Question 4

Differentiate w.r.t. $x$ :

$y = x^{x} - 2 \sin x$

Solution :

The second term is simple to differentiate.
The first term $x^{x}$ requires logarithmic differentiation.

Let:

$u = x^{x}$

Taking log on both sides:

$\log u = x \log x$

Differentiate w.r.t. $x$ :

Left side: $\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} (x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$

So:

$\frac{1}{u} \frac{d u}{d x} = \log x + 1$ $\frac{d u}{d x} = x^{x} (\log x + 1)$

Now differentiate the whole expression:

Given: $y = x^{x} - 2 \sin x$

$\frac{d y}{d x} = x^{x} (\log x + 1) - 2 \cos x$

Question 5

Differentiate w.r.t. $x$ :

$y = (x + 3)^{2} (x + 4)^{3} (x + 5)^{4}$

Solution

Taking logarithm on both sides:

$\log y = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5)$

Differentiate both sides w.r.t. $x$ :

Left side:

$\frac{1}{y} \cdot \frac{d y}{d x}$

Right side:

$2 \cdot \frac{1}{x + 3} + 3 \cdot \frac{1}{x + 4} + 4 \cdot \frac{1}{x + 5}$

So: $\frac{1}{y} \frac{d y}{d x} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = (x + 3)^{2} (x + 4)^{3} (x + 5)^{4} (\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5})$

Question 6

Differentiate w.r.t. $x$ :

$y = (1 + \frac{1}{x}) (1 + \frac{1}{x^{2}}) (1 + \frac{1}{x^{3}})$

Solution

Take logarithm on both sides:

$\log y = \log (1 + \frac{1}{x}) + \log (1 + \frac{1}{x^{2}}) + \log (1 + \frac{1}{x^{3}})$

Differentiate both sides w.r.t. $x$ :

Left side: $\frac{1}{y} \frac{d y}{d x}$

Right side (chain rule):

$\frac{d}{d x} \log (1 + \frac{1}{x}) = \frac{1}{1 + \frac{1}{x}} \cdot (- \frac{1}{x^{2}}) = - \frac{1}{x (x + 1)}$

Similarly:

$\frac{d}{d x} \log (1 + \frac{1}{x^{2}}) = \frac{1}{1 + \frac{1}{x^{2}}} \cdot (- \frac{2}{x^{3}}) = - \frac{2}{x^{2} (x^{2} + 1)}$

$\frac{d}{d x} \log (1 + \frac{1}{x^{3}}) = \frac{1}{1 + \frac{1}{x^{3}}} \cdot (- \frac{3}{x^{4}}) = - \frac{3}{x^{3} (x^{3} + 1)}$

So:

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{x (x + 1)} - \frac{2}{x^{2} (x^{2} + 1)} - \frac{3}{x^{3} (x^{3} + 1)}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = (1 + \frac{1}{x}) (1 + \frac{1}{x^{2}}) (1 + \frac{1}{x^{3}}) [- \frac{1}{x (x + 1)} - \frac{2}{x^{2} (x^{2} + 1)} - \frac{3}{x^{3} (x^{3} + 1)}]$

Final Answer

$\frac{d y}{d x} = (1 + \frac{1}{x}) (1 + \frac{1}{x^{2}}) (1 + \frac{1}{x^{3}}) [- \frac{1}{x (x + 1)} - \frac{2}{x^{2} (x^{2} + 1)} - \frac{3}{x^{3} (x^{3} + 1)}]$

Question 7

Differentiate w.r.t. $x$ :

$y = (\log x)^{x} + x^{\log x}$

Solution :

Since the expression is the sum of two terms, differentiate them separately.

Part 1: Differentiate $u = (\log x)^{x}$

Take logarithm on both sides:

$\log u = x \log (\log x)$

Differentiate:

Left side: $\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} [x \log (\log x)] = \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log (\log x) + \frac{1}{\log x}$

So:

$\frac{d u}{d x} = (\log x)^{x} [\log (\log x) + \frac{1}{\log x}]$

Part 2: Differentiate $v = x^{\log x}$

Let: $v = x^{\log x}$

Take logarithm:

$\log v = \log x \cdot \log x = (\log x)^{2}$

Differentiate:

Left side: $\frac{1}{v} \frac{d v}{d x}$

Right side:

$2 \log x \cdot \frac{1}{x} = \frac{2 \log x}{}$

Thus:

$\frac{d v}{d x} = x^{\log x} \cdot \frac{2 \log x}{x}$

Combine (since $y = u + v$ )

$\frac{d y}{d x} = (\log x)^{x} [\log (\log x) + \frac{1}{\log x}] + x^{\log x} \cdot \frac{2 \log x}{x}$

Question 8

Differentiate w.r.t. $x$ :

$y = (\sin x)^{x} + \sin^{- 1} x$

Solution

Part 1: Differentiate $(\sin x)^{x}$

Let

$u = (\sin x)^{x}$

Take logarithm:

$\log u = x \log (\sin x)$

Differentiate both sides:

Left side: $\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} [x \log (\sin x)] = \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x$

$= \log (\sin x) + x \cot x$

So: $\frac{d u}{d x} = (\sin x)^{x} [\log (\sin x) + x \cot x]$

Part 2: Differentiate $\sin^{- 1} x$

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

Final Derivative

$\frac{d y}{d x} = (\sin x)^{x} [\log (\sin x) + x \cot x] + \frac{1}{\sqrt{1 - x^{2}}}$

Question 9

Differentiate w.r.t. $x$ :

$y = x^{\sin x} + (\sin x)^{\cos x}$

Solution

Split into two parts:

$y = u + v$

where

$u = x^{\sin x}, v = (\sin x)^{\cos x}$

Part 1: Differentiate $u = x^{\sin x}$

Take log on both sides:

$\log u = \sin x \cdot \log x$

Differentiate:

$\frac{1}{u} \frac{d u}{d x} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x}$

So:

$\frac{d u}{d x} = x^{\sin x} (\cos x \log x + \frac{\sin x}{x})$

Part 2: Differentiate $v = (\sin x)^{\cos x}$

Take log:

$\log v = \cos x \cdot \log (\sin x)$

Differentiate (product rule):

$\frac{1}{v} \frac{d v}{d x} = - \sin x \cdot \log (\sin x) + \cos x \cdot \frac{\cos x}{\sin x}$ $= - \sin x \log (\sin x) + \cos x \cot x$

So: $\frac{d v}{d x} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \log (\sin x)]$

Final Derivative

$\frac{d y}{d x} = x^{\sin x} (\cos x \log x + \frac{\sin x}{x}) + (\sin x)^{\cos x} [\cos x \cot x - \sin x \log (\sin x)]$

Question 10

Differentiate w.r.t. $x$ :

$y = x^{x \cos x} + \frac{x^{2} + 1}{x^{2} - 1}$

Solution

Split into two parts:

$y = u + v$

where

$u = x^{x \cos x}, v = \frac{x^{2} + 1}{x^{2} - 1}$

Part 1: Differentiate $u = x^{x \cos x}$

Take log both sides:

$\log u = x \cos x \cdot \log x$

Differentiate:

Left side:

$\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} [x \cos x \cdot \log x]$ $= \cos x \cdot \log x + x (- \sin x) \log x + x \cos x \cdot \frac{1}{x}$

$= \cos x \log x - x \sin x \log x + \cos x$

$= \cos x (\log x + 1) - x \sin x \log x$

So:

$\frac{d u}{d x} = x^{x \cos x} [\cos x (\log x + 1) - x \sin x \log x]$

Part 2: Differentiate $v = \frac{x^{2} + 1}{x^{2} - 1}$

Use quotient rule:

$v^{'} = \frac{(x^{2} - 1) (2 x) - (x^{2} + 1) (2 x)}{(x^{2} - 1)^{2}}$

$= \frac{2 x (x^{2} - 1 - x^{2} - 1)}{(x^{2} - 1)^{2}}$

$= \frac{2 x (- 2)}{(x^{2} - 1)^{2}}$

$= \frac{- 4 x}{(x^{2} - 1)^{2}}$

Final Derivative

$\frac{d y}{d x} = x^{x \cos x} [\cos x (\log x + 1) - x \sin x \log x] + \frac{- 4 x}{(x^{2} - 1)^{2}}$

Question 11

Differentiate w.r.t. $x$ :

$y = (x \cos x)^{x} + (x \sin x)^{1 / x}$

Solution

Part 1: Let

$u = (x \cos x)^{x}$

Taking log:

$\log u = x \log (x \cos x)$

Differentiate:

$\frac{1}{u} \frac{d u}{d x} = \log (x \cos x) + x \cdot \frac{1}{x \cos x} (\cos x - x \sin x)$

$= \log (x \cos x) + \frac{\cos x - x \sin x}{x \cos x}$

$= \log (x \cos x) + \frac{1}{x} - \tan x$

Thus:

$\frac{d u}{d x} = (x \cos x)^{x} [\log (x \cos x) + \frac{1}{x} - \tan x]$

Part 2: Let

$v = (x \sin x)^{1 / x}$

Taking log:

$\log v = \frac{1}{x} \log (x \sin x)$

Differentiate:

$\frac{1}{v} \frac{d v}{d x} = - \frac{1}{x^{2}} \log (x \sin x) + \frac{1}{x} \cdot \frac{\sin x + x \cos x}{x \sin x}$ $= - \frac{\log (x \sin x)}{x^{2}} + \frac{\sin x + x \cos x}{x^{2} \sin x}$

$= - \frac{\log (x \sin x)}{x^{2}} + \frac{1}{x^{2}} + \frac{\cos x}{x \sin x}$

Thus:

$\frac{d v}{d x} = (x \sin x)^{1 / x} [- \frac{\log (x \sin x)}{x^{2}} + \frac{1}{x^{2}} + \frac{\cos x}{x \sin x}]$

Answer

$\frac{d y}{d x} = (x \cos x)^{x} [\log (x \cos x) + \frac{1}{x} - \tan x] + (x \sin x)^{1 / x} [- \frac{\log (x \sin x)}{x^{2}} + \frac{1}{x^{2}} + \frac{\cos x}{x \sin x}]$

Question 12

Differentiate with respect to $x$ :

$x^{y} + y^{x} = 1$

This is an implicit function (y also depends on x), so we will use implicit differentiation + logarithmic differentiation.

Solution :

Differentiate both sides w.r.t.

$x$ :

Step 1: Differentiate $x^{y}$

Write:

$x^{y} = e^{y \log x}$

Differentiate:

$\frac{d}{d x} (x^{y}) = x^{y} \cdot \frac{d}{d x} (y \log x)$

$= x^{y} (\log x \cdot \frac{d y}{d x} + \frac{y}{x})$

Step 2: Differentiate $y^{x}$

Write:

$y^{x} = e^{x \log y}$

Differentiate:

$\frac{d}{d x} (y^{x}) = y^{x} \cdot \frac{d}{d x} (x \log y)$ $= y^{x} (\log y + x \cdot \frac{1}{y} \cdot \frac{d y}{d x})$

$= y^{x} (\log y + \frac{x}{y} \frac{d y}{d x})$

Step 3: Differentiate RHS

$\frac{d}{d x} (1) = 0$

Step 4: Combine all terms

$x^{y} (\log x \cdot \frac{d y}{d x} + \frac{y}{x}) + y^{x} (\log y + \frac{x}{y} \frac{d y}{d x}) = 0$

Now collect $\frac{d y}{d x}$ terms together:

$x^{y} \log x \cdot \frac{d y}{d x} + y^{x} \frac{x}{y} \frac{d y}{d x} = - x^{y} \frac{y}{x} - y^{x} \log y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x^{y} \log x + y^{x} \frac{x}{y}) = - (\frac{y x^{y}}{x} + y^{x} \log y)$

Final Answer

$\frac{d y}{d x} = \frac{- (\frac{y x^{y}}{x} + y^{x} \log y)}{x^{y} \log x + \frac{x}{y} y^{x}}$

Or more neatly:

$\frac{d y}{d x} = \frac{- (\frac{y}{x} x^{y} + y^{x} \log y)}{x^{y} \log x + \frac{x}{y} y^{x}}$

Question 13

Differentiate w.r.t. $x$ :

$y^{x} = x^{y}$

This is an implicit relation involving both $x$ and $y$ , so we will use logarithmic implicit differentiation.

Solution

Given:

$y^{x} = x^{y}$

Take natural log on both sides:

$\log (y^{x}) = \log (x^{y})$

Apply log rule: $\log (a^{b}) = b \log a$

$x \log y = y \log x$

Differentiate both sides w.r.t. $x$

Left side:

$\frac{d}{d x} (x \log y) = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{d y}{d x}$

Right side:

$\frac{d}{d x} (y \log x) = \frac{d y}{d x} \log x + y \cdot \frac{1}{x}$

Arrange terms

$\log y + \frac{x}{y} \frac{d y}{d x} = \log x \cdot \frac{d y}{d x} + \frac{y}{x}$

Bring all $\frac{d y}{d x}$ terms to one side:

$\frac{x}{y} \frac{d y}{d x} - \log x \cdot \frac{d y}{d x} = \frac{y}{x} - \log y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (\frac{x}{y} - \log x) = \frac{y}{x} - \log y$

Final Answer

$\frac{d y}{d x} = \frac{\frac{y}{x} - \log y}{\frac{x}{y} - \log x}$

QUESTION 14

Differentiate with respect to $x$ :

$(\cos x)^{y} = (\cos y)^{x}$

Solution:

Take logarithm on both sides:

$\log ((\cos x)^{y}) = \log ((\cos y)^{x})$

Using the rule $\log (a^{b}) = b \log a$ :

$y \log (\cos x) = x \log (\cos y)$

Differentiate both sides w.r.t. $x$ :

Left side:

$\frac{d}{d x} [y \log (\cos x)] = \frac{d y}{d x} \log (\cos x) + y \cdot \frac{1}{\cos x} (- \sin x)$

$= \frac{d y}{d x} \log (\cos x) - y \tan x$

Right side:

$\frac{d}{d x} [x \log (\cos y)] = 1 \cdot \log (\cos y) + x \cdot \frac{1}{\cos y} (- \sin y) \frac{d y}{d x}$

$= \log (\cos y) - x \tan y \frac{d y}{d x}$

Rearranging terms:

$\frac{d y}{d x} \log (\cos x) - (- x \tan y \frac{d y}{d x}) = \log (\cos y) + y \tan x$

$\frac{d y}{d x} (\log (\cos x) + x \tan y) = \log (\cos y) + y \tan x$

$\frac{d y}{d x} = \frac{\log (\cos y) + y \tan x}{\log (\cos x) + x \tan y}$

QUESTION 15

Differentiate w.r.t. $x$ :

$x y = e^{x - y}$

This is an implicit function, so we differentiate both sides w.r.t. $x$ .

Solution :

Differentiate both sides:

Left side

$\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$

(using product rule)

Right side

$\frac{d}{d x} (e^{x - y}) = e^{x - y} \cdot \frac{d}{d x} (x - y) = e^{x - y} (1 - \frac{d y}{d x})$

Form the equation

$x \frac{d y}{d x} + y = e^{x - y} (1 - \frac{d y}{d x})$

Expand RHS:

$x \frac{d y}{d x} + y = e^{x - y} - e^{x - y} \frac{d y}{d x}$

Bring $\frac{d y}{d x}$ terms together:

$x \frac{d y}{d x} + e^{x - y} \frac{d y}{d x} = e^{x - y} - y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x + e^{x - y}) = e^{x - y} - y$

$\frac{d y}{d x} = \frac{e^{x - y} - y}{x + e^{x - y}}$

Since from the given equation:

$x y = e^{x - y}$

Replace $e^{x - y}$ in the derivative:

$\frac{d y}{d x} = \frac{x y - y}{x + x y}$

Factor numerator and denominator:

$= \frac{y (x - 1)}{x (1 + y)}$

QUESTION 16

Find the derivative of the function

$f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$

and hence find $f^{'} (1)$ .

SOLUTION

Take logarithm on both sides

$\log f (x) = \log (1 + x) + \log (1 + x^{2}) + \log (1 + x^{4}) + \log (1 + x^{8})$

Differentiate w.r.t. $x$ :

$\frac{f^{'} (x)}{f (x)} = \frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}$

Multiply both sides by $f (x)$ :

$f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) [\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}]$

Final derivative

$f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) [\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}]$

Now find $f^{'} (1)$

Substitute $x = 1$ :

First compute $f (1)$ :

$f (1) = (1 + 1) (1 + 1^{2}) (1 + 1^{4}) (1 + 1^{8}) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$

Now compute the bracket part:

$\frac{1}{1 + 1} + \frac{2 \cdot 1}{1 + 1} + \frac{4 \cdot 1^{3}}{1 + 1} + \frac{8 \cdot 1^{7}}{1 + 1}$ $= \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}$

$= \frac{1}{2} + 1 + 2 + 4 = \frac{1}{2} + 7 = \frac{15}{2}$

Therefore:

$f^{'} (1) = f (1) \cdot \frac{15}{2}$

$f^{'} (1) = 16 \cdot \frac{15}{2} = 8 \cdot 15 = 120$

$f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) [\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}]$ $f^{'} (1) = 120$

QUESTION 17

Differentiate the function:

$y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$ by:

(i) Product rule

(ii) Expanding the product

(iii) Logarithmic differentiation

and check that all answers are the same.

(i) DIFFERENTIATION BY PRODUCT RULE

Let:

$u = x^{2} - 5 x + 8, v = x^{3} + 7 x + 9$

$u^{'} = 2 x - 5, v^{'} = 3 x^{2} + 7$

Using product rule:

$y^{'} = u^{'} v + u v^{'}$

$y^{'} = (2 x - 5) (x^{3} + 7 x + 9) + (x^{2} - 5 x + 8) (3 x^{2} + 7)$

(ii) DIFFERENTIATION BY EXPANDING FIRST

Expand: $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$

Multiply:

$= x^{2} (x^{3} + 7 x + 9) - 5 x (x^{3} + 7 x + 9) + 8 (x^{3} + 7 x + 9)$

$= x^{5} + 7 x^{3} + 9 x^{2} - 5 x^{4} - 35 x^{2} - 45 x + 8 x^{3} + 56 x + 72$

Combine like terms:

$y = x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72$

Differentiate term-wise:

$y^{'} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11$

(iii) BY LOGARITHMIC DIFFERENTIATION

$y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$

Take log:

$\log y = \log (x^{2} - 5 x + 8) + \log (x^{3} + 7 x + 9)$

Differentiate:

$\frac{y^{'}}{y} = \frac{2 x - 5}{x^{2} - 5 x + 8} + \frac{3 x^{2} + 7}{x^{3} + 7 x + 9}$

Multiply by $y$ :

$y^{'} = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) [\frac{2 x - 5}{x^{2} - 5 x + 8} + \frac{3 x^{2} + 7}{x^{3} + 7 x + 9}]$

Simplify (cancel terms):

$y^{'} = (2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)$

QUESTION 18

If $u, v, w$ are functions of $x$ , prove that:

$\frac{d}{d x} (u v w) = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

(i) By repeated application of the product rule

Let:

$y = u v w$

First consider:

$y = (u v) w$

Differentiate using product rule:

$\frac{d y}{d x} = \frac{d (u v)}{d x} \cdot w + (u v) \frac{d w}{d x}$

Now apply product rule again to $\frac{d (u v)}{d x}$ :

$\frac{d (u v)}{d x} = \frac{d u}{d x} v + u \frac{d v}{d x}$

Substitute back:

$\frac{d y}{d x} = (\frac{d u}{d x} v + u \frac{d v}{d x}) w + u v \frac{d w}{d x}$

Distribute $w$ :

$\frac{d y}{d x} = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

Result (Method 1 final statement)

$\frac{d}{d x} (u v w) = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

(ii) By logarithmic differentiation

Given:

$y = u v w$

Take logarithm on both sides:

$\log y = \log u + \log v + \log w$

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x}$

Multiply both sides by $y = u v w$ :

$\frac{d y}{d x} = u v w (\frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x})$

Distribute:

$\frac{d y}{d x} = v w \frac{d u}{d x} + u w \frac{d v}{d x} + u v \frac{d w}{d x}$

Final Answer

$\frac{d}{d x} (u v w) = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

This verifies the required identity by both methods.

Conclusion

Yes, both methods give the same derivative result:

$\frac{d}{d x} (u v w) = u^{'} v w + u v^{'} w + u v w^{'}$

November 27, 2025
Exercise-5.4, Class 12th, Maths, Chapter 5, NCERT

Differentiate the following w.r.t. x:

Question 1

Differentiate:

$y = \frac{e^{x}}{\sin x}$ Solution

This is a quotient, so use the Quotient Rule:

$\frac{d}{d x} (\frac{u}{v}) = \frac{v u^{'} - u v^{'}}{v^{2}}$

Let:

$u = e^{x}, v = \sin x$

Then:

$u^{'} = e^{x}, v^{'} = \cos x$

Apply the rule:

$\frac{d y}{d x} = \frac{\sin x \cdot e^{x} - e^{x} \cdot \cos x}{(\sin x)^{2}}$

Factor out $e^{x}$ :

$\frac{d y}{d x} = \frac{e^{x} (\sin x - \cos x)}{\sin^{2} x}$ Final Answer

$\frac{d y}{d x} = \frac{e^{x} (\sin x - \cos x)}{\sin^{2} x}$

Question 2

Differentiate w.r.t. $x$ :

$y = e^{\sin^{- 1} x}$ Solution

Let:

$y = e^{\sin^{- 1} x}$

Use the chain rule:

$\frac{d y}{d x} = e^{\sin^{- 1} x} \cdot \frac{d}{d x} (\sin^{- 1} x)$

We know:

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

Therefore:

$\frac{d y}{d x} = e^{\sin^{- 1} x} \cdot \frac{1}{\sqrt{1 - x^{2}}}$ Final Answer

$\frac{d y}{d x} = \frac{e^{\sin^{- 1} x}}{\sqrt{1 - x^{2}}}$

Question 3

Differentiate:

$y = e^{x^{3}}$ Solution

Use the chain rule:

Let:

$y = e^{u}, u = x^{3}$

Then:

$\frac{d y}{d u} = e^{u}, \frac{d u}{d x} = 3 x^{2}$

Apply chain rule:

$\frac{d y}{d x} = e^{u} \cdot 3 x^{2}$

Substitute $u = x^{3}$ :

$\frac{d y}{d x} = 3 x^{2} e^{x^{3}}$ Final Answer

$\frac{d y}{d x} = 3 x^{2} e^{x^{3}}$

Question 4

Differentiate with respect to $x$ :

$y = \sin (\tan^{- 1} (e^{- x}))$ Solution

Let:

$u = \tan^{- 1} (e^{- x})$

So:

$y = \sin (u)$

Step 1: Differentiate outer function

$\frac{d y}{d u} = \cos u$

Step 2: Differentiate inner function

$u = \tan^{- 1} (e^{- x})$

Derivative of $\tan^{- 1} t$ is $\frac{1}{1 + t^{2}} \cdot t^{'}$

So,

$\frac{d u}{d x} = \frac{1}{1 + (e^{- x})^{2}} \cdot \frac{d}{d x} (e^{- x})$

$\frac{d u}{d x} = \frac{1}{1 + e^{- 2 x}} \cdot (- e^{- x})$

Step 3: Apply chain rule

$\frac{d y}{d x} = \cos (u) \cdot \frac{d u}{d x}$

$\frac{d y}{d x} = \cos (\tan^{- 1} (e^{- x})) \cdot (- \frac{e^{- x}}{1 + e^{- 2 x}})$

We know the identity:

$\cos (\tan^{- 1} t) = \frac{1}{\sqrt{1 + t^{2}}}$

So:

$\cos (\tan^{- 1} (e^{- x})) = \frac{1}{\sqrt{1 + e^{- 2 x}}}$

Final Simplified Answer

$\frac{d y}{d x} = - \frac{e^{- x}}{(1 + e^{- 2 x})^{3 / 2}}$ Final Answer

$\frac{d y}{d x} = - \frac{e^{- x}}{(1 + e^{- 2 x})^{3 / 2}}$

Question 5

Differentiate w.r.t. $x$ :

$y = \log (\cos (e^{x}))$ Solution

Use the chain rule multiple times.

Let:

$y = \log (\cos (e^{x}))$

Step 1: Differentiate outer logarithm

$\frac{d y}{d x} = \frac{1}{\cos (e^{x})} \cdot \frac{d}{d x} (\cos (e^{x}))$

Step 2: Differentiate $\cos (e^{x})$

$\frac{d}{d x} (\cos (e^{x})) = - \sin (e^{x}) \cdot \frac{d}{d x} (e^{x})$ $= - \sin (e^{x}) \cdot e^{x}$

Combine the results

$\frac{d y}{d x} = \frac{1}{\cos (e^{x})} \cdot (- e^{x} \sin (e^{x}))$ $\frac{d y}{d x} = - e^{x} \frac{\sin (e^{x})}{\cos (e^{x})}$ Final Answer

$\frac{d y}{d x} = - e^{x} \tan (e^{x})$

Question 6

Differentiate w.r.t. $x$ :

$y = e^{x} + e^{x^{2}} + e^{x^{3}} + e^{x^{4}} + e^{x^{5}}$

Solution

Differentiate term-by-term:

1. $\frac{d}{d x} (e^{x}) = e^{x}$

2. $\frac{d}{d x} (e^{x^{2}}) = e^{x^{2}} \cdot \frac{d}{d x} (x^{2}) = e^{x^{2}} \cdot 2 x$

3. $\frac{d}{d x} (e^{x^{3}}) = e^{x^{3}} \cdot \frac{d}{d x} (x^{3}) = e^{x^{3}} \cdot 3 x^{2}$

4. $\frac{d}{d x} (e^{x^{4}}) = e^{x^{4}} \cdot \frac{d}{d x} (x^{4}) = e^{x^{4}} \cdot 4 x^{3}$

5. $\frac{d}{d x} (e^{x^{5}}) = e^{x^{5}} \cdot \frac{d}{d x} (x^{5}) = e^{x^{5}} \cdot 5 x^{4}$

Final Answer

$\frac{d y}{d x} = e^{x} + 2 x e^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}}$

Question 7

Differentiate w.r.t. $x$ :

$y = \sqrt{e^{\sqrt{x}}}, x > 0$ Solution

Rewrite the function:

$y = {(e^{\sqrt{x}})}^{1 / 2} = e^{\frac{1}{2} \sqrt{x}}$

Let: $u = \frac{1}{2} \sqrt{x} = \frac{1}{2} x^{1 / 2}$

So:

$y = e^{u}$

Differentiate

$\frac{d y}{d u} = e^{u}$ $\frac{d u}{d x} = \frac{1}{2} \cdot \frac{1}{2} x^{- 1 / 2} = \frac{1}{4 \sqrt{x}}$

Apply chain rule:

$\frac{d y}{d x} = e^{u} \cdot \frac{d u}{d x}$

$\frac{d y}{d x} = e^{\frac{1}{2} \sqrt{x}} \cdot \frac{1}{4 \sqrt{x}}$ Final Answer

$\frac{d y}{d x} = \frac{e^{\frac{1}{2} \sqrt{x}}}{4 \sqrt{x}}, x > 0$

Question 8

Differentiate w.r.t. $x$ :

$y = \log (\log x), x > 1$ Solution

This is a composition of two logarithmic functions, so apply the chain rule.

Let:

$u = \log x$ $y = \log u$

Differentiate step-by-step

$\frac{d y}{d u} = \frac{1}{u}$ $\frac{d u}{d x} = \frac{1}{x}$

Apply chain rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $\frac{d y}{d x} = \frac{1}{\log x} \cdot \frac{1}{x}$ Final Answer

$\frac{d y}{d x} = \frac{1}{x \log x}, x > 1$

Question 9

Differentiate w.r.t. $x$ :

$y = \frac{\cos x}{\log x}, x > 0$ Solution

This is a quotient, so apply the Quotient Rule:

$\frac{d}{d x} (\frac{u}{v}) = \frac{v u^{'} - u v^{'}}{v^{2}}$ Let:

$u = \cos x, v = \log x$ Then:

$u^{'} = - \sin x$

$v^{'} = \frac{1}{x}$

Apply quotient rule

$\frac{d y}{d x} = \frac{(\log x) (- \sin x) - (\cos x) (\frac{1}{x})}{(\log x)^{2}}$ $\frac{d y}{d x} = \frac{- \sin x \log x - \frac{\cos x}{x}}{(\log x)^{2}}$ Final Answer

$\frac{d y}{d x} = \frac{- \sin x \log x - \frac{\cos x}{x}}{(\log x)^{2}}, x > 0$

Question 10

Differentiate w.r.t. $x$ :

$y = \cos (\log x + e^{x}), x > 0$ Solution

Use the chain rule.

Let:

$u = \log x + e^{x}$ So:

$y = \cos (u)$

Differentiate outer function

$\frac{d y}{d u} = - \sin (u)$

Differentiate inner function

$\frac{d u}{d x} = \frac{d}{d x} (\log x) + \frac{d}{d x} (e^{x}) = \frac{1}{x} + e^{x}$

Apply chain rule

$\frac{d y}{d x} = - \sin (u) (\frac{1}{x} + e^{x})$

Substitute back $u = \log x + e^{x}$ :

Final Answer

$\frac{d y}{d x} = - \sin (\log x + e^{x}) (\frac{1}{x} + e^{x}), x > 0$

November 25, 2025
Exercise-5.2, Class 12th, Maths, Chapter 5, NCERT

Differentiate the functions with respect to x in Exercises 1 to 8.

1. Differentiate: $\sin (x^{2} + 5)$ with respect to $x$ .

Solution

We need to differentiate the function:

$y = \sin (x^{2} + 5)$

This is a composite function, so we will apply the Chain Rule

Let:

$u = x^{2} + 5 \Rightarrow y = \sin (u)$

Differentiate both:

$\frac{d y}{d u} = \cos (u)$ $\frac{d u}{d x} = 2 x$

Now apply Chain Rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $\frac{d y}{d x} = \cos (u) \cdot 2 x$

Substitute $u = x^{2} + 5$ :

$\frac{d y}{d x} = 2 x \cos (x^{2} + 5)$

Question 2

Differentiate with respect to $x$ :

$\cos (\sin x)$

Solution

Given:

$y = \cos (\sin x)$

This is a composite function, where:

$u = \sin x \Rightarrow y = \cos (u)$

Now differentiate step-by-step:

Step 1: Differentiate outer function

$\frac{d y}{d u} = - \sin (u)$

Step 2: Differentiate inner function

$\frac{d u}{d x} = \cos x$

Step 3: Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Substitute back $u = \sin x$ :

$\frac{d y}{d x} = - \sin (\sin x) \cos x$

Question 3

Differentiate with respect to $x$ :

$\sin (a x + b)$

Solution

Let:

$y = \sin (a x + b)$

This is again a composite function, so we apply the Chain Rule.

Step 1:

Let the inner function:

$u = a x + b$

Then,

$y = \sin (u)$

Step 2: Differentiate

$\frac{d y}{d u} = \cos (u)$

Step 3: Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Substitute $u = a x + b$ :

$\frac{d y}{d x} = a \cos (a x + b)$

Question 4

Differentiate with respect to $x$ :

$\frac{d}{d x} [\sec (\tan (\sqrt{x}))$

Solution

Let:

$u = \tan (\sqrt{x})$

$y = \sec (u)$

Differentiate outer function

$\frac{d y}{d u} = \sec (u) \tan (u)$

Differentiate inner function

$u = \tan (v), where v = \sqrt{x}$

$\frac{d u}{d v} = \sec^{2} (v)$

Differentiate

$v = \sqrt{x}$ $\frac{d v}{d x} = \frac{1}{2 \sqrt{x}}$

Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x}$ $\frac{d y}{d x} = \sec (u) \tan (u) \cdot \sec^{2} (v) \cdot \frac{1}{2 \sqrt{x}}$

Substitute $u = \tan (\sqrt{x})$ , $v = \sqrt{x}$ :

Final Answer

$\frac{d}{d x} [\sec (\tan (\sqrt{x}))] = \frac{1}{2 \sqrt{x}} \sec (\tan (\sqrt{x})) \tan (\tan (\sqrt{x})) \sec^{2} (\sqrt{x})$

Question 5

Differentiate with respect to $x$ :

$\frac{\sin (a x + b)}{\cos (c x + d)}$ Solution

Let: $y = \frac{\sin (a x + b)}{\cos (c x + d)}$

This is a quotient, so we use the Quotient Rule:

$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$

where

$u = \sin (a x + b), v = \cos (c x + d)$

Step 1: Differentiate $u$

$\frac{d u}{d x} = \cos (a x + b) \cdot a$

Step 2: Differentiate $v$

$\frac{d v}{d x} = - \sin (c x + d) \cdot c$

Step 3: Apply Quotient Rule

$\frac{d y}{d x} = \frac{\cos (c x + d) \cdot (a \cos (a x + b)) - \sin (a x + b) \cdot (- c \sin (c x + d))}{\cos^{2} (c x + d)}$

Simplify the numerator:

$= \frac{a \cos (a x + b) \cos (c x + d) + c \sin (a x + b) \sin (c x + d)}{\cos^{2} (c x + d)}$

Final Answer

$\frac{d}{d x} (\frac{\sin (a x + b)}{\cos (c x + d)}) = \frac{a \cos (a x + b) \cos (c x + d) + c \sin (a x + b) \sin (c x + d)}{\cos^{2} (c x + d)}$

Question 6

Differentiate with respect to $x$ :

$\cos (x^{3}) \cdot \sin^{2} (x^{5})$

Solution

The given function is a product, so we apply the Product Rule:

$\frac{d}{d x} (u \cdot v) = u^{'} v + u v^{'}$

Let:

$u = \cos (x^{3})$

Step 1: Differentiate $u = \cos (x^{3})$

Using chain rule:

$u^{'} = - \sin (x^{3}) \cdot 3 x^{2}$

$u^{'} = - 3 x^{2} \sin (x^{3})$

Step 2: Differentiate $v = \sin^{2} (x^{5})$

Rewrite:

$v = (\sin (x^{5}))^{2}$

Let $t = \sin (x^{5})$ , then $v = t^{2}$

$\frac{d v}{d t} = 2 t = 2 \sin (x^{5})$

$\frac{d t}{d x} = \cos (x^{5}) \cdot 5 x^{4}$

So,

$v^{'} = 2 \sin (x^{5}) \cdot 5 x^{4} \cos (x^{5})$

Step 3: Apply Product Rule

$\frac{d y}{d x} = u^{'} v + u v^{'}$

Substitute:

$\frac{d y}{d x} = (- 3 x^{2} \sin (x^{3})) \cdot \sin^{2} (x^{5}) + \cos (x^{3}) \cdot 10 x^{4} \sin (x^{5}) \cos (x^{5})$

Final Answer

$\frac{d}{d x} [\cos (x^{3}) \sin^{2} (x^{5})] = - 3 x^{2} \sin (x^{3}) \sin^{2} (x^{5}) + 10 x^{4} \cos (x^{3}) \sin (x^{5}) \cos (x^{5})$

$y = 2 cot (x^{2})$

Let’s differentiate step by step.

Question 7

$y = 2 \sqrt{\cot (x^{2})}$

Rewrite using exponent form:

$y = 2 (\cot (x^{2}))^{1 / 2}$ Solution

Let:

$u = \cot (x^{2})$ $y = 2 u^{1 / 2}$

Step 1: Differentiate outer function

$\frac{d y}{d u} = 2 \cdot \frac{1}{2} u^{- 1 / 2} = u^{- 1 / 2}$

Step 2: Differentiate inner function

$u = \cot (x^{2})$

Derivative of $\cot t$ is $- \csc^{2} t$

So, by chain rule:

$\frac{d u}{d x} = - \csc^{2} (x^{2}) \cdot 2 x$

Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

$\frac{d y}{d x} = u^{- 1 / 2} \cdot (- 2 x \csc^{2} (x^{2}))$

Substitute back $u = \cot (x^{2})$ :

$\frac{d y}{d x} = - 2 x \csc^{2} (x^{2}) \cdot (\cot (x^{2}))^{- 1 / 2}$

Rewrite using square root:

$\frac{d y}{d x} = - \frac{2 x \csc^{2} (x^{2})}{\sqrt{\cot (x^{2})}}$

Final Answer

$\frac{d}{d x} [2 \sqrt{\cot (x^{2})}] = - \frac{2 x \csc^{2} (x^{2})}{\sqrt{\cot (x^{2})}}$

Question 8

Differentiate with respect to $x$ :

$\cos (\sqrt{x})$ Solution

Let: $y = \cos (\sqrt{x})$

This is a composite function, so we will apply the Chain Rule.

Step 1: Identify inner and outer functions

$u = \sqrt{x} = x^{1 / 2}$

Step 2: Differentiate each part

$\frac{d y}{d u} = - \sin (u)$

$\frac{d u}{d x} = \frac{1}{2 \sqrt{x}}$

Step 3: Apply Chain Rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Substitute $u = \sqrt{x}$ :

Final Answer

$\frac{d}{d x} [\cos (\sqrt{x})] = - \frac{\sin (\sqrt{x})}{2 \sqrt{x}}$

Question 9

Prove that the function

$f (x) = ∣ x - 1 ∣, x \in R$

is not differentiable at $x = 1$ .

Solution

First, write the function in piecewise form:

$f (x) = {\begin{cases} 1 - x, & if x < 1 \\ x - 1, & if x > 1 \end{cases}$

Also,

$f (1) = ∣ 1 - 1 ∣ = 0$

To check differentiability at $x = 1$ , evaluate the left-hand derivative and the right-hand derivative.

Left-hand derivative (LHD) at $x = 1$

For $x < 1$ , $f (x) = 1 - x$

$\frac{d}{d x} (1 - x) = - 1$ So,

$LHD at x = 1 = \lim_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} = - 1$

Right-hand derivative (RHD) at $x = 1$

For $x > 1$ , $f (x) = x - 1$

$\frac{d}{d x} (x - 1) = 1$

So,

$RHD at x = 1 = \lim_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h} = 1$

Conclusion

$LHD = - 1, RHD = 1$

Since: $LHD \neq RHD$

$∴ f (x) = ∣ x - 1 ∣ is not differentiable at x = 1.$

Reason

The graph of $∣ x - 1 ∣$ has a sharp corner (cusp) at $x = 1$ , which makes the slope undefined there.

Question 10

Prove that the greatest integer function defined by

$f (x) = [x], 0 < x < 3$

is not differentiable at $x = 1$ and $x = 2$ .

Solution:

Understanding the function

The function $[x]$ defines the greatest integer less than or equal to $x$ .

For $0 < x < 3$ , the function behaves as follows:

$f (x) = {\begin{cases} 0, & 0 < x < 1 \\ 1, & 1 < x < 2 \\ 2, & 2 < x < 3 \end{cases}$

Continuity & Differentiability at $x = 1$

Left-hand limit approaching 1:

For $x < 1$ , $f (x) = 0$

$\lim_{x \to 1^{-}} f (x) = 0$

Right-hand limit approaching 1:

For $x > 1$ , $f (x) = 1$

$\lim_{x \to 1^{+}} f (x) = 1$

Since

$\lim_{x \to 1^{-}} f (x) \neq \lim_{x \to 1^{+}} f (x),$ $f (x) is discontinuous at x = 1.$ Differentiability result

A function must be continuous at a point to be differentiable there.

Since $f (x)$ is not continuous at $x = 1$ , it cannot be differentiable at $x = 1$ .

$f (x) is not differentiable at x = 1.$

Similarly at $x = 2$

Left-hand limit approaching 2

For $x < 2$ , $f (x) = 1$

$\lim_{x \to 2^{-}} f (x) = 1$

Right-hand limit approaching 2

For $x > 2$ , $f (x) = 2$

$\lim_{x \to 2^{+}} f (x) = 2$

Since

$\lim_{x \to 2^{-}} f (x) \neq \lim_{x \to 2^{+}} f (x),$ $f (x) is discontinuous at x = 2.$ Thus,

$f (x) is not differentiable at x = 2.$

November 25, 2025
Exercise-5.1, Class 12th, Maths, Chapter 5, NCERT
Question 25

Examine the continuity of the function

$f (x) = {\begin{cases} \sin x - \cos x, & if x \neq 0 \\ - 1, & if x = 0 \end{cases}$

Answer

We are given the function:

$f (x) = {\begin{cases} \sin x - \cos x, & x \neq 0 \\ - 1, & x = 0 \end{cases}$

We need to examine the continuity at $x = 0$ .

Step 1: Value of the function at $x = 0$

$f (0) = - 1$

Step 2: Find the limit of $f (x)$ as $x \to 0$

$\lim_{x \to 0} (\sin x - \cos x)$

We know standard limits:

$\sin 0 = 0, \cos 0 = 1$

So,

$\lim_{x \to 0} (\sin x - \cos x) = 0 - 1 = - 1$

Step 3: Compare Limit and Function Value

Expression Value

$\lim_{x \to 0} f (x)$ $- 1$

$f (0)$ $- 1$

Both values are equal.

Conclusion

Since:

$\lim_{x \to 0} f (x) = f (0)$ The function $f (x)$ is continuous at $x = 0$ .

Question 26

Find the value of $k$ so that the function $f$ is continuous at $x = \frac{π}{2}$ :

$f (x) = {\begin{cases} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{cases}$

Solution

For continuity at $x = \frac{π}{2}$ , we need:

$\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$

Given:

$f (\frac{π}{2}) = 3$

Now compute the limit:

$\lim_{x \to \frac{π}{2}} \frac{k \cos x}{π - 2 x}$

Substitute $x = \frac{π}{2}$ :

$\frac{k \cos (\frac{π}{2})}{π - 2 (\frac{π}{2})} = \frac{k (0)}{0}$

This is the indeterminate form $\frac{0}{0}$ , so apply L’Hôpital’s Rule.

Differentiate numerator and denominator:

$\lim_{x \to \frac{π}{2}} \frac{- k \sin x}{- 2} = \lim_{x \to \frac{π}{2}} \frac{k \sin x}{2}$

Now substitute $x = \frac{π}{2}$ :

$\frac{k \sin (\frac{π}{2})}{2} = \frac{k (1)}{2} = \frac{k}{2}$

For continuity: $\frac{k}{2} = 3$ $k = 6$

Final Answer

$k = 6$

Question 27

Find the value of $k$ so that the function $f$ is continuous at $x = 2$ :

$f (x) = {\begin{cases} k x^{2}, & x \leq 2 \\ 3, & x > 2 \end{cases}$

Solution

For continuity at $x = 2$ :

$\lim_{x \to 2} f (x) = f (2)$

Step 1: Value at the point

$f (2) = k (2^{2}) = 4 k$

Step 2: Limit as $x \to 2$

Since the function changes definition at $x = 2$ , we use left-hand limit (LHL) and right-hand limit (RHL):

Left-hand limit (LHL)

$\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} k x^{2} = k (2^{2}) = 4 k$

Right-hand limit (RHL)

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} 3 = 3$

Condition for continuity

$LHL = RHL = f (2)$

So, $4 k = 3$

$k = \frac{3}{4}$

Final Answer

$k = \frac{3}{4}$

Thus, the function is continuous at $x = 2$ when $k = \frac{3}{4}$ .

Question 28

Find the value of $k$ so that the function $f$ is continuous at $x = π$ :

$f (x) = {\begin{cases} k x + 1, & x \leq π \\ \cos x, & x > π \end{cases}$

Solution

For continuity at $x = π$ , we require:

$\lim_{x \to π} f (x) = f (π)$

Step 1: Value of the function at $x = π$

Since $x = π$ falls in the first part $(x \leq π)$ :

$f (π) = k π + 1$ Step 2: Left-hand limit (LHL)

$\lim_{x \to π^{-}} f (x) = \lim_{x \to π^{-}} (k x + 1) = k π + 1$

Step 3: Right-hand limit (RHL)

$\lim_{x \to π^{+}} f (x) = \lim_{x \to π^{+}} \cos x = \cos π$ $\cos π = - 1$

Condition for Continuity

$LHL = RHL = f (π)$ $k π + 1 = - 1$

Solve for $k$

$k π = - 1 - 1 = - 2$

$k = \frac{- 2}{π}$

Question 29

Find the value of $k$ so that the function $f$ is continuous at $x = 5$ :

$f (x) = {\begin{cases} k x + 1, & x \leq 5 \\ 3 x - 5, & x > 5 \end{cases}$

Solution

For continuity at $x = 5$ , we need:

$\lim_{x \to 5} f (x) = f (5)$

Step 1: Function value at the point

Since $x \leq 5$ :

$f (5) = k (5) + 1 = 5 k + 1$

Step 2: Left-hand limit (LHL) as $x \to 5^{-}$

$\lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{-}} (k x + 1) = 5 k + 1$

Step 3: Right-hand limit (RHL) as $x \to 5^{+}$

$\lim_{x \to 5^{+}} f (x) = \lim_{x \to 5^{+}} (3 x - 5) = 3 (5) - 5 = 15 - 5 = 10$

Step 4: Apply continuity condition

$LHL = RHL = f (5)$ $5 k + 1 = 10$

$5 k = 9$

$k = \frac{9}{5}$

Question 30

Find the values of $a$ and $b$ such that the function

$f (x) = {\begin{cases} 5, & x \leq 2 \\ a x + b, & 2 < x < 10 \\ 21, & x \geq 10 \end{cases}$

is continuous.

Solution

For continuity, the function must be continuous at:
- $x = 2$ (where left part meets the middle part)
- $x = 10$ (where middle part meets the right part)
Continuity at $x = 2$

$\lim_{x \to 2^{-}} f (x) = f (2) = 5$ $\lim_{x \to 2^{+}} f (x) = a (2) + b = 2 a + b$

For continuity:

$2 a + b = 5 (Equation 1)$

Continuity at $x = 10$

$\lim_{x \to 10^{-}} f (x) = a (10) + b = 10 a + b$ $f (10) = 21$

For continuity:

$10 a + b = 21 (Equation 2)$

Solve Equations (1) and (2)

$2 a + b = 5$ $10 a + b = 21$

Subtract (1) from (2):

$(10 a + b) - (2 a + b) = 21 - 5$ $8 a = 16$

$a = 2$

Substitute $a = 2$ into Equation (1):

$2 (2) + b = 5$

$4 + b = 5$

$b = 1$

Question 31

Show that the function

$f (x) = \cos (x^{2})$

is a continuous function.

Solution

We know that:
- $x^{2}$ is a polynomial → continuous everywhere
- $\cos x$ is a continuous function for all real numbers
A composition of two continuous functions is also continuous.

Here,

$f (x) = \cos (x^{2}) = \cos [g (x)] where g (x) = x^{2}$

Since both $g (x)$ and $\cos x$ are continuous, therefore:

$f (x) = \cos (x^{2}) is continuous for all x \in R$

Answer:

$\cos (x^{2}) is continuous everywhere.$

Question 32

Show that the function

$f (x) = ∣ \cos x ∣$

is a continuous function.

Solution
- $\cos x$ is continuous for all real numbers
- The absolute value function $∣ x ∣$ is also continuous
- The composition of continuous functions is continuous
$f (x) = ∣ \cos x ∣ = ∣ g (x) ∣, where g (x) = \cos x$

Since both $g (x)$ and $∣ x ∣$ are continuous,

$f (x) = ∣ \cos x ∣ is continuous for all x$

Answer:

$∣ \cos x ∣ is continuous for all real x .$

Question 33

Examine whether

$f (x) = \sin ∣ x ∣$

is a continuous function.

Solution
- $∣ x ∣$ is continuous for all real numbers
- $\sin x$ is continuous for all real numbers
- Composition of continuous functions is continuous
Thus:

$f (x) = \sin (∣ x ∣) = \sin (g (x)), g (x) = ∣ x ∣$

Since both are continuous,

$\sin ∣ x ∣ is continuous for all x$

Answer:

$\sin ∣ x ∣ is continuous everywhere.$

Question 34

Find all the points of discontinuity of

$f (x) = ∣ x ∣ - ∣ x + 1 ∣$

Solution

Check where each absolute expression changes form.

Break points occur where inside values become zero:

$x = 0 and x = - 1$

So evaluate function in intervals:

Case 1: $x < - 1$

$∣ x ∣ = - x, ∣ x + 1 ∣ = - (x + 1)$ $f (x) = - x - (- (x + 1)) = - x + x + 1 = 1$

Case 2: $- 1 \leq x < 0$

$∣ x ∣ = - x, ∣ x + 1 ∣ = x + 1$ $f (x) = - x - (x + 1) = - 2 x - 1$

Case 3: $x \geq 0$

$∣ x ∣ = x, ∣ x + 1 ∣ = x + 1$ $f (x) = x - (x + 1) = - 1$

Check continuity at critical points

At $x = - 1$

$\lim_{x \to - 1^{-}} f (x) = 1$ $\lim_{x \to - 1^{+}} f (x) = - 2 (- 1) - 1 = 2 - 1 = 1$ $f (- 1) = - 2 (- 1) - 1 = 1$

So continuous at $x = - 1$

At $x = 0$

$\lim_{x \to 0^{-}} f (x) = - 2 (0) - 1 = - 1$ $\lim_{x \to 0^{+}} f (x) = - 1$ $f (0) = - 1$

So continuous at $x = 0$

Final Answer

$The function f (x) = ∣ x ∣ - ∣ x + 1 ∣ is continuous everywhere.$
November 24, 2025
Exercise-5.1, Class 12th, Maths, Chapter 5, NCERT
Question 13.

Is the function defined by

$f (x) = {\begin{cases} x + 5, & if x \leq 1 \\ x - 5, & if x > 1 \end{cases}$

a continuous function?

Answer

To check the continuity of the function at $x = 1$ , evaluate the following:

Value of the function at $x = 1$

Since $x \leq 1$ :

$f (1) = 1 + 5 = 6$

Left-hand limit (LHL) as $x \to 1^{-}$

$\lim_{x \to 1^{-}} (x + 5) = 6$

Right-hand limit (RHL) as $x \to 1^{+}$

$\lim_{x \to 1^{+}} (x - 5) = - 4$

Comparison

$LHL = 6, RHL = - 4, f (1) = 6$

Since

$LHL \neq RHL$

Final Answer

The function is not continuous at $x = 1$ because the left-hand limit and right-hand limit are not equal.

Discuss the continuity of the function f, where f is defined by

Question 14.

Discuss the continuity of the function

$f (x) = {\begin{cases} 3, & if 0 \leq x \leq 1 \\ 4, & if 1 < x < 3 \\ 5, & if 3 \leq x \leq 10 \end{cases}$

Answer

To check continuity, we examine the points where the definition of the function changes:
at $x = 1$ and $x = 3$ .

Continuity on the intervals
- On $[0, 1]$ : $f (x) = 3$ (a constant function → continuous).
- On $(1, 3)$ : $f (x) = 4$ (constant function → continuous).
- On $[3, 10]$ : $f (x) = 5$ (constant function → continuous).
So the only possible discontinuities are at the endpoints where the pieces join.

Check continuity at $x = 1$

Left-hand limit (as $x \to 1^{-}$ )

$\lim_{x \to 1^{-}} f (x) = 3$

Right-hand limit (as $x \to 1^{+}$ )

$\lim_{x \to 1^{+}} f (x) = 4$

Value of the function

$f (1) = 3$

Since

$\lim_{x \to 1^{-}} f (x) = 3 and \lim_{x \to 1^{+}} f (x) = 4$

Left-hand limit ≠ Right-hand limit
→ function is discontinuous at $x = 1$

Check continuity at $x = 3$

Left-hand limit (as $x \to 3^{-}$ )

$\lim_{x \to 3^{-}} f (x) = 4$

Right-hand limit (as $x \to 3^{+}$ )

$\lim_{x \to 3^{+}} f (x) = 5$

Value of the function

$f (3) = 5$

Since

$4 \neq 5$

→ function is discontinuous at $x = 3$ .

Final Conclusion

The function $f (x)$ is:
- Continuous on each open interval $(0, 1)$ , $(1, 3)$ , and $(3, 10)$
- Discontinuous at the points $x = 1$ and $x = 3$
Question 15

Discuss the continuity of the function $f$ defined by:

$f (x) = {\begin{cases} 2 x, & if x < 0 \\ 0, & if 0 < x < 1 \\ 4 x, & if x > 1 \end{cases}$

Answer

This is a piecewise function, and we need to check continuity at the points where the definition changes, i.e., at $x = 0$ and $x = 1$ .

Continuity for $x < 0$ , $0 < x < 1$ , and $x > 1$

In each interval, the function is a polynomial (linear function), and polynomials are continuous everywhere.
So, $f (x)$ is continuous within each interval.

Check continuity at $x = 0$

Left-hand limit (LHL) as $x \to 0^{-}$

Using $2 x$ :

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} 2 x = 0$

Right-hand limit (RHL) as $x \to 0^{+}$

Using 0:

$\lim_{x \to 0^{+}} f (x) = 0$

Value of the function at $x = 0$

Notice: The function definition does not include $x = 0$ in any case, so:

$f (0) does not exist$

Conclusion at $x = 0$

Even though

$LHL = RHL = 0,$

the value of the function at $x = 0$ is not defined.
So, the function is not continuous at $x = 0$ .

Check continuity at $x = 1$

Left-hand limit (LHL) as $x \to 1^{-}$

Using 0:

$\lim_{x \to 1^{-}} f (x) = 0$

Right-hand limit (RHL) as $x \to 1^{+}$

Using $4 x$ :

$\lim_{x \to 1^{+}} f (x) = 4 (1) = 4$

Value of the function at $x = 1$

Not defined in any case, so:

$f (1) does not exist$

Conclusion at $x = 1$

Since

$LHL = 0 \neq 4 = RHL$

The limit does not exist. Therefore, the function is not continuous at $x = 1$ .

Final Conclusion

The function $f (x)$ is:
- Continuous within each open interval $(- \infty, 0)$ , $(0, 1)$ , and $(1, \infty)$
- Not continuous at $x = 0$ (because $f (0)$ is not defined)
- Not continuous at $x = 1$ (because left-hand and right-hand limits are not equal and $f (1)$ is not defined)
Question 16

Discuss the continuity of the function $f (x)$ , where

$f (x) = {\begin{cases} - 2, & if x \leq - 1 \\ 2 x, & if - 1 < x \leq 1 \\ 2, & if x > 1 \end{cases}$

Answer:

To check continuity, we must examine the possible points of discontinuity—here the function changes its definition at $x = - 1$ and $x = 1$ .
So, we check continuity at these points.

Continuity at $x = - 1$

Value of the function at $x = - 1$

Since $x \leq - 1$ :

$f (- 1) = - 2$

Left-hand limit (LHL) as $x \to - 1^{-}$

Using $- 2$ :

$\lim_{x \to - 1^{-}} f (x) = - 2$

Right-hand limit (RHL) as $x \to - 1^{+}$

Using $2 x$ :

$\lim_{x \to - 1^{+}} 2 x = 2 (- 1) = - 2$

Conclusion

$L H L = R H L = f (- 1) = - 2$

So, the function is continuous at $x = - 1$ .

Continuity at $x = 1$

Value of the function at $x = 1$

Using $2 x$ :

$f (1) = 2 (1) = 2$

Left-hand limit (LHL) as $x \to 1^{-}$

Using $2 x$ :

$\lim_{x \to 1^{-}} 2 x = 2$

Right-hand limit (RHL) as $x \to 1^{+}$

Using constant $2$ :

$\lim_{x \to 1^{+}} f (x) = 2$

Conclusion

$L H L = R H L = f (1) = 2$

Thus, the function is continuous at $x = 1$ .

Final Conclusion

Since the function is continuous at both points where it changes its definition ( $x = - 1$ and $x = 1$ ), and there are no other breaks, gaps, or jumps:

$The function f (x) is continuous for all real values of x .$

Question 17

Find the relationship between $a$ and $b$ so that the function $f (x)$ defined by

$f (x) = {\begin{cases} a x + 1, & if x \leq 3 \\ b x + 3, & if x > 3 \end{cases}$

is continuous at $x = 3$ .

Answer

To ensure continuity at $x = 3$ , we require:

$Left-hand limit = Right-hand limit = f (3)$

Value of the function at $x = 3$

Using the first expression since $x \leq 3$ :

$f (3) = a (3) + 1 = 3 a + 1$

Left-hand Limit (LHL) as $x \to 3^{-}$

$\lim_{x \to 3^{-}} f (x) = 3 a + 1$

Right-hand Limit (RHL) as $x \to 3^{+}$

Using $b x + 3$ :

$\lim_{x \to 3^{+}} f (x) = b (3) + 3 = 3 b + 3$

Condition for continuity

$3 a + 1 = 3 b + 3$

Solving

$3 a - 3 b = 2$

Final Answer

$a - b = \frac{2}{3}$

This is the required relationship between $a$ and $b$ for $f (x)$ to be continuous at $x = 3$ .

Question 18

For what value of $λ$ is the function defined by

$f (x) = {\begin{cases} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{cases}$

continuous at $x = 0$ ? What about at $x = 1$ ?

Answer

Checking continuity at $x = 0$

Value of the function at $x = 0$

Since $x \leq 0$ :

$f (0) = λ (0^{2} - 2 \cdot 0) = λ (0) = 0$

Left-hand limit (LHL)

$\lim_{x \to 0^{-}} λ (x^{2} - 2 x) = λ (0 - 0) = 0$

Right-hand limit (RHL)

$\lim_{x \to 0^{+}} (4 x + 1) = 4 (0) + 1 = 1$

Condition for continuity

$L H L = R H L = f (0)$

So,

$0 = 1$

This statement is never true, so no real value of $λ$ can make the function continuous at $x = 0$ .

Checking continuity at $x = 1$

At $x = 1$ , the value is taken from the second piece $4 x + 1$ , and there is no matching left-hand expression at 1, since the first part ends at $x = 0$ .
Thus, the function is not defined in a neighborhood around 1 from the left side.

Therefore, the function cannot be continuous at $x = 1$ .

Final Conclusion

$There is no value of λ that makes the function continuous at x = 0.$ $The function is not continuous at x = 1 because continuity cannot be checked.$

Question 19

Show that the function $g (x) = x - [x]$ is discontinuous at all integral points.
Here $[x]$ denotes the greatest integer less than or equal to $x$ .

Answer

Given Function

$g (x) = x - [x]$

The expression $x - [x]$ represents the fractional part of $x$ , denoted as ${x}$ .
Thus,

$g (x) = {x}$

The fractional part function always satisfies:

$0 \leq {x} < 1$

To show discontinuity at integer points

Let $x = n$ , where $n$ is any integer.

Left-hand limit (LHL) as $x \to n^{-}$

When $x$ approaches $n$ from the left, $x = n - h$ where $h \to 0^{+}$ .
Then the greatest integer function gives:

$[x] = n - 1$

So,

$g (x) = x - [x] = (n - h) - (n - 1) = 1 - h$

Taking limit,

$\lim_{x \to n^{-}} g (x) = \lim_{h \to 0^{+}} (1 - h) = 1$

Right-hand limit (RHL) as $x \to n^{+}$

When approaching from the right, $x = n + h$ , $h \to 0^{+}$ , then

$[x] = n$

So,

$g (x) = (n + h) - n = h$

Taking limit,

$\lim_{x \to n^{+}} g (x) = \lim_{h \to 0^{+}} h = 0$

Value of the function at $x = n$

$g (n) = n - [n] = n - n = 0$

Comparison

$\lim_{x \to n^{-}} g (x) = 1, \lim_{x \to n^{+}} g (x) = 0, g (n) = 0$

Since,

$\lim_{x \to n^{-}} g (x) \neq \lim_{x \to n^{+}} g (x)$

The limit does not exist at $x = n$ , and therefore, the function is discontinuous at every integer $n$ .

Final Conclusion

$g (x) = x - [x] is discontinuous at all integral points.$

Question 20

Is the function defined by

$f (x) = x^{2} - \sin x + 5$

continuous at $x = π$ ?

Answer

The function $f (x) = x^{2} - \sin x + 5$ is composed of the following functions:
- $x^{2}$ → a polynomial function (continuous for all real $x$ )
- $\sin x$ → a trigonometric function (continuous for all real $x$ )
- Constant $5$ → continuous everywhere
The sum or difference of continuous functions is also continuous for all real numbers.

Therefore, $f (x)$ is continuous everywhere, including at $x = π$ .

Check using limits

Value at $x = π$ :

$f (π) = π^{2} - \sin π + 5 = π^{2} - 0 + 5 = π^{2} + 5$

Limit as $x \to π$ :

$\lim_{x \to π} f (x) = \lim_{x \to π} (x^{2} - \sin x + 5) = π^{2} - 0 + 5 = π^{2} + 5$

Comparison

$\lim_{x \to π} f (x) = f (π)$

So the function is continuous at $x = π$ .

Final Answer

$Yes, the function is continuous at x = π .$

Question 21

Discuss the continuity of the following functions:

(a) $f (x) = \sin x + \cos x$
(b) $f (x) = \sin x - \cos x$
(c) $f (x) = \sin x \cdot \cos x$

Answer

To discuss continuity, recall that:
- $\sin x$ and $\cos x$ are continuous functions for all real values of $x$ .
- Sum, difference, and product of continuous functions are also continuous.
So, we analyze each function:

(a) $f (x) = \sin x + \cos x$
- $\sin x$ is continuous for all $x \in R$
- $\cos x$ is continuous for all $x \in R$
The sum of two continuous functions is continuous.

Conclusion

$f (x) = \sin x + \cos x is continuous for all real x .$

(b) $f (x) = \sin x - \cos x$
- Difference of continuous functions remains continuous.
Conclusion

$f (x) = \sin x - \cos x is continuous for all real x .$

(c) $f (x) = \sin x \cdot \cos x$
- Product of two continuous functions is continuous.
Conclusion

$f (x) = \sin x \cdot \cos x is continuous for all real x .$

Question 22

Discuss the continuity of the cosine, cosecant, secant, and cotangent functions.

Answer

To discuss the continuity of these trigonometric functions, recall:
- A function is continuous at a point if the limit exists and equals the value of the function at that point.
- Discontinuity occurs where the function is not defined.
Cosine Function

$f (x) = \cos x$
- $\cos x$ is defined for all real values of $x$ .
- It is smooth and has no breaks or gaps.
Conclusion

$\cos x is continuous for all x \in R .$

Cosecant Function

$f (x) = \csc x = \frac{1}{\sin x}$
- $\csc x$ is not defined where $\sin x = 0$ .
- $\sin x = 0$ at $x = n π$ , where $n$ is any integer.
Conclusion

$\csc x is discontinuous at x = n π .$ $It is continuous for all other real values of x .$

Secant Function

$f (x) = \sec x = \frac{1}{\cos x}$
- $\sec x$ is not defined where $\cos x = 0$ .
- $\cos x = 0$ at $x = \frac{(2 n + 1) π}{2}$ , where $n$ is any integer.
Conclusion

$\sec x is discontinuous at x = \frac{(2 n + 1) π}{2} .$ $It is continuous everywhere else.$

Cotangent Function

$f (x) = \cot x = \frac{\cos x}{\sin x}$
- $\cot x$ is not defined where $\sin x = 0$ .
- Similar to cosecant, discontinuity occurs at $x = n π$ .
Conclusion

$\cot x is discontinuous at x = n π .$ $It is continuous for all other real values of x .$

Question 23

Find all points of discontinuity of

$f (x) = {\begin{cases} \frac{\sin x}{x}, & if x < 0 \\ x + 1, & if x \geq 0 \end{cases}$

Answer

The function changes its definition at $x = 0$ , so discontinuity (if any) must be checked at $x = 0$ .

Step 1: Compute limit from the left side as $x \to 0^{-}$

Consider

$\lim_{x \to 0^{-}} \frac{\sin x}{x}$

We know the standard limit identity:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

So,

$\lim_{x \to 0^{-}} f (x) = 1$

Step 2: Compute right-hand limit as $x \to 0^{+}$

From the second part $f (x) = x + 1$ when $x \geq 0$ :

$\lim_{x \to 0^{+}} (x + 1) = 0 + 1 = 1$

Step 3: Value of the function at $x = 0$

Since $x \geq 0$ ,

$f (0) = 0 + 1 = 1$

Comparison

$\lim_{x \to 0^{-}} f (x) = 1, \lim_{x \to 0^{+}} f (x) = 1, f (0) = 1$

Since all three are equal:

$\lim_{x \to 0} f (x) = f (0)$

Final Conclusion

$The function is continuous at x = 0.$

There are no other points where the definition changes, and both components of the function are continuous in their respective intervals.

$Therefore, f (x) is continuous for all x .$

Question 24

Determine if the function defined by

$f (x) = {\begin{cases} x^{2} \sin (\frac{1}{x}), & if x \neq 0 \\ 0, & if x = 0 \end{cases}$

is continuous at $x = 0$ .

Answer

To check continuity at $x = 0$ , we need to verify:

$\lim_{x \to 0} f (x) = f (0)$

Given:

$f (0) = 0$

Find the limit of $f (x)$ as $x \to 0$

$f (x) = x^{2} \sin (\frac{1}{x})$

We know that:

$- 1 \leq \sin (\frac{1}{x}) \leq 1$

Multiplying the entire inequality by $x^{2} \geq 0$ :

$- x^{2} \leq x^{2} \sin (\frac{1}{x}) \leq x^{2}$

Now take the limit as $x \to 0$ :

$\lim_{x \to 0} - x^{2} = 0 and \lim_{x \to 0} x^{2} = 0$

By Squeeze (Sandwich) Theorem:

$\lim_{x \to 0} x^{2} \sin (\frac{1}{x}) = 0$

Therefore:

$\lim_{x \to 0} f (x) = 0 = f (0)$

Final Conclusion

$Yes, the function f (x) is continuous at x = 0.$
November 23, 2025

Expression	Value
$\lim_{x \to 0} f (x)$	$- 1$
$f (0)$	$- 1$

Tag: CONTINUITY AND DIFFERENTIABILITY

Exercise-Miscellaneous, Class 12th, Maths, Chapter 5, NCERT (11-22)

Formula Used

Solution

Final Answer

Question 12

Solution:

Step 1: Differentiate y w.r.t. t

Step 2: Differentiate x w.r.t. t

Step 3: Substitute in formula

Use trigonometric identities

Substitute these in:dydx=6(2sin⁡t2cos⁡t2)5⋅2sin⁡2t2

Final Answer

Question 13

Solution

Term 1: sin⁡−1x

Term 2: sin⁡−11−x2

Combine both derivatives

Final Answer

x1+y+y1+x=0, −1<x<1

ddx(x1+y)+ddx(y1+x)=0

Use product rule for each term

First term:

1+y+x⋅121+y⋅dydx

Second term:

dydx1+x+y⋅121+x

1+y+x21+ydydx+1+xdydx+y21+x=0

dydx(x21+y+1+x)=−1+y−y21+x

x1+y+y1+x=0y1+x=−x1+y

y21+x=−x1+y2(1+x)

−1+y+x1+y2(1+x)=−1+y(1−x2(1+x))=−1+y2+x2(1+x)

x21+y+1+x=x+2(1+x)(1+y)21+y

(1+x)(1+y)=−x1+yy

Question 15

First derivative

Second derivative

Compute 1+(dydx)2

Now evaluate the required expression

Conclusion

Question 16

LHS

RHS

Now equate derivatives

Use original equation for substitution

Final step

Question 17

Step 1: First derivatives w.r.t. t

Step 2: First derivative dydx

Step 3: Second derivative d2ydx2

Final Answer

Question 18

First, rewrite the function without modulus

Question 19

Differentiate LHS

Differentiate RHS

Equate derivatives of both sides

Question 20

Example

Check Continuity

Check Differentiability

For ∣x∣:

For ∣x−1∣:

Question 21

Question 22

Multiply both sides by 1−x2

Exercise-Miscellaneous, Class 12th, Maths, Chapter 5, NCERT(1-10)

Question 1

Solution

Question 2

Solution

Question 3

Solution

Question 4

Solution

Step 1: Differentiate u=cos⁡−1(x/2)

Step 2: Differentiate v=(2x+7)−1/2

Apply Product Rule

Question 6

Simplify the expression

Question 7

Substitute these in: $\frac{d y}{d x} = \frac{6 (2 \sin \frac{t}{2} \cos \frac{t}{2})}{5 \cdot 2 \sin^{2} \frac{t}{2}}$

Term 1: $\sin^{- 1} x$

Term 2: $\sin^{- 1} \sqrt{1 - x^{2}}$

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0, - 1 < x < 1$

$\frac{d}{d x} (x \sqrt{1 + y}) + \frac{d}{d x} (y \sqrt{1 + x}) = 0$

$\sqrt{1 + y} + x \cdot \frac{1}{2 \sqrt{1 + y}} \cdot \frac{d y}{d x}$

$\frac{d y}{d x} \sqrt{1 + x} + y \cdot \frac{1}{2 \sqrt{1 + x}}$

$\sqrt{1 + y} + \frac{x}{2 \sqrt{1 + y}} \frac{d y}{d x} + \sqrt{1 + x} \frac{d y}{d x} + \frac{y}{2 \sqrt{1 + x}} = 0$

$\frac{d y}{d x} (\frac{x}{2 \sqrt{1 + y}} + \sqrt{1 + x}) = - \sqrt{1 + y} - \frac{y}{2 \sqrt{1 + x}}$

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ $y \sqrt{1 + x} = - x \sqrt{1 + y}$

$\frac{y}{2 \sqrt{1 + x}} = - \frac{x \sqrt{1 + y}}{2 (1 + x)}$

$- \sqrt{1 + y} + \frac{x \sqrt{1 + y}}{2 (1 + x)} = - \sqrt{1 + y} (1 - \frac{x}{2 (1 + x)})$ $= - \sqrt{1 + y} \frac{2 + x}{2 (1 + x)}$

$\frac{x}{2 \sqrt{1 + y}} + \sqrt{1 + x} = \frac{x + 2 \sqrt{(1 + x) (1 + y)}}{2 \sqrt{1 + y}}$

$\sqrt{(1 + x) (1 + y)} = - \frac{x \sqrt{1 + y}}{y}$

Compute $1 + {(\frac{d y}{d x})}^{2}$

Step 1: First derivatives w.r.t. $t$

Step 2: First derivative $\frac{d y}{d x}$

Step 3: Second derivative $\frac{d^{2} y}{d x^{2}}$

For $∣ x ∣$ :

For $∣ x - 1 ∣$ :

Multiply both sides by $1 - x^{2}$

Step 1: Differentiate $u = \cos^{- 1} (x / 2)$

Step 2: Differentiate $v = (2 x + 7)^{- 1 / 2}$

Differentiate both sides w.r.t. $x$

Substitute $y = (\log x)^{\log x}$

Substitute $y = (\sin x - \cos x)^{(\sin x - \cos x)}$

1. $x^{x}$

2. $x^{a}$

3. $a^{x}$

4. $a^{a}$

$Find the second order derivatives of the functions given in Exercises 1 to 10.$

Q1. $y = x^{2} + 3 x + 2$

Q2. $y = x^{20}$

Q3. $y = x \cos x$

Q4. $y = \log x$

Q5. $y = x^{3} \log x$

Q6. $y = e^{x} \sin 5 x$

Q7. $y = e^{6 x} \cos 3 x$

Solution $\frac{d y}{d x} = e^{6 x} (6 \cos 3 x - 3 \sin 3 x)$ $\frac{d^{2} y}{d x^{2}} = e^{6 x} (27 \cos 3 x - 36 \sin 3 x)$

Q8. $y = \tan^{- 1} x$

Solution $\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ $\frac{d^{2} y}{d x^{2}} = - \frac{2 x}{(1 + x^{2})^{2}}$

Q9. $y = \log (\log x)$

Solution $\frac{d y}{d x} = \frac{1}{x \log x}$ $\frac{d^{2} y}{d x^{2}} = - \frac{\log x + 1}{x^{2} (\log x)^{2}}$

Q10. $y = \sin (\log x)$

Solution $\frac{d y}{d x} = \frac{\cos (\log x)}{x}$ $\frac{d^{2} y}{d x^{2}} = \frac{- \cos (\log x) - \sin (\log x)}{x^{2}}$

Q11. If $y = 5 \cos x - 3 \sin x$ , show that $\frac{d^{2} y}{d x^{2}} + y = 0$

Differentiate $x$

Differentiate $y$

Differentiate $x$

Differentiate $y$

Differentiate $y$

Differentiate $x$

Differentiate $x$

Differentiate $y$

Differentiate $x$

Differentiate $y$