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Question 1

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

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

Solution

Given:

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

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

$$(uvw{)}^{\mathrm{\prime }}=u^{\mathrm{\prime }}vw+u{v}^{\mathrm{\prime }}w+uv{w}^{\mathrm{\prime }}$$

Let:

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

Then:

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

Applying product rule:

$$y^{\mathrm{\prime }}=u^{\mathrm{\prime }}vw+u{v}^{\mathrm{\prime }}w+uv{w}^{\mathrm{\prime }}$$
$$y^{\mathrm{\prime }}=(-\sin x)(\cos 2x)(\cos 3x)+(\cos x)(-2\sin 2x)(\cos 3x)$$

$$+(\cos x)(\cos 2x)(-3\sin 3x)$$
$$\boxed{{\displaystyle y^{\mathrm{\prime }}=-\sin x\cos 2x\cos 3x-2\cos x\sin 2x\cos 3x-3\cos x\cos 2x\sin 3x}}$$

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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:

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

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. $x$:

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

Multiply both sides by $y$:

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

Final Answer 

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}]}}$$

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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:

$$\frac{1}{y}\cdot \frac{dy}{d\mathrm{x<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}$$

Right side (product rule):

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

So:

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

Multiply both sides by $y$:

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

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

$$\boxed{{\displaystyle \frac{dy}{dx}=(\log x{)}^{\cos x}[\frac{\cos x}{x\log x}-\sin x\log (\log x)]}}$$

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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:$$\frac{1}{u}\frac{du}{dx}$$

Right side (product rule):

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

So:

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

Now differentiate the whole expression:

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

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

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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:

$$\frac{1}{y}\cdot \frac{dy}{dx}$$

Right side:

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

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

Multiply both sides by $y$:

$$\frac{dy}{dx}=(x+3{)}^2(x+4{)}^3(x+5{)}^4(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5})$$

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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:$$\frac{1}{y}\frac{dy}{dx}$$

Right side (chain rule):

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

Similarly:

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

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

So:

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

Multiply both sides by $y$:

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=(1+\frac{1}{x})(1+\frac{1}{x^2})(1+\frac{1}{x^3})[-\frac{1}{x(x+1)}-\frac{2}{x^2(x^2+1)}-\frac{3}{x^3(x^3+1)}]}}$$

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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:$$\frac{1}{u}\frac{du}{dx}$$

Right side (product rule):

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

So:

$$\frac{du}{dx}=(\log x{)}^x[\log (\log x)+\frac{1}{\log x}]$$

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:$$\frac{1}{v}\frac{dv}{dx}$$

Right side:

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

Thus:

$$\frac{dv}{dx}=x^{\log x}\cdot \frac{2\log x}{x}$$

Combine (since $y=u+v$)

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

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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:$$\frac{1}{u}\frac{du}{dx}$$

Right side (product rule):

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

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

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

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

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

Final Derivative

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

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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:

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

So:

$$\frac{du}{dx}=x^{\sin x}(\cos x\log x+\frac{\sin x}{x})$$

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

Take log:

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

Differentiate (product rule):

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

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

Final Derivative

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

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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:

$$\frac{1}{u}\frac{du}{dx}$$

Right side (product rule):

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

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

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

So:

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

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

Use quotient rule:

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

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

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

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

Final Derivative

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

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Question 11 

Differentiate w.r.t. $x$:

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

Solution

Part 1: Let

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

Taking log:

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

Differentiate:

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

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

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

Thus:

$$\boxed{{\displaystyle \frac{du}{dx}=(x\cos x{)}^x[\log (x\cos x)+\frac{1}{x}-\tan x]}}$$

Part 2: Let

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

Taking log:

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

Differentiate:

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

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

Thus:

$$\boxed{{\displaystyle \frac{dv}{dx}=(x\sin x{)}^{1\mathrm{/}x}[-\frac{\log (x\sin x)}{x^2}+\frac{1}{x^2}+\frac{\cos x}{x\sin x}]}}$$

Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=(x\cos x{)}^x[\log (x\cos x)+\frac{1}{x}-\tan x]+(x\sin x{)}^{1\mathrm{/}x}[-\frac{\log (x\sin x)}{x^2}+\frac{1}{x^2}+\frac{\cos x}{x\sin x}]}}$$

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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:

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

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

Step 2: Differentiate $y^x$

Write:

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

Differentiate:

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

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

Step 3: Differentiate RHS

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

Step 4: Combine all terms

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

Now collect $\frac{dy}{dx}$ terms together:

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

Factor out $\frac{dy}{dx}$:

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-(\frac{y{x}^y}{x}+y^x\log y)}{x^y\log x+\frac{x}{y}{y}^x}}}$$

Or more neatly:

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-(\frac{y}{x}{x}^y+y^x\log y)}{x^y\log x+\frac{x}{y}{y}^x}}}$$

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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:

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

Right side:

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

Arrange terms

$$\log y+\frac{x}{y}\frac{dy}{dx}=\log x\cdot \frac{dy}{dx}+\frac{y}{x}$$

Bring all ${\displaystyle \frac{dy}{dx}}$ terms to one side:

$$\frac{x}{y}\frac{dy}{dx}-\log x\cdot \frac{dy}{dx}=\frac{y}{x}-\log y$$

Factor out $\frac{dy}{dx}$:

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}}}$$

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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:

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

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

Right side:

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

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

Rearranging terms:

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

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

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{\log (\cos y)+y\tan x}{\log (\cos x)+x\tan y}}}$$

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QUESTION 15

Differentiate w.r.t. $x$:

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

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

Solution :

Differentiate both sides:

Left side

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

(using product rule)

Right side

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

Form the equation

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

Expand RHS:

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

Bring $\frac{dy}{dx}$ terms together:

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

Factor out $\frac{dy}{dx}$:

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

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{e^{x-y}-y}{x+e^{x-y}}}}$$

Since from the given equation:

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

Replace $e^{x-y}$ in the derivative:

$$\frac{dy}{dx}=\frac{xy-y}{x+xy}$$

Factor numerator and denominator:

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

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QUESTION 16

Find the derivative of the function

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

and hence find $f^{\mathrm{\prime }}(1)$.

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. $x$:

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

Multiply both sides by $f(x)$:

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

Final derivative 

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

Now find $f^{\mathrm{\prime }}(1)$

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:

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

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

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

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QUESTION 17

Differentiate the function:

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

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

$$u^{\mathrm{\prime }}=2x-5,v^{\mathrm{\prime }}=3x^2+7$$

Using product rule:

$$y^{\mathrm{\prime }}=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$

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

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(ii) DIFFERENTIATION BY EXPANDING FIRST

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

Multiply:

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

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

Combine like terms:

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

Differentiate term-wise:

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

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(iii) BY LOGARITHMIC DIFFERENTIATION

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

Take log:

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

Differentiate:

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

Multiply by $y$:

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

Simplify (cancel terms):

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

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QUESTION 18

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

$$\frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}$$

(i) By repeated application of the product rule

Let:

$$y=uvw$$

First consider:

$$y=(uv)w$$

Differentiate using product rule:

$$\frac{dy}{dx}=\frac{d(uv)}{dx}\cdot w+(uv)\frac{dw}{dx}$$

Now apply product rule again to $\frac{d(uv)}{dx}$:

$$\frac{d(uv)}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}$$

Substitute back:

$$\frac{dy}{dx}=(\frac{du}{dx}v+u\frac{dv}{dx})w+uv\frac{dw}{dx}$$

Distribute $w$:

$$\frac{dy}{dx}=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}$$

Result (Method 1 final statement)

$$\boxed{{\displaystyle \frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}}}$$

(ii) By logarithmic differentiation

Given:

$$y=uvw$$

Take logarithm on both sides:

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

Differentiate:

$$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}$$

Multiply both sides by $y=uvw$:

$$\frac{dy}{dx}=uvw(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx})$$

Distribute:

$$\frac{dy}{dx}=vw\frac{du}{dx}+uw\frac{dv}{dx}+uv\frac{dw}{dx}$$

Final Answer

$$\boxed{{\displaystyle \frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}}}$$

This verifies the required identity by both methods.

Conclusion

Yes, both methods give the same derivative result:

$$\boxed{{\displaystyle \frac{d}{dx}(uvw)=u^{\mathrm{\prime }}vw+u{v}^{\mathrm{\prime }}w+uv{w}^{\mathrm{\prime }}}}$$