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Find $\frac{d\mathrm{y }}{d\mathrm{x }}$in the following:

Question 1

Differentiate:

$$2x+3y=\sin x$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin x)$$

$$2+3\frac{dy}{dx}=\cos x$$

Now solve for $\frac{dy}{dx}$:

$$3\frac{dy}{dx}=\cos x-2$$

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{\cos x-2}{3}}}$$

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Question 2

Differentiate:

$$2x+3y=\sin y$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin y)$$
$$2+3\frac{dy}{dx}=\cos y\cdot \frac{dy}{dx}$$

(using chain rule on $\sin y$)

Now collect $\frac{dy}{dx}$ terms on one side:

$$3\frac{dy}{dx}-\cos y\frac{dy}{dx}=-2$$

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

$$\frac{dy}{dx}(3-\cos y)=-2$$

So,

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-2}{3-\cos y}}}$$

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Question 3

Find $\frac{dy}{dx}$ if:

$$ax+b{y}^2=\cos y$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(ax)+\frac{d}{dx}(b{y}^2)=\frac{d}{dx}(\cos y)$$

Differentiate term by term

$$a+b\cdot 2y\frac{dy}{dx}=-\sin y\cdot \frac{dy}{dx}$$

(using chain rule on cos y)

So,

$$a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}$$

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

$$2by\frac{dy}{dx}+\sin y\frac{dy}{dx}=-a$$

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

$$\frac{dy}{dx}(2by+\sin y)=-a$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-a}{2by+\sin y}}}$$

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Question 4

Find $\frac{dy}{dx}$ if:

$$xy+y^2=\tan x+y$$

Solution

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

$$\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(\tan x)+\frac{d}{dx}(y)$$

Differentiate term by term

1. $\frac{d}{dx}(xy)=x\frac{dy}{dx}+y$
   (product rule)

2. $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

3. $\frac{d}{dx}(\tan x)={\sec }^{2}x$

4. $\frac{d}{dx}(y)=\frac{dy}{dx}$

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Substitute all derivatives

$$x\frac{dy}{dx}+y+2y\frac{dy}{dx}={\sec }^{2}x+\frac{dy}{dx}$$

Collect $\frac{dy}{dx}$ terms on one side:

$$x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}={\sec }^{2}x-y$$

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

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{{\sec }^{2}x-y}{x+2y-1}}}$$

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Find $\frac{d\mathrm{y }}{d\mathrm{x }}$in the following:

Question 1

Differentiate:

$$2x+3y=\sin x$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin x)$$

$$2+3\frac{dy}{dx}=\cos x$$

Now solve for $\frac{dy}{dx}$:

$$3\frac{dy}{dx}=\cos x-2$$

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{\cos x-2}{3}}}$$

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Question 2

Differentiate:

$$2x+3y=\sin y$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(2x)+\frac{d}{dx}(3y)=\frac{d}{dx}(\sin y)$$
$$2+3\frac{dy}{dx}=\cos y\cdot \frac{dy}{dx}$$

(using chain rule on $\sin y$)

Now collect $\frac{dy}{dx}$ terms on one side:

$$3\frac{dy}{dx}-\cos y\frac{dy}{dx}=-2$$

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

$$\frac{dy}{dx}(3-\cos y)=-2$$

So,

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-2}{3-\cos y}}}$$

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Question 3

Find $\frac{dy}{dx}$ if:

$$ax+b{y}^2=\cos y$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(ax)+\frac{d}{dx}(b{y}^2)=\frac{d}{dx}(\cos y)$$

Differentiate term by term

$$a+b\cdot 2y\frac{dy}{dx}=-\sin y\cdot \frac{dy}{dx}$$

(using chain rule on cos y)

So,

$$a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}$$

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

$$2by\frac{dy}{dx}+\sin y\frac{dy}{dx}=-a$$

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

$$\frac{dy}{dx}(2by+\sin y)=-a$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-a}{2by+\sin y}}}$$

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Question 4

Find $\frac{dy}{dx}$ if:

$$xy+y^2=\tan x+y$$

Solution

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

$$\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(\tan x)+\frac{d}{dx}(y)$$

Differentiate term by term

1. $\frac{d}{dx}(xy)=x\frac{dy}{dx}+y$
   (product rule)

2. $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

3. $\frac{d}{dx}(\tan x)={\sec }^{2}x$

4. $\frac{d}{dx}(y)=\frac{dy}{dx}$

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Substitute all derivatives

$$x\frac{dy}{dx}+y+2y\frac{dy}{dx}={\sec }^{2}x+\frac{dy}{dx}$$

Collect $\frac{dy}{dx}$ terms on one side:

$$x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}={\sec }^{2}x-y$$

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

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{{\sec }^{2}x-y}{x+2y-1}}}$$

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Question 5

Find $\frac{dy}{dx}$ if:

$$x^2+xy+y^2=100$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(100)$$

Now apply differentiation rules:

Term-by-term differentiation

1. $\frac{d}{dx}(x^2)=2x$

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

3. (product rule)

4. $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

5. $\frac{d}{dx}(100)=0$

Substitute into the equation

$$2x+(x\frac{dy}{dx}+y)+2y\frac{dy}{dx}=0$$

Group $\frac{dy}{dx}$ terms:

$$x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-y$$

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

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-2x-y}{x+2y}}}$$

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Question 6

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$x^3+x^{2}y+x{y}^2+y^3=81$$

Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x^3)+\frac{d}{dx}(x^{2}y)+\frac{d}{dx}(x{y}^2)+\frac{d}{dx}(y^3)=\frac{d}{dx}(81)$$

Differentiate term-by-term

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

2. $\frac{d}{dx}(x^{2}y)=2xy+x^2\frac{dy}{dx}$
   (product rule)

3. $\frac{d}{dx}(x{y}^2)=y^2+2xy\frac{dy}{dx}$
   (product rule)

4. $\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}$
   (chain rule)

5. $\frac{d}{dx}(81)=0$

Substitute everything back

$$3x^2+2xy+x^2\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$$

Group the terms containing $\frac{dy}{dx}$:

$$x^2\frac{dy}{dx}+2xy\frac{dy}{dx}+3y^2\frac{dy}{dx}=-3x^2-2xy-y^2$$

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

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

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{3x^2+2xy+y^2}{x^2+2xy+3y^2}}}$$

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Question 7

Find ${\displaystyle \frac{dy}{dx}}$ if:

$${\sin }^{2}y+\cos xy=k$$Solution

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}({\sin }^{2}y)+\frac{d}{dx}(\cos xy)=\frac{d}{dx}(k)$$

Differentiate term by term

      1. ${\sin }^{2}y$

Let $u=\sin y$, so expression = $u^2$

$$\frac{d}{dx}({\sin }^{2}y)=2\sin y\cdot \cos y\cdot \frac{dy}{dx}$$

(because $\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}$

So:$$2\sin y\cos y\frac{dy}{dx}$$

     2. $\cos (xy)$

Apply chain rule and product rule inside:

Derivative of $\cos A$ is $-\sin A$:

$$\frac{d}{dx}(\cos (xy))=-\sin (xy)\cdot \frac{d}{dx}(xy)$$

Now:

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

(using product rule)

So:

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

Right-hand side

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

Combine all results

$$2\sin y\cos y\frac{dy}{dx}-\sin (xy)(x\frac{dy}{dx}+y)=0$$

Expand:

$$2\sin y\cos y\frac{dy}{dx}-x\sin (xy)\frac{dy}{dx}-y\sin (xy)=0$$

Group the terms with $\frac{dy}{dx}$:

$$\frac{dy}{dx}(2\sin y\cos y-x\sin (xy))=y\sin (xy)$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{y\sin (xy)}{2\sin y\cos y-x\sin (xy)}}}$$

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Question 8

Find ${\displaystyle \frac{dy}{dx}}$ if:

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

Solution

Differentiate both sides with respect to $x$:

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

Differentiate term by term

1. ${\sin }^{2}x$

Let $u=\sin x$, then $u^2={\sin }^{2}x$

$$\frac{d}{dx}({\sin }^{2}x)=2\sin x\cdot \cos x$$

    2. ${\cos }^{2}y$

Let $v=\cos y$, so $v^2={\cos }^{2}y$

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

(using chain rule: derivative of $\cos y$ is $-\sin y\frac{dy}{dx}$)

So:

$$\frac{d}{dx}({\cos }^{2}y)=-2\cos y\sin y\frac{dy}{dx}$$

Right-hand side

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

Put everything together

$$2\sin x\cos x-2\cos y\sin y\frac{dy}{dx}=0$$

Move the second term to the other side:

$$2\sin x\cos x=2\cos y\sin y\frac{dy}{dx}$$

Divide both sides by $2\cos y\sin y$:

$$\frac{dy}{dx}=\frac{\sin x\cos x}{\cos y\sin y}$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{\sin x\cos x}{\sin y\cos y}}}$$

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Question 9

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$$

Solution

Let:$$y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$$

We know the identity:

$$\sin (2\theta )=\frac{2\tan \theta }{1+{\tan }^2\theta }$$Comparing:

$$\frac{2x}{1+x^2}=\sin (2\theta )\text{if }x=\tan \theta$$So let:$$x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$$

Then:$$y={\sin }^{-1}(\sin (2\theta ))=2\theta =2{\tan }^{-1}x$$

Now differentiate:

$$y=2{\tan }^{-1}x$$

$$\frac{dy}{dx}=2\cdot \frac{1}{1+x^2}$$

$$\frac{2}{1+x^2}$$

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Question 10

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$$

Solution

We use the trigonometric identity:

$$\tan (3\theta )=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$$

Compare with the given expression:

$$\frac{3x-x^3}{1-3x^2}=\tan (3\theta )\text{if }x=\tan \theta$$

So let:

$$x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$$

Then:$$y={\tan }^{-1}(\tan (3\theta ))=3\theta =3{\tan }^{-1}x$$

The given interval:$$-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$$ensures:

$$\mathrm{\mid }x\mathrm{\mid }<\frac{1}{\sqrt{3}}\Rightarrow \mathrm{\mid }\theta \mathrm{\mid }<\frac{\pi }{6}\Rightarrow \mathrm{\mid }3\theta \mathrm{\mid }<\frac{\pi }{2}$$

Thus, ${\tan }^{-1}(\tan (3\theta ))=3\theta$ is valid (i.e., no ambiguity or discontinuity).

Differentiate

$$y=3{\tan }^{-1}x$$
$$\frac{dy}{dx}=3\cdot \frac{1}{1+x^2}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{3}{1+x^2},-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}}}$$

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Question 11

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$$Solution

Let:

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

We use a trigonometric substitution.

Let:

$$x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$$

Compute the inside expression:

$$\frac{1-x^2}{1+x^2}=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$$

We know the identity:

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

Thus:

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

So:$$y={\cos }^{-1}(\cos (2\theta ))=2\theta$$

Since $0<x<1\Rightarrow 0<\theta <\frac{\pi }{4}$, thus $0<2\theta <\frac{\pi }{2}$, ensuring the inverse cosine gives principal value.

Therefore:

$$y=2\theta =2{\tan }^{-1}x$$

Differentiate

$$\frac{dy}{dx}=2\cdot \frac{1}{1+x^2}$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{2}{1+x^2},0<x<1}}$$

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Question 12

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$$Solution

Let:

$$y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$$

We use trigonometric substitution similar to the previous problem.

Let:

$$x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$$

Then:

$$\frac{1-x^2}{1+x^2}=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$$

Using the identity:

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

So:$$y={\sin }^{-1}(\cos (2\theta ))$$

We also know:

$$\cos (2\theta )=\sin (\frac{\pi }{2}-2\theta )$$

Thus:$$y={\sin }^{-1}(\sin (\frac{\pi }{2}-2\theta ))=\frac{\pi }{2}-2\theta$$

Since $0<x<1$, $0<\theta <\frac{\pi }{4}$, so:

$$0<2\theta <\frac{\pi }{2}\Rightarrow \mathrm{\mid }\frac{\pi }{2}-2\theta \mathrm{\mid }\le \frac{\pi }{2}$$

Hence the principal value is correct.

Differentiate

$$y=\frac{\pi }{2}-2\theta =\frac{\pi }{2}-2{\tan }^{-1}x$$

$$\frac{dy}{dx}=-2\cdot \frac{1}{1+x^2}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{2}{1+x^2},0<x<1}}$$

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Question 13

$$y={\cos }^{-1}\left(\frac{2x}{1+x^2}\right),-1<x<1$$

Now we will solve this correctly.

Solution

Let:

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

Use substitution:

$$x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$$

Then:

$$\frac{2x}{1+x^2}=\frac{2\tan \theta }{1+{\tan }^2\theta }$$

We know the identity:

$$\sin (2\theta )=\frac{2\tan \theta }{1+{\tan }^2\theta }$$

So:

$$\frac{2x}{1+x^2}=\sin (2\theta )$$

Thus:

$$y={\cos }^{-1}(\sin (2\theta ))$$

We also know:$$\sin (2\theta )=\cos (\frac{\pi }{2}-2\theta )$$

So:$$y=\frac{\pi }{2}-2\theta =\frac{\pi }{2}-2{\tan }^{-1}x$$

Differentiate

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

$$\frac{dy}{dx}=-2\cdot \frac{1}{1+x^2}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{2}{1+x^2},-1<x<1}}$$

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Question 14

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$Solution

Let:$$y={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$$

We use a substitution to simplify.

Let:

$$x=\sin \theta \Rightarrow \theta ={\sin }^{-1}x$$

Then:

$$\sqrt{1-x^2}=\sqrt{1-{\sin }^2\theta }=\cos \theta$$

So:

$$2x\sqrt{1-x^2}=2\sin \theta \cos \theta =\sin (2\theta )$$

Thus:$$y={\sin }^{-1}(\sin (2\theta ))=2\theta$$

The given interval:

$$-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$

gives:

$$-\frac{\pi }{4}<\theta <\frac{\pi }{4}\Rightarrow \mathrm{\mid }2\theta \mathrm{\mid }<\frac{\pi }{2}$$

Therefore:$$y=2\theta =2{\sin }^{-1}x$$

Differentiate

$$y=2{\sin }^{-1}x$$

$$\frac{dy}{dx}=2\cdot \frac{1}{\sqrt{1-x^2}}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}},-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}}}$$

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Question 15

Find ${\displaystyle \frac{dy}{dx}}$ if:

$$y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right),0<x<\frac{1}{\sqrt{2}}$$Solution

Let:

$$y={\sec }^{-1}(u),\text{where }u=\frac{1}{2x^2-1}$$

We know:

$$\frac{d}{dx}({\sec }^{-1}u)=\frac{1}{\mathrm{\mid }u\mathrm{\mid }\sqrt{u^2-1}}\cdot \frac{du}{dx}$$

Step 1: Differentiate the inside function

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

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

Step 2: Apply inverse secant derivative formula

$$\frac{dy}{dx}=\frac{1}{\mid \frac{1}{2x^2-1}\mid \sqrt{{\left(\frac{1}{2x^2-1}\right)}^2-1}}\cdot (-\frac{4x}{(2x^2-1{)}^2})$$

Since $0<x<\frac{1}{\sqrt{2}}$, we have $2x^2-1<0$, so:

$$\mid \frac{1}{2x^2-1}\mid =-\frac{1}{2x^2-1}$$

Thus:

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

Simplify inner root:

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

So:

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

Since $2x^2-1<0$, $\mathrm{\mid }2x^2-1\mathrm{\mid }=1-2x^2$:

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

Final Simplification

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

Cancel $x$:

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

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{2}{(2x^2-1)\sqrt{1-x^2}},0<x<\frac{1}{\sqrt{2}}}}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{2}{(2x^2-1)\sqrt{1-x^2}}}}$$