Q: Find $\frac{d y}{d x}: x y+y^{2}=\tan x+y$

Given, $x y+y^{2}=\tan x+y$ Differentiating with respect to $x$, we get $\frac{d}{d x}\left(x y+y^{2}\right)=\frac{d}{d x}(\tan x+y)$ $\Rightarrow \frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}$ $\Rightarrow\left[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x} \quad$ [using product rule and chain rule] $\Rightarrow y .1+x \frac{d y}{d x}+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x} \Rightarrow(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$ $\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$

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

Given, $y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ $\Rightarrow \sin y=\frac{2 x}{1+x^{2}}$ Differentiating with respect to $x$, we get $$ \begin{equation*} \frac{d}{d x}(\sin y)=\frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right) \tag{1} \end{equation*} $$ $\Rightarrow \cos y \frac{d y}{d x}=\frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)$ The function $\frac{2 x}{1+x^{2}}$, is of the form of $\frac{u}{v}$ By quotient rule, we get $$ \begin{aligned} \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right) & =\frac{\left(1+x^{2}\right) \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \\ & =\frac{\left(1+x^{2}\right) \cdot 2-2 x \cdot[0+2 x]}{\left(1+x^{2}\right)^{2}} \\ & =\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}} \\ & =\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} \end{aligned} $$ Also, $\sin y=\frac{2 x}{1+x^{2}}$ $$ \begin{aligned} \cos y & =\sqrt{1-\sin ^{2} y}=\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}} \\ & =\sqrt{\frac{\left(1+x^{2}\right)^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}} \\ & =\sqrt{\frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}} \\ & =\frac{1-x^{2}}{1+x^{2}} \end{aligned} $$ From (1), (2) and (3), we get $\frac{1-x^{2}}{1+x^{2}} \times \frac{d y}{d x}=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$ $\Rightarrow \frac{d y}{d x}=\frac{2}{1+x^{2}}$

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

Given, $y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$ $$ \begin{equation*} \Rightarrow \tan y=\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) \tag{1} \end{equation*} $$ Since, we know that $$ \begin{equation*} \Rightarrow \tan y=\left(\frac{3 \tan \frac{y}{3}-\tan ^{3} \frac{y}{3}}{1-3 \tan ^{2} \frac{y}{3}}\right) \tag{2} \end{equation*} $$ Comparing (1) and (2) we get, $x=\tan \frac{y}{3}$ Differentiating with respect to $x$, we get $\frac{d}{d x}(x)=\frac{d}{d x}\left(\tan \frac{y}{3}\right)$ $\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{d}{d x}\left(\frac{y}{3}\right)$ $\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{1}{3} \cdot \frac{d y}{d x}$ $\Rightarrow \frac{d y}{d x}=\frac{3}{\sec ^{2} \frac{y}{3}}=\frac{3}{1+\tan ^{2} \frac{y}{3}}$ $\therefore \frac{d y}{d x}=\frac{3}{1+x^{2}}$

Question 1

Find $\frac{d y}{d x}: 2 x+3 y=\sin x$

Accepted Answer

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

Differentiating with respect to $x$, we get

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

$\Rightarrow \frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\cos x$

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

$\Rightarrow 3 \frac{d y}{d x}=\cos x-2$

$	herefore \frac{d x}{d y}=\frac{\cos x-2}{3}$

Question 2

Find $\frac{d y}{d x}: 2 x+3 y=\sin y$

Accepted Answer

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

Differentiating with respect to $x$, we get

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

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

[By using chain rule]

$\Rightarrow 2=(\cos y-3) \frac{d y}{d x}$

$	herefore \frac{d y}{d x}=\frac{2}{\cos y-3}$

Question 3

Find $\frac{d y}{d x}: a x+b y^{2}=\cos y$

Accepted Answer

Given, $a x+b y^{2}=\cos y$

Differentiating with respect to $x$, we get

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

$$

\begin{equation*}

\Rightarrow a+b \frac{d}{d x}\left(y^{2}ight)=\frac{d}{d x}(\cos y) 	ag{1}

\end{equation*}

$$

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

From (1) and (2), we obtain

$a+b 	imes 2 y \frac{d y}{d x}=-\sin y \frac{d y}{d x}$

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

$	herefore \frac{d y}{d x}=\frac{-a}{2 b y+\sin y}$

Question 4

Find $\frac{d y}{d x}: x y+y^{2}=	an x+y$

Accepted Answer

Given, $x y+y^{2}=	an x+y$

Differentiating with respect to $x$, we get

$\frac{d}{d x}\left(x y+y^{2}ight)=\frac{d}{d x}(	an x+y)$

$\Rightarrow \frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}ight)=\frac{d}{d x}(	an x)+\frac{d y}{d x}$

$\Rightarrow\left[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}ight]+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x} \quad$ [using product rule and chain rule]

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

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

Question 5

Find $\frac{d y}{d x}: x^{2}+x y+y^{2}=100$

Accepted Answer

Given, $x^{2}+x y+y^{2}=100$

Differentiating with respect to $x$, we get

$\frac{d}{d x}\left(x^{2}+x y+y^{2}ight)=\frac{d}{d x}(100)$

$\Rightarrow \frac{d}{d x}\left(x^{2}ight)+\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}ight)=0$

$\Rightarrow 2 x+\left[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}ight]+2 y \frac{d y}{d x}=0$

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

$\Rightarrow 2 x+y+(x+2 y) \frac{d y}{d x}=0$

$	herefore \frac{d y}{d x}=-\frac{2 x+y}{x+2 y}$

Question 6

Find $\frac{d y}{d x}: x^{3}+x^{2} y+x y^{2}+y^{3}=81$

Accepted Answer

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

Differentiating with respect to $x$, we get

$\frac{d}{d x}\left(x^{3}+x^{2} y+x y^{2}+y^{3}ight)=\frac{d}{d x}(8$

$\Rightarrow \frac{d}{d x}\left(x^{3}ight)+\frac{d}{d x}\left(x^{2} yight)+\frac{d}{d x}\left(x y^{2}ight)+\frac{d}{d x}\left(y^{3}ight)=0$

$\Rightarrow 3 x^{2}+\left[y \frac{d}{d x}\left(x^{2}ight)+x^{2} \frac{d y}{d x}ight]+\left[y^{2} \frac{d}{d x}(x)+x \frac{d}{d x}\left(y^{2}ight)ight]+3 y^{2} \frac{d y}{d x}=0$

$\Rightarrow 3 x^{2}+\left[y \cdot 2 x+x^{2} \frac{d y}{d x}ight]+\left[y^{2} \cdot 1+x \cdot 2 y \cdot \frac{d y}{d x}ight]+3 y^{2} \frac{d x}{d y}=0$

$\Rightarrow\left(x^{2}+2 x y+3 y^{2}ight) \frac{d y}{d x}+\left(3 x^{2}+2 x y+y^{2}ight)=0$

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

Question 7

Find $\frac{d x}{d y}: \sin ^{2} y+\cos x y=\pi$

Accepted Answer

Given, $\sin ^{2} y+\cos x y=\pi$

Differentiating with respect to $x$, we get

$$

\begin{align*}

& \frac{d}{d x}\left(\sin ^{2} y+\cos x yight)=\frac{d}{d x}(\pi) \

& \Rightarrow \frac{d}{d x}\left(\sin ^{2} yight)+\frac{d}{d x}(\cos x y)=0 	ag{1}

\end{align*}

$$

Using chain rule, we obtain

$$

\begin{align*}

& \frac{d}{d x}\left(\sin ^{2} yight)=2 \sin y \frac{d}{d x}(\sin y)=2 \sin y \cos y \frac{d y}{d x}  	ag{2}\

& \frac{d}{d x}(\cos x y)=-\sin x y \frac{d}{d x}(x y)=-\sin x y\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}ight] \

& =-\sin x y\left[y \cdot 1+x \frac{d y}{d x}ight]=-y \sin x y-x \sin x y \frac{d y}{d x} 	ag{3}

\end{align*}

$$

From (1), (2) and (3), we obtain

$2 \sin y \cos y \frac{d y}{d x}+\left(-y \sin x y-x \sin x y \frac{d y}{d x}ight)=0$

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

$\Rightarrow(\sin 2 y-x \sin x y) \frac{d x}{d y}=y \sin x y$

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

Question 8

Find $\frac{d y}{d x}: \sin ^{2} x+\cos ^{2} y=1$

Accepted Answer

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

Differentiating with respect to $x$, we get

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

$\Rightarrow \frac{d}{d x}\left(\sin ^{2} xight)+\frac{d}{d x}\left(\cos ^{2} yight)=0$

$\Rightarrow 2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cos y \cdot \frac{d}{d x}(\cos y)=0$

$\Rightarrow 2 \sin x \cos x+2 \cos y(-\sin y) \cdot \frac{d y}{d x}=0$

$\Rightarrow \sin 2 x-\sin 2 y \frac{d y}{d x}=0$

$	herefore \frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}$

Question 9

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

Accepted Answer

Given,

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

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

Differentiating with respect to $x$, we get

$$

\begin{equation*}

\frac{d}{d x}(\sin y)=\frac{d}{d x}\left(\frac{2 x}{1+x^{2}}ight) 	ag{1}

\end{equation*}

$$

$\Rightarrow \cos y \frac{d y}{d x}=\frac{d}{d x}\left(\frac{2 x}{1+x^{2}}ight)$

The function $\frac{2 x}{1+x^{2}}$, is of the form of $\frac{u}{v}$

By quotient rule, we get

$$

\begin{aligned}

\frac{d}{d x}\left(\frac{2 x}{1+x^{2}}ight) & =\frac{\left(1+x^{2}ight) \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}ight)}{\left(1+x^{2}ight)^{2}} \

& =\frac{\left(1+x^{2}ight) \cdot 2-2 x \cdot[0+2 x]}{\left(1+x^{2}ight)^{2}} \

& =\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}ight)^{2}} \

& =\frac{2\left(1-x^{2}ight)}{\left(1+x^{2}ight)^{2}}

\end{aligned}

$$

Also, $\sin y=\frac{2 x}{1+x^{2}}$

$$

\begin{aligned}

\cos y & =\sqrt{1-\sin ^{2} y}=\sqrt{1-\left(\frac{2 x}{1+x^{2}}ight)^{2}} \

& =\sqrt{\frac{\left(1+x^{2}ight)^{2}-4 x^{2}}{\left(1+x^{2}ight)^{2}}} \

& =\sqrt{\frac{\left(1-x^{2}ight)^{2}}{\left(1+x^{2}ight)^{2}}} \

& =\frac{1-x^{2}}{1+x^{2}}

\end{aligned}

$$

From (1), (2) and (3), we get

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

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

Question 10

Find $\frac{d y}{d x}: y=	an ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}ight),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Accepted Answer

Given, $y=	an ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}ight)$

$$

\begin{equation*}

\Rightarrow 	an y=\left(\frac{3 x-x^{3}}{1-3 x^{2}}ight) 	ag{1}

\end{equation*}

$$

Since, we know that

$$

\begin{equation*}

\Rightarrow 	an y=\left(\frac{3 	an \frac{y}{3}-	an ^{3} \frac{y}{3}}{1-3 	an ^{2} \frac{y}{3}}ight) 	ag{2}

\end{equation*}

$$

Comparing (1) and (2) we get,

$x=	an \frac{y}{3}$

Differentiating with respect to $x$, we get

$\frac{d}{d x}(x)=\frac{d}{d x}\left(	an \frac{y}{3}ight)$

$\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{d}{d x}\left(\frac{y}{3}ight)$

$\Rightarrow 1=\sec ^{2} \frac{y}{3} \cdot \frac{1}{3} \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{3}{\sec ^{2} \frac{y}{3}}=\frac{3}{1+	an ^{2} \frac{y}{3}}$

$	herefore \frac{d y}{d x}=\frac{3}{1+x^{2}}$

Question 11

Find $\frac{d y}{d x}: y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}ight), 0<x<1$

Solution:

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

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

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

Comparing LHS and RHS, we get

$	an \frac{y}{2}=x$

Differentiating with respect to $x$, we get

$\sec ^{2} \frac{y}{2} \cdot \frac{d}{d x}\left(\frac{y}{2}ight)=\frac{d}{d x}(x)$

$\Rightarrow \sec ^{2} \frac{y}{2} 	imes \frac{1}{2} \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sec ^{2} \frac{y}{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{2}{1+	an ^{2} \frac{y}{2}}$

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

Accepted Answer

(No solution provided.)

Question 12

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

Accepted Answer

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

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

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

Differentiating with respect to $x$, we get

$$

\begin{equation*}

\frac{d}{d x}(\sin y)=\frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}ight) 	ag{1}

\end{equation*}

$$

Using chain rule, we get

$$

\begin{aligned}

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

& \cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\left(\frac{1-x^{2}}{1+x^{2}}ight)^{2}} \

& \quad=\sqrt{\frac{\left(1+x^{2}ight)^{2}-\left(1-x^{2}ight)^{2}}{\left(1+x^{2}ight)^{2}}}=\sqrt{\frac{4 x^{2}}{\left(1+x^{2}ight)^{2}}}=\frac{2 x}{1+x^{2}}

\end{aligned}

$$

Therefore,

$$

\begin{align*}

\frac{d}{d x}(\sin y) & =\frac{2 x}{1+x^{2}} \frac{d y}{d x}  	ag{2}\

\frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}ight) & =\frac{\left(1+x^{2}ight) \cdot \frac{d}{d x}\left(1-x^{2}ight)-\left(1-x^{2}ight) \cdot \frac{d}{d x}\left(1+x^{2}ight)}{\left(1+x^{2}ight)^{2}} \

& =\frac{\left(1+x^{2}ight)(-2 x)-\left(1-x^{2}ight)(2 x)}{\left(1+x^{2}ight)^{2}} \

& =\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1+x^{2}ight)^{2}} \

& =\frac{-4 x}{\left(1+x^{2}ight)^{2}} 	ag{3}

\end{align*}

$$

[using quotient rule]

From equation (1), (2) and (3), we get

$\frac{2 x}{1+x^{2}} \frac{d y}{d x}=\frac{-4 x}{\left(1+x^{2}ight)^{2}}$

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

Question 13

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

Accepted Answer

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

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

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

Differentiating with respect to $x$, we get

$$

\begin{aligned}

& \frac{d}{d x}(\cos y)=\frac{d}{d x}\left(\frac{2 x}{1+x^{2}}ight) \

& \Rightarrow-\sin y \cdot \frac{d y}{d x}=\frac{\left(1+x^{2}ight) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}ight)}{\left(1+x^{2}ight)^{2}} \

& \Rightarrow-\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{\left(1+x^{2}ight) 	imes 2-2 x 	imes 2 x}{\left(1+x^{2}ight)^{2}} \

& \left.\Rightarrow \sqrt{1-\left(\frac{2 x}{1+x^{2}}ight)^{2}}ight] \frac{d y}{d x}=-\left[\frac{2\left(1-x^{2}ight)}{\left(1+x^{2}ight)^{2}}ight] \

& \Rightarrow \sqrt{\frac{\left(1+x^{2}ight)^{2}-4 x^{2}}{\left(1+x^{2}ight)^{2}}} \cdot \frac{d y}{d x}=\frac{-2\left(1-x^{2}ight)}{\left(1+x^{2}ight)} \

& \Rightarrow \sqrt{\frac{\left(1-x^{2}ight)^{2}}{\left(1+x^{2}ight)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}ight)}{\left(1-x^{2}ight)^{2}} \

& \Rightarrow \frac{1-x^{2}}{1+x^{2}} \cdot \frac{d y}{d x}=\frac{-2\left(1-x^{2}ight)}{\left(1+x^{2}ight)^{2}} \

& \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}}

\end{aligned}

$$

Question 14

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

Accepted Answer

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

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

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

Differentiating with respect to $x$, we get

$\cos y \cdot \frac{d y}{d x}=2\left[x \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)+\sqrt{1-x^{2}} \frac{d x}{d x}\right]$

$\Rightarrow \sqrt{1-\sin ^{2} y} \frac{d y}{d x}=2\left[\frac{x}{2} \cdot \frac{-2 x}{\sqrt{1-x^{2}}}+\sqrt{1-x^{2}}\right]$

$\Rightarrow \sqrt{1-\left(2 x \sqrt{1-x^{2}}\right)^{2}} \cdot \frac{d y}{d x}=2\left[\frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}}\right]$

$\Rightarrow \sqrt{1-4 x^{2}\left(1-x^{2}\right)} \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]$

$\Rightarrow \sqrt{\left(1-2 x^{2}\right)^{2}} \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]$

$\Rightarrow\left(1-2 x^{2}\right) \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]$

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

Question 15

Find $\frac{d y}{d x}: y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}ight), 0<x<\frac{1}{\sqrt{2}}$

Solution:

Given, $y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}ight)$

$$

\begin{aligned}

& \Rightarrow y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}ight) \

& \Rightarrow \sec y=\left(\frac{1}{2 x^{2}-1}ight) \

& \Rightarrow \cos y=2 x^{2}-1 \

& \Rightarrow 2 x^{2}=1+\cos y \

& \Rightarrow 2 x^{2}=2 \cos ^{2} \frac{y}{2} \

& \Rightarrow x=\cos \frac{y}{2}

\end{aligned}

$$

Differentiating with respect to $x$, we get

$$

\begin{aligned}

& \frac{d}{d x}(x)=\frac{d}{d x}\left(\cos \frac{y}{2}ight) \

& \Rightarrow 1=\sin \frac{y}{2} \cdot \frac{d}{d x}\left(\frac{y}{2}ight) \

& \Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x} \

& \Rightarrow \frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}} \

& \Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-\cos ^{2} \frac{y}{2}}} \

& \Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}

\end{aligned}

$$

\section*{EXERCISE 5.4}

Accepted Answer

(No solution provided.)