Solution

The given relationship is $2 x+3 y=\sin x$

Differentiating this relationship with respect to $x$, we obtain

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

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

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

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

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

Solution

The given relationship is $2 x+3 y=\sin y$

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\frac{d}{d x}(\sin y)$ $\Rightarrow 2+3 \frac{d y}{d x}=\cos y \frac{d y}{d x} \quad$ [By using chain rule]

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

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

Solution

The given relationship is $a x+b y^{2}=\cos y$

Differentiating this relationship with respect to $x$, we obtain

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

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

Using chain rule, we obtain $\frac{d}{d x}(y^{2})=2 y \frac{d y}{d x}$ and $\frac{d}{d x}(\cos y)=-\sin y \frac{d y}{d x}$

From (1) and (2), we obtain

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

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

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

Solution

The given relationship is $x y+y^{2}=\tan x+y$

Differentiating this relationship with respect to $x$, we obtain

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

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

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

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

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

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

Solution

The given relationship is $x^{2}+x y+y^{2}=100$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(x^{2}+x y+y^{2})=\frac{d}{d x}(100) \\ & \Rightarrow \frac{d}{d x}(x^{2})+\frac{d}{d x}(x y)+\frac{d}{d x}(y^{2})=0 \end{aligned} $

$\Rightarrow 2 x+[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}]+2 y \frac{d y}{d x}=0 \quad$ [Using product rule and chain rule]

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

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

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

Solution

The given relationship is $x^{3}+x^{2} y+x y^{2}+y^{3}=81$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(x^{3}+x^{2} y+x y^{2}+y^{3})=\frac{d}{d x}(81) \\ & \Rightarrow \frac{d}{d x}(x^{3})+\frac{d}{d x}(x^{2} y)+\frac{d}{d x}(x y^{2})+\frac{d}{d x}(y^{3})=0 \\ & \Rightarrow 3 x^{2}+[y \frac{d}{d x}(x^{2})+x^{2} \frac{d y}{d x}]+[y^{2} \frac{d}{d x}(x)+x \frac{d}{d x}(y^{2})]+3 y^{2} \frac{d y}{d x}=0 \\ & \Rightarrow 3 x^{2}+[y \cdot 2 x+x^{2} \frac{d y}{d x}]+[y^{2} \cdot 1+x \cdot 2 y \cdot \frac{d y}{d x}]+3 y^{2} \frac{d y}{d x}=0 \\ & \Rightarrow(x^{2}+2 x y+3 y^{2}) \frac{d y}{d x}+(3 x^{2}+2 x y+y^{2})=0 \\ & \therefore \frac{d y}{d x}=\frac{-(3 x^{2}+2 x y+y^{2})}{(x^{2}+2 x y+3 y^{2})} \end{aligned} $

Solution

The given relationship is $\sin ^{2} y+\cos x y=\pi$

Differentiating this relationship with respect to $x$, we obtain

$$ \begin{align*} & \frac{d}{d x}(\sin ^{2} y+\cos x y)=\frac{d}{d x}(\pi) \\ & \Rightarrow \frac{d}{d x}(\sin ^{2} y)+\frac{d}{d x}(\cos x y)=0 \tag{1} \end{align*} $$

Using chain rule, we obtain

$$ \begin{align*} \frac{d}{d x}(\sin ^{2} y) & =2 \sin y \frac{d}{d x}(\sin y)=2 \sin y \cos y \frac{d y}{d x} \tag{2}\\ \frac{d}{d x}(\cos x y) & =-\sin x y \frac{d}{d x}(x y)=-\sin x y[y \frac{d}{d x}(x)+x \frac{d y}{d x}] \\ & =-\sin x y[y \cdot 1+x \frac{d y}{d x}]=-y \sin x y-x \sin x y \frac{d y}{d x} \tag{3} \end{align*} $$

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

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

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

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

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

Solution

The given relationship is $\sin ^{2} x+\cos ^{2} y=1$

Differentiating this relationship with respect to $x$, we obtain $\frac{d}{d x}(\sin ^{2} x+\cos ^{2} y)=\frac{d}{d x}(1)$

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

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

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

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

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

Solution

The given relationship is $y=\sin ^{-1}(\frac{2 x}{1+x^{2}})$

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

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

Differentiating this relationship with respect to $x$, we obtain

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

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

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

Therefore, by quotient rule, we obtain

$$ \begin{align*} & \frac{d}{d x}(\frac{2 x}{1+x^{2}})=\frac{(1+x^{2}) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}(1+x^{2})}{(1+x^{2})^{2}} \\ & =\frac{(1+x^{2}) \cdot 2-2 x \cdot[0+2 x]}{(1+x^{2})^{2}}=\frac{2+2 x^{2}-4 x^{2}}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}} \tag{2} \end{align*} $$

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

$$ \begin{align*} \Rightarrow \cos y & =\sqrt{1-\sin ^{2} y}=\sqrt{1-(\frac{2 x}{1+x^{2}})^{2}}=\sqrt{\frac{(1+x^{2})^{2}-4 x^{2}}{(1+x^{2})^{2}}} \\ & =\sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\frac{1-x^{2}}{1+x^{2}} \tag{3} \end{align*} $$

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

$ \begin{aligned} & \frac{1-x^{2}}{1+x^{2}} \times \frac{d y}{d x}=\frac{2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{d y}{d x}=\frac{2}{1+x^{2}} \end{aligned} $

Solution

The given relationship is $y=\tan ^{-1}(\frac{3 x-x^{3}}{1-3 x^{2}})$ $y=\tan ^{-1}(\frac{3 x-x^{3}}{1-3 x^{2}})$

$\Rightarrow \tan y=\frac{3 x-x^{3}}{1-3 x^{2}}$

It is known that, $\tan y=\frac{3 \tan \frac{y}{3}-\tan ^{3} \frac{y}{3}}{1-3 \tan ^{2} \frac{y}{3}}$

Comparing equations (1) and (2), we obtain

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

Differentiating this relationship with respect to $x$, we obtain

$\frac{d}{d x}(x)=\frac{d}{d x}(\tan \frac{y}{3})$

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

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

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

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

Solution

The given relationship is, $y=\cos ^{-1}(\frac{1-x^{2}}{1+x^{2}})$

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

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

On comparing L.H.S. and R.H.S. of the above relationship, we obtain

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

Differentiating this relationship with respect to $x$, we obtain

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

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

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

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

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

Solution

The given relationship is $y=\sin ^{-1}(\frac{1-x^{2}}{1+x^{2}})$ $y=\sin ^{-1}(\frac{1-x^{2}}{1+x^{2}})$

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

Differentiating this relationship with respect to $x$, we obtain

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

Using chain rule, we obtain

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

$ \begin{aligned} & \cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-(\frac{1-x^{2}}{1+x^{2}})^{2}} \\ & =\sqrt{\frac{(1+x^{2})^{2}-(1-x^{2})^{2}}{(1+x^{2})^{2}}}=\sqrt{\frac{4 x^{2}}{(1+x^{2})^{2}}}=\frac{2 x}{1+x^{2}} \\ & \therefore \frac{d}{d x}(\sin y)=\frac{2 x}{1+x^{2}} \frac{d y}{d x} \end{aligned} $

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

[Using quotient rule]

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

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

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

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

$ \begin{aligned} & \frac{2 x}{1+x^{2}} \frac{d y}{d x}=\frac{-4 x}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned} $

Alternate method

$ \begin{aligned} & y=\sin ^{-1}(\frac{1-x^{2}}{1+x^{2}}) \\ & \Rightarrow \sin y=\frac{1-x^{2}}{1+x^{2}} \end{aligned} $

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

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

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

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

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

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

$\Rightarrow x=\tan (\frac{\pi}{4}-\frac{y}{2})$

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(x)=\frac{d}{d x} \cdot[\tan (\frac{\pi}{4}-\frac{y}{2})] \\ & \Rightarrow 1=\sec ^{2}(\frac{\pi}{4}-\frac{y}{2}) \cdot \frac{d}{d x}(\frac{\pi}{4}-\frac{y}{2}) \\ & \Rightarrow 1=[1+\tan ^{2}(\frac{\pi}{4}-\frac{y}{2})] \cdot(-\frac{1}{2} \frac{d y}{d x}) \\ & \Rightarrow 1=(1+x^{2})(-\frac{1}{2} \frac{d y}{d x}) \\ & \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned} $

Solution

The given relationship is $y=\cos ^{-1}(\frac{2 x}{1+x^{2}})$

$ \begin{aligned} & y=\cos ^{-1}(\frac{2 x}{1+x^{2}}) \\ & \Rightarrow \cos y=\frac{2 x}{1+x^{2}} \end{aligned} $

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}(\cos y)=\frac{d}{d x} \cdot(\frac{2 x}{1+x^{2}}) \\ & \Rightarrow-\sin y \cdot \frac{d y}{d x}=\frac{(1+x^{2}) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}(1+x^{2})}{(1+x^{2})^{2}} \end{aligned} $

$ \begin{aligned} & \Rightarrow-\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{(1+x^{2}) \times 2-2 x \cdot 2 x}{(1+x^{2})^{2}} \\ & \Rightarrow[\sqrt{1-(\frac{2 x}{1+x^{2}})^{2}}] \frac{d y}{d x}=-[\frac{2(1-x^{2})}{(1+x^{2})^{2}}] \\ & \Rightarrow \sqrt{\frac{(1+x^{2})^{2}-4 x^{2}}{(1+x^{2})^{2}}} \frac{d y}{d x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \sqrt{\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}} \frac{d y}{d x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{1-x^{2}}{1+x^{2}} \cdot \frac{d y}{d x}=\frac{-2(1-x^{2})}{(1+x^{2})^{2}} \\ & \Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned} $

Solution

The given relationship is $y=\sin ^{-1}(2 x \sqrt{1-x^{2}})$

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

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

Differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & \cos y \frac{d y}{d x}=2[x \frac{d}{d x}(\sqrt{1-x^{2}})+\sqrt{1-x^{2}} \frac{d x}{d x}] \\ & \Rightarrow \sqrt{1-\sin ^{2} y} \frac{d y}{d x}=2[\frac{x}{2} \cdot \frac{-2 x}{\sqrt{1-x^{2}}}+\sqrt{1-x^{2}}] \\ & \Rightarrow \sqrt{1-(2 x \sqrt{1-x^{2}})^{2}} \frac{d y}{d x}=2[\frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow \sqrt{1-4 x^{2}(1-x^{2})} \frac{d y}{d x}=2[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow \sqrt{(1-2 x^{2})^{2}} \frac{d y}{d x}=2[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow(1-2 x^{2}) \frac{d y}{d x}=2[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}] \\ & \Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}} \end{aligned} $

Solution

The given relationship is $y=\sec ^{-1}(\frac{1}{2 x^{2}-1})$

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

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

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

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

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

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

Differentiating this relationship with respect to $x$, we obtain

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

$\Rightarrow 1=-\sin \frac{y}{2} \cdot \frac{d}{d x}(\frac{y}{2})$

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

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

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