Exercise 10.4 - Chapter 10 Differentiability and Methods of Differentiation 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 10.4
Find the derivatives of the following functions
Question 1.
$
y=x^{\cos x}
$
Solution:
$
y=x^{\cos x}
$
Taking log on both sides
$
\begin{aligned}
& \log y=\log x^{\cos x}=\cos x \log x \\
& \text { differentiating w.r.to } \mathrm{x} \text { we get } \\
& \frac{1}{y} \frac{d y}{d x}=\cos x\left(\frac{1}{x}\right)+\log x(-\sin x) \\
& \text { (i.e.) } \\
& \frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \cos x-\sin x \log x \\
& \frac{d y}{d x}=y\left[\frac{1}{x} \cos x-\sin x \log x\right] \\
& =x^{\cos x}\left[\frac{1}{x} \cos x-\sin x \log x\right] \\
&
\end{aligned}
$
Question 2.
$
\mathrm{y}=\mathrm{x}^{\log \mathrm{x}}+(\log \mathrm{x})^{\mathrm{x}}
$
Solution:
$
y=x^{\log x}+(\log \mathrm{x})^{\mathrm{x}}
$
Let $\mathrm{y}=\mathrm{u}+\mathrm{v}$
Then $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\mathrm{u}=\mathrm{x}^{\log \mathrm{x}}$
Taking log on both sides
$\log \mathrm{u}=\log \mathrm{x} \log \mathrm{x}=\log (\mathrm{x})^2$
differentiating w.r.to $\mathrm{x}$

$
\begin{aligned}
\frac{1}{u} \frac{d u}{d x} & =2(\log x) \frac{1}{x}=\frac{2}{x} \log x \\
\frac{d u}{d x} & =u\left[\frac{2}{x} \log x\right] \\
& =x^{\log x}\left(\frac{2}{x} \log x\right) \\
& =\frac{2}{x}\left(x^{\log x}\right) \log x \\
v & =(\log x)
\end{aligned}
$
Taking $\log$ on both sides
$\log u=\log (\log x)^x=x \log (\log x)$
differentiating w.r.to $\mathrm{x}$
$
\begin{aligned}
\frac{1}{v} \frac{d v}{d x} & =x\left[\frac{1}{\log x} \times \frac{1}{x}\right]+1[\log (\log x)] \\
& =\frac{1}{\log x}+\log (\log x) \\
\frac{d v}{d x} & =v\left[\frac{1}{\log x}+\log (\log x)\right] \\
& =(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]
\end{aligned}
$
Substituting (2) and (3) in (1) we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d u}{d x}+\frac{d \nu}{d x} \\
& =\frac{2}{x}\left(x^{\log x}\right) \log x+(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right] \\
& =\frac{2 \log x}{x}\left(x^{\log x}\right)+(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]
\end{aligned}
$
Question 3.
$
\sqrt{x y}=\mathrm{e}^{(x-y)}
$
Solution:

(i.e.)
$
\begin{aligned}
\sqrt{x y} & =e^{x-y} \\
(x y)^{1 / 2} & =e^{x-y}
\end{aligned}
$
Taking log on both sides we get
$
\begin{aligned}
& \log (x y)^{1 / 2}=\log e^{x-y} \\
& \text { (i.e.) } \frac{1}{2}(\log x+\log y)=x-y \\
& \Rightarrow \quad \log x+\log y=2 x-2 y \\
& \frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=2-2 \frac{d y}{d x} \\
& \text { Now } \quad \frac{d y}{d x}\left(\frac{1}{y}+2\right)=2-1 / x \\
& \Rightarrow \text { (i.e.) } \quad \frac{d y}{d x}\left(\frac{1+2 y}{y}\right)=\frac{2 x-1}{x} \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{2 x-1}{x} / \frac{1+2 y}{y}=\frac{y(2 x-1)}{x(1+2 y)} \\
&
\end{aligned}
$
Question 4.
$
\mathrm{x}^{\mathrm{y}}=\mathrm{y}^{\mathrm{x}}
$
Solution:
$
x^y=y^x
$
Taking log on both sides
$
\log x=\log y
$
(i.e.) $y \log x=x \log y$

differentiating w.r.to $\mathrm{x}$
$
y\left(\frac{1}{x}\right)+\log x\left(\frac{d y}{d x}\right)=x\left(\frac{1}{y} \frac{d y}{d x}\right)+\log y(1)
$
(i.e.) $\frac{d y}{d x}\left(\log x-\frac{x}{y}\right)=\log y-\frac{y}{x}$
(i.e.) $\quad \frac{d y}{d x}\left(\frac{y \log x-x}{y}\right)=\frac{x \log y-y}{x}$
$\therefore \quad \frac{d y}{d x}=\frac{x \log y-y}{x}$
$
=\frac{y(x \log y-y)}{x(y \log x-x)}
$
Question 5 .
$(\cos x)^{\log x}$
Solution:
$
y=(\cos x)^{\log x}
$
Taking log on both sides
$
\begin{aligned}
& \log \mathrm{y}=\log (\cos \mathrm{x})^{\log \mathrm{x}}=\log \mathrm{x}(\log \cos \mathrm{x}) \\
& \text { differentiating w.r.to } \quad \frac{1}{y} \frac{d y}{d x}=\log x\left(\frac{1}{\cos x}(-\sin x)\right)+(\log \cos x) \frac{1}{x} \\
& \text { (i.e.) } \quad \frac{1}{y} \frac{d y}{d x}=-\tan x \log x+\frac{1}{x} \log \cos x \\
& \therefore \quad \frac{d y}{d x}=y\left[-\tan x \log x+\frac{1}{x} \log \cos x\right] \\
& \qquad=(\cos x)^{\log x}\left[-\tan x \log x+\frac{1}{x} \log \cos x\right]
\end{aligned}
$
Question 6.
$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$
Solution:
$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$

Differentiating w.r.to $\mathrm{x}$
$
\begin{aligned}
\frac{1}{a^2}(2 x)+\frac{1}{b^2}\left(2 y \frac{d y}{d x}\right) & =0 \\
\frac{d y}{d x}\left(\frac{2 y}{b^2}\right) & =-\frac{2 x}{a^2} \\
\frac{d y}{d x} & =-\frac{2 x}{a^2} / \frac{2 y}{b^2}=-\frac{2 x}{a^2} \times \frac{b^2}{2 y} \\
\therefore \quad & =\frac{-b^2 x}{a^2 y}
\end{aligned}
$
Question 7.
$
\sqrt{x^2+y^2}=\tan ^{-1}\left(\frac{y}{x}\right)
$
Solution:
$
\sqrt{x^2+y^2}=\tan ^{-1}\left(\frac{y}{x}\right)
$
Differentiating w.r.to $x$

$\begin{aligned}
& \frac{1}{2 \sqrt{x^2+y^2}}\left(2 x+2 y \frac{d y}{d x}\right)=\frac{1}{1+\left(\frac{y}{x}\right)^2}\left[\frac{x \frac{d y}{d x}-y(1)}{x^2}\right] \\
& \text { (i.e.) } \frac{\not 2\left(x+y \frac{d y}{d x}\right)}{2 \sqrt{x^2+y^2}}=\frac{x^2}{x^2+y^2}\left[\frac{\left(\frac{x d y}{d x}-y\right)}{x^2}\right] \\
& \text { (i.e.) }(x+y) \frac{d y}{d x}=\frac{\sqrt{x^2+y^2}}{x^2+y^2}\left(x \frac{d y}{d x}-y\right) \\
& \Rightarrow x+y \frac{d y}{d x}=\frac{1}{\sqrt{x^2+y^2}}\left(x \frac{d y}{d x}-y\right) \\
& \frac{d y}{d x}\left(y-\frac{x}{\sqrt{x^2+y^2}}\right)=\frac{-y}{\sqrt{x^2+y^2}}-x \\
& \frac{d y}{d x}\left[\frac{y \sqrt{x^2+y^2}-x}{\sqrt{x^2+y^2}}\right]=-\left(\frac{y+x \sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{y+x \sqrt{x^2+y^2}}{x-y \sqrt{x^2+y^2}} \\
&
\end{aligned}$

Question 8.
$
\tan (x+y)+\tan (x-y)=x
$
Solution:
$
\tan (\mathrm{x}+\mathrm{y})+\tan (\mathrm{x}-\mathrm{y})=\mathrm{x}
$
Differentiating w.r.to $\mathrm{x}$ we get
$
\begin{aligned}
& \therefore \quad\left[\sec ^2(x+y)\right]\left[1+\frac{d y}{d x}\right]+\left[\sec ^2(x-y)\right]\left(1-\frac{d y}{d x}\right)=1 \\
& \therefore \quad \frac{d y}{d x}=\frac{1-\sec ^2(x+y)-\sec ^2(x-y)}{\sec ^2(x+y)-\sec ^2(x-y)}
\end{aligned}
$
Question 9.
If $\cos (\mathrm{xy})=\mathrm{x}$, show that $\frac{d y}{d x}=\frac{-(1+y \sin (x y))}{x \sin x y}$

Solution:
$
\cos (x y)=x
$
Differentiating w.r.to $\mathrm{x}$
$
\begin{aligned}
& {[-\sin (x y)]\left[x \frac{d y}{d x}+y(1)\right]=1 } \\
& \frac{d y}{d x}(-x \sin (x y))=1+y \sin (x y) \\
\therefore \quad & \frac{d y}{d x}=\frac{1+y \sin (x y)}{-x \sin x y}=\frac{-(1+y \sin x y)}{(x \sin x y)}
\end{aligned}
$
Question 10 .
$
\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}
$
Solution:
$
y=\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}
$
Let
$
x=2 \theta
$
Now
$
\frac{1-\cos x}{1+\cos x}=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}=\tan ^2 \theta
$
So
$
\Rightarrow \quad \frac{d y}{d \theta}=1 ; x=2 \theta \Rightarrow \frac{d x}{d \theta}=2
$
So $\frac{d y}{d x}=\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{1}{2}$
Question 11.
$
\tan ^{-1}\left(\frac{6 x}{1-9 x^2}\right)
$

Solution:
$
\begin{aligned}
& y=\tan ^{-1}\left(\frac{6 x}{1-9 x^2}\right) \\
& \frac{6 x}{1-9 x^2}=\frac{2(3 x)}{1-(3 x)^2} \\
& 3 x=\tan \theta \Rightarrow x=\frac{\tan \theta}{3} ; \frac{d x}{d \theta}=\frac{1}{3}\left(\sec ^2 \theta\right) \\
& =\frac{1}{3}\left(1+(3 x)^2\right)=\frac{1}{3}\left(1+9 x^2\right) \\
& \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{2}{\frac{1}{3}\left(1+9 x^2\right)} \\
& =\frac{6}{1+9 x^2} \\
&
\end{aligned}
$
Question 12 .
$
\cos \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]
$

Solution:
$
\begin{aligned}
& y=\cos \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right] \\
& \text { Put } x=\cos 2 \theta ; \frac{d x}{d \theta}=(-\sin 2 \theta) 2 \\
& \frac{1-x}{1+x}=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}=\tan ^2 \theta \\
& \sqrt{\frac{1-x}{1+x}}=\sqrt{\tan ^2 \theta}=\tan \theta \\
& y=\cos \left[2 \tan ^{-1}(\tan \theta)\right] \\
& \Rightarrow \quad \frac{d y}{d \theta}=(-\sin 2 \theta)(2)=-2 \sin 2 \theta \\
& \frac{d y}{d x}=\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{-2 \sin 2 \theta}{-2 \sin 2 \theta}=1 \\
&
\end{aligned}
$
$
\text { (i.e.) } y=\cos [2 \theta]
$
Question 13.
$
x=a \cos ^3 t ; y=a \sin ^2 t
$
Solution:
$
\begin{aligned}
x & =a \cos ^3 t=a(\cos t)^3 \\
\frac{d x}{d t} & =a\left[3 \cos ^2 t\right](-\sin t)=-3 a \cos ^2 t \sin t \\
y & =a \sin ^3 t=a(\sin t)^3 \\
\frac{d y}{d t} & =a\left[3 \sin ^2 t \cos t\right]=3 a \sin ^2 t \cos t \\
\frac{d y}{d x} & =\frac{d y}{d t} / \frac{d x}{d t}=\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}=-\tan t
\end{aligned}
$
Question 14
$
x=a(\cos t+t \sin t) ; y=a[\sin t-t \cos t]
$
Solution:

$
\begin{aligned}
& x=a[\cos t+t \sin t] \\
& \frac{d x}{d t}=a[-\sin t+(t \cos t+\sin t(1)] \\
& =a[-\sin t+t \cos t+\sin t]=a t \cos t \\
& y=a[\sin t-t \cos t] \\
& \frac{d y}{d t}=a[\cos t-(t(-\sin t)+\cos t(1)] \\
& =a[\cos t+t \sin t-\cos t]=a t \sin t \\
& \frac{d y}{d x}=\frac{d y}{d t} / \frac{d x}{d t}=\frac{a t \sin t}{a t \cos t}=\tan t \\
&
\end{aligned}
$
Question 15 .
$
\mathrm{x}=\frac{1-t^2}{1+t^2} ; \mathrm{y}=\frac{2 t}{1+t^2}
$
Solution:
$
\begin{aligned}
x & =\frac{1-t^2}{1+t^2} ; \frac{d x}{d t}=\frac{\left(1+t^2\right)(-2 t)-\left(1-t^2\right)(2 t)}{\left(1+t^2\right)^2} \\
& =\frac{-2 t-2 t^3-2 t+2 t^3}{\left(1+t^2\right)^2}=\frac{-4 t}{\left(1+t^2\right)^2} \\
y & =\frac{2 t}{1+t^2} \\
\frac{d y}{d t} & =\frac{\left(1+t^2\right)(2)-2 t(2 t)}{\left(1+t^2\right)^2}=\frac{2+2 t^2-4 t^2}{\left(1+t^2\right)^2} \\
& =\frac{2-2 t^2}{\left(1+t^2\right)^2}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \\
\frac{d y}{d x} & =\frac{d y}{d t} / \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} / \frac{-4 t}{\left(1+t^2\right)^2}=\frac{t^2-1}{2 t}
\end{aligned}
$

Question 16 .
$
\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)
$
Solution:
$
\begin{aligned}
y & =\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\
x & =\tan \theta \Rightarrow \frac{d x}{d \theta}=\sec ^2 \theta=1+\tan ^2 \theta=1+x^2 \\
y & =\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)=\cos ^{-1}(\cos 2 \theta)=2 \theta \\
y & =2 \theta \Rightarrow \frac{d y}{d \theta}=2 \\
\frac{d y}{d x} & =\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{2}{1+x^2}
\end{aligned}
$

Question 17.
$\sin ^{-1}\left(3 x-4 x^3\right)$
Solution:
Now
$
\begin{aligned}
y & =\sin ^{-1}\left(3 x-4 x^3\right) \\
x & =\sin \theta \Rightarrow \frac{d x}{d \theta}=\cos \theta=\sqrt{\cos ^2 \theta}=\sqrt{1-\sin ^2 \theta}=\sqrt{1-x^2} \\
y & =\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right) \\
& =\sin ^{-1}(\sin 3 \theta)=3 \theta \\
y & =3 \theta \Rightarrow \frac{d y}{d \theta}=3 \\
\frac{d y}{d x} & =\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{3}{\sqrt{1-x^2}}
\end{aligned}
$
Question 18.
$
\tan ^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)
$

Solution:
So
$
\text { Now } \quad \begin{aligned}
y & =\tan ^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) \\
\frac{\cos x+\sin x}{\cos x-\sin x} & =\frac{\cos x(1+\sin x / \cos x)}{\cos x(1-\sin x / \cos x)} \\
& =\frac{1+\tan x}{1-\tan x}=\tan \left(\frac{\pi}{4}+x\right) \\
y & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+x\right)\right)=\frac{\pi}{4}+x \\
& \frac{d y}{d x}=1
\end{aligned}
$
Question 19.
Find the derivative of $\sin x^2$ with respect to $x^2$
Solution:
Here $\quad u=\sin x^2$ and $v=x^2$
Now we have to find $\quad \frac{d u}{d v}=\frac{d u / d x}{d v / d x}$
$
\frac{d u}{d x}=\left[\cos \left(x^2\right)\right](2 x)=2 x \cos \left(x^2\right)
$
$
\frac{d v}{d x}=2 x
$
$
\therefore \quad \frac{d u}{d \nu}=\frac{d u}{d x} / \frac{d v}{d x}=\frac{2 x \cos \left(x^2\right)}{2 x}=\cos \left(x^2\right)
$
Question 20.
Find the derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} \mathrm{x}$.
Solution:
Let $\mathrm{u}=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $\mathrm{v}=\tan ^{-1} \mathrm{x}$

Now we have to find $\frac{d u}{d v}$
To find $\frac{d u}{d x}:$ put $x=\tan \theta$
$
\Rightarrow \frac{d x}{d \theta}=\sec ^2 \theta=1+\tan ^2 \theta
$
Now
$
\begin{aligned}
u & =\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta) \\
& =2 \theta \\
u & =2 \theta \Rightarrow \frac{d u}{d \theta}=2 \\
v & =\tan ^{-1}(\tan \theta)=\theta \\
\frac{d v}{d \theta} & =1 \\
\frac{d u}{d v} & =\frac{d u / d \theta}{d v / d \theta}=\frac{2}{1}=2
\end{aligned}
$
Question 21.
If $u=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}$ and $v=\tan ^{-1} x$, find $\frac{d u}{d v}$

Solution:
$
\begin{aligned}
u & =\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \text { and } v=\tan ^{-1} x \\
\frac{d u}{d v} & =\frac{d u}{d x} / \frac{d v}{d x}
\end{aligned}
$
To find $\frac{d v}{d x}$ :
$
\begin{aligned}
& x=\tan \theta \Rightarrow \frac{d x}{d \theta}=\sec ^2 \theta=1+\tan \theta \\
& \frac{\sqrt{1+x^2}-1}{x}=\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta} \\
& =\frac{\sec \theta-1}{\tan \theta}=\frac{1}{\cos \theta}-1 / \frac{\sin \theta}{\cos \theta} \\
& =\frac{1-\cos \theta}{\sin \theta}=\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2} \\
& \therefore \quad \text { So } u=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2} \\
& \Rightarrow \quad \frac{d u}{d \theta}=\frac{1}{2} \text { : } \\
& v=\tan ^{-1} x \Rightarrow v=\tan ^{-1}(\tan \theta)=1 \\
& \frac{d v}{d \theta}=1 \frac{d u}{d v}=\frac{d u / d \theta}{d v / d \theta}=\frac{1 / 2}{1}=\frac{1}{2} \\
&
\end{aligned}
$
Question 22.
Find the derivative with
$
\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right) \text { with respect to } \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)
$
Solution:

$
\begin{aligned}
& u=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right) \text { and } v=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right) \\
& \frac{d u}{d v}=\frac{d u}{d x} / \frac{d v}{d x} \\
& u=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right) \\
& \text { Now } \frac{\sin x}{1+\cos x}=\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}=\tan \frac{x}{2} \\
& \text { So } u=\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2} \\
& \Rightarrow \quad \frac{d u}{d x}=\frac{1}{2} \\
& v=\tan ^{-1} \frac{\cos x}{1+\sin x} \\
& \cos x=\sin \left(\frac{\pi}{2}-x\right) \text { and } \sin x=\cos \left(\frac{\pi}{2}-x\right) \\
& \frac{\cos x}{1+\sin x}=\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}=\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \\
& =\tan \left(\frac{\pi}{4}-\frac{x}{2}\right) \\
& v=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]=\frac{\pi}{4}-\frac{x}{2} \\
& \frac{d v}{d x}=-\frac{1}{2} \\
& \Rightarrow \quad \frac{d u}{d v}=\frac{d u}{d x} / \frac{d v}{d x}=\frac{1 / 2}{-1 / 2}=-1 \\
&
\end{aligned}
$
To find
Question 23.
If $y=\sin ^{-1}$ then find $y$ ".
Solution:

$
\begin{aligned}
y & =\sin ^{-1} x \\
y^{\prime} & =\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}=\left(1-x^2\right)^{-1 / 2} \\
y^{\prime \prime} & =\frac{d^2 y}{d x^2}=\frac{-1}{2}\left(1-x^2\right)^{\frac{-3}{2}}(-2 x) \\
& =\frac{-(-2 x)}{2\left(1-x^2\right)^{3 / 2}}=\frac{x}{\left(1-x^2\right)^{3 / 2}}
\end{aligned}
$
Question 24 .
If $y=e^{\tan ^{-1} x}$, show that $\left(1+x^2\right) y^{\prime \prime}+(2 x-1) y^{\prime}=0$
Solution:
$
\begin{aligned}
& \mathrm{y}=\mathrm{e}^{\tan ^{-1} \mathrm{x}} \\
& \mathrm{y}=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\left(\frac{1}{1+x^2}\right) \\
& \Rightarrow \mathrm{y}^{\prime}=\frac{y}{1+x^2} \Rightarrow \mathrm{y}^{\prime}\left(1+\mathrm{x}^2\right)=\mathrm{y}
\end{aligned}
$
differentiating w.r.to $\mathrm{x}$
$y^{\prime}(2 x)+\left(1+x^2\right)\left(y^{\prime \prime}\right)=y^{\prime}$
(i.e.) $\left(1+x^2\right) y^{\prime \prime}+y^{\prime}(2 x)-y^{\prime}=0$
(i.e.) $\left(1+x^2\right) y^{\prime \prime}+(2 x-1) y^{\prime}=0$
Question 25.
If $\mathrm{y}=\frac{\sin ^{-1} x}{\sqrt{1-x^2}}$ show that $\left(1-x^2\right) y_2-3 \mathrm{xy}_1-\mathrm{y}=0$
Solution:
$
\begin{aligned}
y & =\frac{\sin ^{-1} x}{\sqrt{1-x^2}} \\
\Rightarrow \quad y \sqrt{1-x^2} & =\sin ^{-1} x
\end{aligned}
$
differentiating w.r.to $x$ we get
$
\Rightarrow y\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)+\left(\sqrt{1-x^2}\right) y_1=\frac{1}{\sqrt{1-x^2}}
$
(i.e.) $y\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)+\left(\sqrt{1-x^2}\right) y_1=\frac{1}{\sqrt{1-x^2}}$
multiplying both sides by $\sqrt{1-x^2}$ we get

$
\begin{aligned}
y & =\sin ^{-1} x \\
y^{\prime} & =\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}=\left(1-x^2\right)^{-1 / 2} \\
y^{\prime \prime} & =\frac{d^2 y}{d x^2}=\frac{-1}{2}\left(1-x^2\right)^{\frac{-3}{2}}(-2 x) \\
& =\frac{-(-2 x)}{2\left(1-x^2\right)^{3 / 2}}=\frac{x}{\left(1-x^2\right)^{3 / 2}}
\end{aligned}
$
Question 24 .
If $y=e^{\tan ^{-1} x}$, show that $\left(1+x^2\right) y^{\prime \prime}+(2 x-1) y^{\prime}=0$
Solution:
$
\begin{aligned}
& \mathrm{y}=\mathrm{e}^{\tan ^{-1} \mathrm{x}} \\
& \mathrm{y}=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\left(\frac{1}{1+x^2}\right) \\
& \Rightarrow \mathrm{y}^{\prime}=\frac{y}{1+x^2} \Rightarrow \mathrm{y}^{\prime}\left(1+\mathrm{x}^2\right)=\mathrm{y}
\end{aligned}
$
differentiating w.r.to $\mathrm{x}$
$y^{\prime}(2 x)+\left(1+x^2\right)\left(y^{\prime \prime}\right)=y^{\prime}$
(i.e.) $\left(1+x^2\right) y^{\prime \prime}+y^{\prime}(2 x)-y^{\prime}=0$
(i.e.) $\left(1+x^2\right) y^{\prime \prime}+(2 x-1) y^{\prime}=0$
Question 25.
If $\mathrm{y}=\frac{\sin ^{-1} x}{\sqrt{1-x^2}}$ show that $\left(1-x^2\right) y_2-3 \mathrm{xy}_1-\mathrm{y}=0$
Solution:
$
\begin{aligned}
y & =\frac{\sin ^{-1} x}{\sqrt{1-x^2}} \\
\Rightarrow \quad y \sqrt{1-x^2} & =\sin ^{-1} x
\end{aligned}
$
differentiating w.r.to $x$ we get
$
\Rightarrow y\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)+\left(\sqrt{1-x^2}\right) y_1=\frac{1}{\sqrt{1-x^2}}
$
(i.e.) $y\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)+\left(\sqrt{1-x^2}\right) y_1=\frac{1}{\sqrt{1-x^2}}$
multiplying both sides by $\sqrt{1-x^2}$ we get

$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{a \sin \theta}{a(1+\cos \theta)}=\frac{\sin \theta}{1+\cos \theta} \\
\frac{d y}{d x} & =\frac{\sin \theta}{1+\cos \theta} \\
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) / \frac{d x}{d \theta} \\
\frac{d}{d \theta}\left(\frac{d y}{d x}\right) & =\frac{(1+\cos \theta) \cos \theta-\sin \theta(-\sin \theta)}{(1+\cos \theta)^2} \\
& =\frac{\cos \theta+\cos ^2 \theta+\sin ^2 \theta}{\left(1+\cos ^2 \theta\right)^2} \\
& =\frac{1+\cos \theta}{(1+\cos \theta)^2}=\frac{1}{1+\cos \theta} \\
\frac{d x}{d \theta} & =a(1+\cos \theta) \\
\frac{d^2 y}{d x^2} & =\frac{1}{1+\cos \theta} / a(1+\cos \theta)=\frac{1}{a(1+\cos \theta)^2} \\
\frac{d^2 y}{d x^2} & =y^{\prime \prime} \operatorname{at} \theta=\frac{\pi}{2}=\frac{1}{a(1+0)^2}=\frac{1}{a}
\end{aligned}
$
Question 27.
If $\sin \mathrm{y}=\mathrm{x} \sin (\mathrm{a}+\mathrm{y})$ Then prove that $\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}, \mathrm{a} \neq \mathrm{n} \pi$

Solution:
$
\begin{aligned}
& \sin y=x \sin (a+y) \\
& \Rightarrow \quad \frac{\sin y}{\sin (a+y)}=x \\
&
\end{aligned}
$
Differentiating both sides w.r.to $x$
$
\begin{aligned}
& \frac{\sin (a+y) \cos y \frac{d y}{d x}-\sin y\left(\cos (a+y) \frac{d y}{d x}\right)}{\sin ^2(a+y)} \\
& \text { (i.e.) } \frac{d y}{d x}[\sin (a+y) \cos y-\cos (a+y) \sin y]=\sin ^2(a+y) \\
& \Rightarrow \quad \frac{d y}{d x} \sin (a+y-y)=\sin ^2(a+y) \\
& \text { (i.e.) } \frac{d y}{d x}(\sin a)=\sin ^2(a+y) \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a} \\
&
\end{aligned}
$
Question 28.
If $\mathrm{y}=\left(\cos ^{-1} \mathrm{x}\right)^2$, prove that $\left(1-\mathrm{x}^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-2=0$. Hence find $\mathrm{y}_2$ when $\mathrm{x}=0$.

Solution:
$
\begin{aligned}
y & =\left(\cos ^{-1} x\right)^2 \\
y_1 & =\frac{d y}{d x}=\left(2 \cos ^{-1} x\right)\left(\frac{-1}{\sqrt{1-x^2}}\right) \\
y_1 & =\frac{-2 \cos ^{-1} x}{\sqrt{1-x^2}} \\
\Rightarrow \quad\left(\text { i.e.) } \quad\left(\sqrt{1-x^2}\right)_{y_1}\right. & =-2 \cos ^{-1} x
\end{aligned}
$
Squaring on both sides $\left(1-x^2\right)\left(y_1^2\right)=4\left(\cos ^{-1} x\right)^2=4 y$
$
\Rightarrow \quad\left(1-x^2\right)\left(y_1^2\right)=4 y
$
differentiating again w.r.to $x$ we get
$
\begin{aligned}
& \left(1-x^2\right)\left(2 y_1 y_2\right)+\left(y_1^2\right)(-2 x)=4 y_1 \\
& \Rightarrow \quad\left(1-x^2\right)\left(2 y_1 y_2\right)=4 y_1+2 x y_1^2 \\
& \text { (i.e.) } \quad\left(1-x^2\right)\left(2 y_1 y_2\right)=2 y_1\left(2+x y_1\right) \\
& \left(\div \text { by } 2 y_1\right)\left(1-x^2\right) y_2=2+x y_1 \\
& \text { So } \quad\left(1-x^2\right) y_2-x y_1-2=0 \\
&
\end{aligned}
$

When $x=0$
$
\begin{aligned}
& (1-0) \quad y_2-0 y_1-2=0 \\
& y_2-2=0 \\
& y_2=2
\end{aligned}
$