Exercise 10.3 - Chapter 10 Differentiability and Methods of Differentiation 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 10.3
Differentiate the following
Question 1.
$
y=\left(x^2+4 x+6\right)^5
$
Solution:
$
\begin{aligned}
& \text { Let }=\mathrm{u}=\mathrm{x}^2+4 \mathrm{x}+6 \\
& \Rightarrow \frac{d u}{d x}=2 \mathrm{x}+4 \\
& \text { Now } \mathrm{y}=\mathrm{u}^5 \Rightarrow \frac{d y}{d x}=5 \mathrm{u}^4 \\
& \therefore \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}=5 \mathrm{u}^4(2 \mathrm{x}+4) \\
& =5\left(\mathrm{x}^2+4 \mathrm{x}+6\right)^4(2 \mathrm{x}+4) \\
& =5(2 \mathrm{x}+4)\left(\mathrm{x}^2+4 \mathrm{x}+6\right)^4
\end{aligned}
$
Question 2.
$
y=\tan 3 x
$
Solution:
$
\begin{aligned}
& \mathrm{y}=\tan 3 \mathrm{x} \\
& \text { put } \mathrm{u}=3 \mathrm{x} \\
& \frac{d u}{d x}=3 \\
& \text { Now } \mathrm{y}=\tan \mathrm{u} \\
& \Rightarrow \frac{d u}{d x}=\sec ^2 \mathrm{u} \\
& \text { So } \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}=\left(\sec ^2 \mathrm{u}\right) \text { (3) } \\
& =3 \sec ^2 3 \mathrm{x}
\end{aligned}
$
Question 3.
$
y=\cos (\tan x)
$
Solution:
Put $u=\tan x$
$
\frac{d u}{d x}=\sec ^2 \mathrm{x}
$
Now $\mathrm{y}=\cos \mathrm{u} \Rightarrow \frac{d u}{d x}=-\sin \mathrm{u}$
Now $\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$
$=(-\sin u)\left(\sec ^2 x\right)$
$=-\sec ^2(\sin (\tan x))$

Question 4.
$
y=\sqrt[3]{1+x^3}
$
Solution:
Put
$
\begin{aligned}
y & =\sqrt[3]{1+x^3}=\left(1+x^3\right)^{\frac{1}{3}} \\
t & =1+x^3 \\
\therefore \frac{d t}{d x} & =3 x^2
\end{aligned}
$

So
$
\begin{aligned}
y & =t^{1 / 3} \\
\frac{d y}{d t} & =\frac{1}{3} t^{-2 / 3} \\
\frac{d y}{d x} & =\frac{d y}{d t} \times \frac{d t}{d x}=\frac{1}{3} t^{-\frac{2}{3}}\left(3 x^2\right) \\
& =\frac{1}{3}\left(3 x^2\right) t^{-2 / 3}=x^2\left(1+x^3\right)^{\frac{-2}{3}}
\end{aligned}
$
Question 5.
$
\mathrm{y}=e^{\sqrt{x}}
$
Solution:
So,
$
\text { Let } \begin{aligned}
y & =e^{\sqrt{x}} \Rightarrow u=\sqrt{x} \\
\frac{d u}{d x} & =\frac{1}{2 \sqrt{x}}, \frac{d y}{d u}=e^{\sqrt{x}} \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=e^{\sqrt{x}}\left(\frac{1}{2 \sqrt{x}}\right) \\
& =\frac{e^{\sqrt{x}}}{2 \sqrt{x}}
\end{aligned}
$

Question 6.
$
y=\sin \left(e^x\right)
$
Solution:
$
y=\sin \left(e^x\right)
$

Let $u=e^x$
Now
$
\begin{aligned}
\frac{d u}{d x} & =e^x \\
y & =\sin u \Rightarrow \frac{d y}{d u}=\cos u \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=\left(\cos u\left(e^x\right)\right. \\
& =\left(\cos e^x\right) e^x=e^x \cos \left(e^x\right)
\end{aligned}
$
Question 7.
$
F(x)=\left(x^3+4 x\right)^7
$
Solution:
$
\begin{aligned}
& \mathrm{F}(\mathrm{x})=\left(\mathrm{x}^3+4 \mathrm{x}\right)^7 \\
& \text { Put } u=x^3+4 x \\
& \Rightarrow \quad \frac{d u}{d x}=3 x^2+4 \\
& \frac{d \mathrm{~F}(x)}{d x}=\frac{d \mathrm{~F}(x)}{d u} \times \frac{d u}{d x} \\
& =7 u^6\left(3 x^2+4\right)=7\left(x^3+4 x\right)^6\left(3 x^2+4\right) \\
& =7\left(3 x^2+4\right)\left(x^3+4 x\right)^6 \\
&
\end{aligned}
$
Question 8.
$
\mathrm{h}(\mathrm{t})=\left(t-\frac{1}{t}\right)^{\frac{3}{2}}
$

Solution:
Put
$
\begin{aligned}
h(t) & =\left(t-\frac{1}{t}\right)^{\frac{3}{2}} \\
u & =t-1 / t \Rightarrow \frac{d u}{d t}=1-(-1) t^{-2} \\
& =1-\left(-\frac{1}{t^2}\right)=1+\frac{1}{t^2}
\end{aligned}
$
Now
$
\begin{aligned}
h & =u^{3 / 2} \\
\frac{d h}{d u} & =\frac{3}{2} u^{\frac{3}{2}-1}=\frac{3}{2} u^{1 / 2}=\frac{3}{2} \sqrt{u} \\
\frac{d h}{d t} & =\frac{d h}{d u} \times \frac{d u}{d t}=\frac{3}{2}\left(t-\frac{1}{t}\right)^{1 / 2}\left(1+\frac{1}{t^2}\right)
\end{aligned}
$
Question 9.
$f(t)=\sqrt[3]{1+\tan t}$
Solution:
Now
$
\begin{aligned}
f & =(1+\tan t)^{1 / 3} \\
u & =1+\tan t \\
\frac{d u}{d t} & =\sec ^2 t \\
f & =u^{1 / 3} \Rightarrow \frac{d f}{d u}=\frac{1}{3} u^{\frac{1}{3}-1}=\frac{1}{3} u^{-\frac{2}{3}} \\
\frac{d f}{d t} & =\frac{d f}{d u} \times \frac{d u}{d t} \\
& =\frac{1}{3} \sec ^2 t(1+\tan t)^{\frac{-2}{3}}
\end{aligned}
$
Question 10.
$
y=\cos \left(a^3+x^3\right)
$

Solution:
Let
$
\begin{aligned}
y & =\cos \left(a^3+x^3\right) \\
u & =a^3+x^3 \Rightarrow \frac{d u}{d x}=3 x^2 \\
y & =\cos u \Rightarrow \frac{d y}{d u}=-\sin u \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\
& =-\sin u\left(3 x^2\right) \\
& =\left[-\sin \left(a^3+x^3\right)\right]\left(3 x^2\right) \\
& =-3 x^2 \sin \left(a^3+x^3\right)
\end{aligned}
$
$
\Rightarrow \quad \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}
$

Question 11.
$\mathrm{y}=\mathrm{e}^{-\mathrm{mx}}$
Solution:
Now
$
\begin{aligned}
& y=e^{-m x} \\
& u=-m x \Rightarrow \frac{d u}{d x}=-m \\
& y=e^u \Rightarrow \frac{d y}{d u}=\mathrm{e}^u
\end{aligned}
$
So
$
\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}=\left(e^u\right)(-m)=-m e^{-m x}=-m y
$
Question 12.
$y=4 \sec 5 x$
Solution:
$
\begin{aligned}
y & =4 \sec 5 x \\
t & =5 x \Rightarrow \frac{d t}{d x}=5 \\
y & =4 \sec t \Rightarrow \frac{d y}{d t}=4(\sec t \tan t) \\
\frac{d y}{d x} & =\frac{d y}{d t} \times \frac{d t}{d x}=(4 \sec t \tan t)(5) \\
& =20 \sec t \tan t \\
& =20 \sec 5 x \tan 5 x
\end{aligned}
$

Question 13.
$
y=(2 x-5)^4\left(8 x^2-5\right)^{-3}
$
Solution:
$
\begin{aligned}
& y=(2 x-5)^4\left(8 x^2-5\right)^{-3} \\
& u=2 x-5 \Rightarrow \frac{d u}{d x}=2 \\
& v=8 x^2-5 \Rightarrow \frac{d v}{d x}=16 x
\end{aligned}
$
$
\begin{aligned}
& y=u^4 v^{-3} \\
& \frac{d y}{d x}=u^4 \frac{d}{d x}\left(v^{-3}\right)+v^{-3} \frac{d}{d x}\left(u^4\right) \\
&=u^4\left[-3 v^{-4} \frac{d v}{d x}\right]+v^{-3}\left[4 u^3 \frac{d u}{d x}\right] \\
&=-3 u^4 \frac{1}{v^4}(16 x)+\frac{4 u^3}{v^3}(2) \\
&=-\frac{48 x(2 x-5)^4}{\left(8 x^2-5\right)^4}+8 \frac{(2 x-5)^3}{\left(8 x^2-5\right)^3} \\
&=\frac{8(2 x-5)^3}{\left(8 x^2-5\right)^4}\left[-6 x(2 x-5)+\left(8 x^2-5\right)\right] \\
&= \frac{8(2 x-5)^3}{\left(8 x^2-5\right)^4}\left[-12 x^2+30 x+8 x^2-5\right] \\
&=\frac{8(2 x-5)^3}{\left(8 x^2-5\right)^4}\left(-4 \mathrm{x}^2+30 \mathrm{x}-5\right)
\end{aligned}
$

Question 14.
$
\mathrm{y}=\left(\mathrm{x}^2+1\right) \sqrt[3]{x^2+2}
$
Solution:
$
\begin{aligned}
y & =\left(x^2+1\right) \sqrt[3]{x^2+2} \\
y & =\left(x^2+1\right)\left(x^2+2\right)^{1 / 3} \\
u & =x^2+1 \Rightarrow \frac{d u}{d x}=2 x \\
v & =\left(x^2+2\right)^{1 / 3} \Rightarrow v^{\prime}=\frac{d v}{d x}=\frac{1}{3}\left(x^2+2\right)^{-\frac{2}{3}} \\
& =\frac{-2 x}{3}\left(x^2+2\right)^{-\frac{2}{3}}
\end{aligned}
$
Now
$
\Rightarrow
$
$
\begin{aligned}
y & =u v \\
\frac{d y}{d x} & =y^{\prime}=u v^{\prime}+v u^{\prime}
\end{aligned}
$
(i.e.)
$
\begin{aligned}
\frac{d y}{d x} & =\left(x^2+1\right)\left(\frac{2 x}{3}\right)\left(x^2+2\right)^{-\frac{2}{3}}+\left(x^2+2\right)^{\frac{1}{3}}(2 x) \\
& =\frac{2 x}{3}\left(x^2+1\right) \frac{1}{\left(x^2+2\right)^{\frac{2}{3}}}+2 x\left(x^2+2\right)^{\frac{1}{3}} \\
& =\frac{2 x}{3\left(x^2+2\right)^{\frac{2}{3}}}\left[\left(x^2+1\right)+3\left(x^2+2\right)\right] \\
& =\frac{2 x}{3\left(x^2+2\right)^{\frac{2}{3}}}\left[x^2+1+3 x^2+6\right] \\
& =\frac{2 x}{3\left(x^2+2\right) \frac{2}{3}}\left[4 x^2+7\right]=\frac{8 x^3+14 x}{3\left(x^2+2\right)^{\frac{2}{3}}}
\end{aligned}
$

Question 15 .
$
\mathrm{y}=\mathrm{xe}^{-\mathrm{x}^2}
$
Solution:
$
y=\mathrm{xe}^{-\mathrm{x}^2}
$
$y=u v$ where $u=x$ and $v=e^{-x^2}$
Now 'u' $=1$ and $\mathrm{v}^{\prime}=\mathrm{e}^{-\mathrm{x}^2}(-2 \mathrm{x})$ $\mathrm{v}^{\prime}=-2 \mathrm{xe}^{-\mathrm{x}^2}$
Now $y=u v \Rightarrow y^{\prime}=u v^{\prime}+v u^{\prime}$

$
\begin{aligned}
& \text { (i.e.) } \frac{d y}{d x}=\mathrm{x}\left[-2 \mathrm{xe}^{-\mathrm{x}^2}\right]+\mathrm{e}^{-\mathrm{x}^2} \text { (1) } \\
& =\mathrm{e}^{-\mathrm{x}^2}\left(1-2 \mathrm{x}^2\right)
\end{aligned}
$
Question 16 .
$
\mathrm{s}(\mathrm{t})=\sqrt[4]{\frac{t^3+1}{t^3-1}}
$
Solution:

$
\text { Let } \begin{aligned}
s(t) & =\left(\frac{t^3+1}{t^3-1}\right)^{\frac{1}{4}} \\
\Rightarrow \quad \frac{d u}{d t} & =\frac{t^3+1}{t^3-1} \\
\Rightarrow \quad & =\frac{\left(t^3-1\right)\left(3 t^2\right)-\left(t^3+1\right)\left(3 t^2\right)}{\left(t^3-1\right)^2} \\
& =\frac{-6 t^2-3 t^5-3 t^2}{\left(t^3-1\right)^2}
\end{aligned}
$
Now
$
\begin{aligned}
s & =u^{1 / 4} \\
\frac{d s}{d u} & =\frac{1}{4} u^{\frac{-3}{4}}
\end{aligned}
$
$
\begin{aligned}
\therefore \quad \frac{d s}{d t} & =\frac{d s}{d u} \times \frac{d u}{d t}=\frac{1}{4} u^{\frac{-3}{4}}\left(\frac{-6 t^2}{\left(t^3-1\right)^2}\right) \\
& =\frac{-6 t^2}{4\left(t^3-1\right)^2}\left[\frac{t^3+1}{t^3-1}\right]^{-3 / 4} \\
& =\frac{-3 t^2}{2\left(t^3-1\right)^2} \times \frac{\left(t^3-1\right)^{\frac{3}{4}}}{\left(t^3+1\right)^{\frac{3}{4}}} \\
& =\frac{-3 t^2}{2\left(t^3-1\right)^{\frac{5}{4}}\left(t^3+1\right)^{\frac{3}{4}}}
\end{aligned}
$

Question 17.
$
f(x)=\frac{x}{\sqrt{7-3 x}}
$
Solution:
$
\begin{aligned}
& f=\frac{x}{\sqrt{7-3 x}} \\
& f^{\prime}=\frac{u}{v} \Rightarrow f^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& u=x \Rightarrow u^{\prime}=1 \\
& v=\sqrt{7-3 x} \Rightarrow v^{\prime}=\frac{1}{2 \sqrt{7-3 x}}(-3)=\frac{-3}{2 \sqrt{7-3 x}} \\
& f^{\prime}=\frac{d f}{d x}=\frac{\sqrt{7-3 x}(1)-x\left(\frac{-3}{2 \sqrt{7-3 x}}\right)}{(\sqrt{7-3 x})^2} \\
& =\frac{\sqrt{7-3 x}+\frac{3 x}{2 \sqrt{7-3 x}}}{7-3 x}=\frac{2(7-3 x)+3 x}{2(7-3 x) \sqrt{7-3 x}} \\
& =\frac{14-6 x+3 x}{2(7-3 x) \sqrt{7-3 x}}=\frac{14-3 x}{2(7-3 x) \sqrt{7-3 x}} \\
&
\end{aligned}
$

Question 18.
$
y=\tan (\cos x)
$
Solution:
$
y=\tan (\cos x)
$
$
y=\tan (\cos x)
$
Let
$
u=\cos x \Rightarrow \frac{d u}{d x}=-\sin x
$
Now
$
y=\tan (u)
$
$
\begin{aligned}
& \Rightarrow \quad \frac{d y}{d u}=\sec ^2 u \\
& \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}=\left(\sec ^2 u\right)(-\sin x) \\
& =-\sin x \sec ^2(\cos x) \\
&
\end{aligned}
$
Question 19.
$
\mathrm{y}=\frac{\sin ^2 x}{\cos x}
$

Solution:
$
\begin{aligned}
& y=\frac{\sin ^2 x}{\cos x}=\frac{u}{v}(\text { say }) \\
& \Rightarrow \quad \frac{d y}{d x}=y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& \text { Now } \\
& u=\sin ^2 x=(\sin x)^2 \\
& \Rightarrow \quad \frac{d u}{d x}=u^{\prime}=2 \sin x \cos x \\
& v=\cos x \Rightarrow v^{\prime}=-\sin x \\
& \text { Now } \\
& \frac{d y}{d x}=\frac{(\cos x)(2 \sin x \cos x)-\sin ^2 x(-\sin x)}{\cos ^2 x} \\
& =\frac{2 \sin x \cos ^2 x+\sin ^3 x}{\cos ^2 x} \\
& =\sin x \frac{\left(2 \cos ^2 x+\sin ^2 x\right)}{\cos ^2 x} \\
& =\sin x\left(2+\tan ^2 x\right) \\
& =\sin x\left(1+1+\tan ^2 x\right) \\
& =\sin x\left(1+\sec ^2 x\right) \\
&
\end{aligned}
$
Question 20.
$
y=5^{-\frac{1}{x}}
$

Solution:
$
y=5^{-\frac{1}{x}}=\left(\frac{1}{5}\right)^{\frac{1}{x}}
$
Taking log on both sides
$
\log y=\log \left(\frac{1}{5}\right)^{\frac{1}{x}}=\frac{1}{x} \log \left(\frac{1}{5}\right)
$
Differentiating w.r.to $x$
$
\begin{aligned}
\frac{1}{y} \frac{d y}{d x} & =\log \left(\frac{1}{5}\right)\left(\frac{-1}{x^2}\right) \\
& =(\log 1-\log 5)\left(\frac{-1}{x^2}\right) \\
& =0+\frac{\log 5}{x^2} \\
\therefore \quad \frac{d y}{d x} & =y\left(\frac{\log 5}{x^2}\right)=\frac{5^{\frac{-1}{x}}(\log 5)}{x^2}
\end{aligned}
$

Question 21.
$\mathrm{y}=\sqrt{1+2 \tan x}$
Solution:
L'et
$
\begin{aligned}
y & =\sqrt{1+2 \tan x} \\
u & =1+2 \tan x \\
\frac{d u}{d x} & =2 \sec ^2 x
\end{aligned}
$
$
\Rightarrow \quad \frac{d u}{d x}=2 \sec ^2 x
$
Now

$
\begin{aligned}
y & =\sqrt{u} \Rightarrow \frac{d y}{d u}=\frac{1}{2 \sqrt{u}} \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=\frac{1}{2 \sqrt{u}} 2 \sec ^2 x \\
& =\frac{\sec ^2 x}{\sqrt{1+2 \tan x}}
\end{aligned}
$

Question 22.
$
y=\sin ^3 x+\cos ^3 x
$
Solution:
$
\begin{aligned}
& y=\sin ^3 x+\cos ^3 x \\
& \text { Here } \mathrm{u}=\sin ^3 \mathrm{x}=(\sin \mathrm{x})^3 \\
& \Rightarrow \frac{d u}{d x}=3(\sin \mathrm{x})^2(\cos \mathrm{x}) \\
& =3 \sin ^2 \mathrm{x} \cos \mathrm{x} \\
& \mathrm{v}=\cos ^3 \mathrm{x}=(\cos \mathrm{x})^3 \\
& \Rightarrow \frac{d v}{d x}=3(\cos \mathrm{x})^2(-\sin \mathrm{x})=-3 \sin \mathrm{x} \cos ^2 \mathrm{x} \\
& \text { Now } \mathrm{y}=\mathrm{u}+\mathrm{v} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\
& =3 \sin ^2 x \cos x-3 \sin x \cos ^2 x \\
& =3 \sin x \cos x(\sin x-\cos x) \\
&
\end{aligned}
$
Question 23.
$
\mathrm{y}=\sin ^2(\cos \mathrm{kx})
$
Solution:
Now
$
\begin{aligned}
y & =\sin ^2(\cos k x)=[\sin (\cos k x)]^2 \\
u & =\sin (\cos k x) \\
\frac{d u}{d x} & =[\cos (\cos k x)][-\sin k x](k) \\
& =-k \sin k x \cos (\cos k x) \\
y & =u^2 \Rightarrow \frac{d y}{d u}=2 u \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=-2 u[k \sin k x \cos (\cos k x)] \\
& =-2(\sin (\cos k x)[k \sin k x \cos (\cos k x)] \\
& =-k \sin k x \sin (2 \cos k x)
\end{aligned}
$
Question 24 .
$
y=\left(1+\cos ^2 x\right)^6
$

Solution:
Now
$
\Rightarrow \quad \frac{d y}{d u}=6 u^5
$
So
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=6 u^5(-2 \sin x \cos x)=-6 u^5(\sin 2 x) \\
& =-6 \sin 2 x\left(1+\cos ^2 x\right)^5
\end{aligned}
$
Question 25.
$
\mathrm{y}=\frac{e^{3 z}}{1+e^x}
$
Solution:
$
\begin{aligned}
y & =\frac{e^{3 x}}{1+e^x} \\
y & =\frac{u}{v} ; \quad y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
u & =e^{3 x} \Rightarrow u^{\prime}=\frac{d u}{d x}=e^{3 x}(3)=3 e^{3 x} \\
v & =1+e^x \Rightarrow v^{\prime}=\frac{d v}{d x}=e^x \\
y^{\prime} & =\frac{d y}{d x}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& =\frac{\left(1+e^x\right)\left(3 e^{3 x}\right)-e^{3 x}\left(e^x\right)}{\left(1+e^x\right)^2} \\
& =\frac{3 e^{3 x}+3 e^{4 x}-e^{4 x}}{\left(1+e^x\right)^2}=\frac{3 e^{3 x}+2 e^{4 x}}{\left(1+e^x\right)^2}
\end{aligned}
$

Question 26.
$
\mathrm{y}=\sqrt{x+\sqrt{x}}
$
Solution:
Put
$
\begin{aligned}
y & =\sqrt{x+\sqrt{x}} \\
u & =\sqrt{x}\left(\Rightarrow x=u^2\right) \text { So } \frac{d u}{d x}=\frac{1}{2 \sqrt{x}} \\
y & =\sqrt{u^2+u} \\
\frac{d y}{d u} & =\frac{1}{2 \sqrt{u^2+u}}(2 u+1) \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=\frac{2 u+1}{2 \sqrt{u^2+u}}\left(\frac{1}{2 \sqrt{x}}\right) \\
& =\frac{1}{4 \sqrt{x}}\left(\frac{2 \sqrt{x}+1}{\sqrt{x+\sqrt{x}}}\right)
\end{aligned}
$

Question 27.
$
y=e^{x \cos x}
$
Solution:
$
\text { Let } \begin{aligned}
y & =e^x \cos x \\
u & =x \cos x \\
\Rightarrow \quad \frac{d u}{d x} & =x[-\sin x]+\cos x(1) \\
& =-x \sin x+\cos x
\end{aligned}
$
So
$
\begin{aligned}
y & =e^u \Rightarrow \frac{d y}{d u}=e^u \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\
& =e^u[-x \sin x+\cos x] \\
& =(-x \sin x+\cos x) e^{x \cos x}=e^{x \cos x}(\cos x-x \sin x)
\end{aligned}
$
Question 28.
$
\mathrm{y}=\sqrt{x+\sqrt{x+\sqrt{x}}}
$
Solution:
So
$
\begin{aligned}
y & =\sqrt{u} \Rightarrow \frac{d y}{d u}=\frac{1}{2 \sqrt{u}} \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=\frac{1}{2 \sqrt{u}}\left[1+\frac{1}{4 \sqrt{x}}\left(\frac{2 \sqrt{x}+1}{\sqrt{x+\sqrt{x}}}\right)\right] \\
& =\frac{1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}\left[\frac{4 \sqrt{x} \sqrt{x+\sqrt{x}}+2 \sqrt{x}+1}{4 \sqrt{x} \sqrt{x+\sqrt{x}}}\right] \\
& =\frac{4 \sqrt{x} \sqrt{x+\sqrt{x}}+2 \sqrt{x}+1}{8 \sqrt{x} \sqrt{x+\sqrt{x}} \sqrt{x+\sqrt{x+\sqrt{x}}}}
\end{aligned}
$

Question 29.
$
\mathrm{y}=\sin (\tan (\sqrt{\sin x}))
$
Solution:
Put $u=\sqrt{\sin x} \Rightarrow \frac{d u}{d x}=\frac{1}{2 \sqrt{\sin x}}(\cos x)$
Now $y=\sin (\tan u)$
Put $\quad v=\tan u \Rightarrow \frac{d v}{d u}=\sec ^2 u$
Now $y=\sin v$
$
\begin{aligned}
& \frac{d y}{d v}=\cos v \\
& \frac{d y}{d x}=\frac{d y}{d v} \times \frac{d v}{d u} \times \frac{d u}{d x}=\cos v\left[\sec ^2 u\right] \frac{\cos x}{2 \sqrt{\sin x}} \\
&=\frac{\cos (\tan \sqrt{\sin x}) \sec ^2(x \sqrt{\sin x}) \cos x}{2 \sqrt{\sin x}}
\end{aligned}
$

Question 30 .
$
y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)
$
Solution:
Put $t=\frac{1-x^2}{1+x^2}=\frac{u}{v}$
$
\therefore \frac{d t}{d x}=\frac{v u^{\prime}-u v^{\prime}}{v^2}
$
$
\begin{aligned}
\therefore \frac{d t}{d x} & =\frac{\left(1+x^2\right)(-2 x)-\left(1-x^2\right)(2 x)}{\left(1+x^2\right)^2} \\
& =\frac{-2 x-2 x^3-2 x+2 x^3}{\left(1+x^2\right)^2} \\
& =\frac{-4 x}{\left(1+x^2\right)^2}
\end{aligned}
$
Now $y=\sin ^{-1} t \Rightarrow \frac{d y}{d t}=\frac{1}{\sqrt{1-t^2}}$
So, $\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{1}{\sqrt{1-t^2}}\left[\frac{-4 x}{\left(1+x^2\right)^2}\right]$
$
=\frac{1}{\sqrt{1-\left(\frac{1-x^2}{1+x^2}\right)^2}}\left[\frac{-4 x}{\left(1+x^2\right)^2}\right]
$
$=\frac{\left(1+x^2\right)(-4 x)}{\left(1+x^2\right)^2 \sqrt{\left(1+x^2\right)^2-\left(1-x^2\right)^2}}$
$=\frac{-4 x}{\left(1+x^2\right) \sqrt{1+x^4+2 x^2-\left(1+x^4-2 x^2\right)}}=\frac{-4 x}{\left(1+x^2\right) \sqrt{4 x^2}}$
$=\frac{-4 x}{\left(1+x^2\right)(2 x)}=\frac{-2}{1+x^2}$