Exercise 10.2 - Chapter 10 Differentiability and Methods of Differentiation 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 10.2
Find the derivatives of the following functions with respect to corresponding independent variables.
Question 1.
$f(x)=x-3 \sin x$
Solution:
$
\begin{aligned}
& f(x)=x-3 \sin x \\
& =f^{\prime}(x)=1-3(\cos x) \\
& =1-3 \cos x
\end{aligned}
$
Question 2.
$
\mathrm{y}=\sin \mathrm{x}+\cos \mathrm{x}
$
Solution:
$
\frac{d y}{d x}=\cos \mathrm{x}+(-\sin \mathrm{x})=\cos \mathrm{x}-\sin \mathrm{x}
$
Question 3.
$
f(x)=x \sin \mathrm{x}
$
Solution:
$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=\mathrm{uv} \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{uv}^{\prime}+\mathrm{vu}^{\prime}=\mathrm{u} \frac{d u}{d x}+\mathrm{v}^{\frac{d v}{d x}} \\
& \text { Now } \mathrm{u}=\mathrm{x} \Rightarrow \mathrm{u}^{\prime}=1 \\
& \mathrm{v}^{\prime}=\sin \mathrm{x} \Rightarrow \mathrm{v}^{\prime} \cos \mathrm{x} \\
& \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}(\cos \mathrm{x})+\sin \mathrm{x}(1) \\
& =\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}
\end{aligned}
$
Question 4.
$
y=\cos x-2 \tan x
$
Solution:
$
\begin{aligned}
& \frac{d y}{d x}=-\sin \mathrm{x}=2\left(\sec ^2 \mathrm{x}\right) \\
& =-\sin \mathrm{x}-2 \sec ^2 \mathrm{x}
\end{aligned}
$
Question 5 .
$
g(t)=t^3 \cos t
$
Solution:
$
g(t)=t^3 \text { cost (i.e.) } u=t^3 \text { and } v=\cos t
$
let $\mathrm{u}^{\prime}=\frac{d u}{d x}$ and $\mathrm{v}^{\prime}=\frac{d v}{d x}=(-\sin t)$
$
\begin{aligned}
& g^{\prime}(t)=u v^{\prime}+v u^{\prime} \\
& g^{\prime}(t)=t^3(-\sin t)+\cos t\left(3 t^2\right) \\
& =-t^3 \sin t+3 t^2 \cos t
\end{aligned}
$

Question 6.
$
g(t)=4 \sec t+\tan t
$
Solution:
$
\begin{aligned}
& g\{t)=4 \sec t+\tan t \\
& g^{\prime}(t)=4(\sec t \tan t)+\sec ^2 t \\
& =4 \sec t \tan t+\sec ^2 t
\end{aligned}
$
Question 7.
$
\mathrm{y}=\mathrm{e}^{\mathrm{x}} \sin \mathrm{x}
$
Solution:
$
\begin{aligned}
& \mathrm{y}=\mathrm{e}^{\mathrm{x}} \sin \mathrm{x} \\
& \Rightarrow \mathrm{y}^{\mathrm{x}} \mathrm{u}^{\prime}+\mathrm{vu}^{\prime} \\
& \text { Now } \mathrm{u}=\mathrm{e}^{\mathrm{x}} \Rightarrow \mathrm{u}^{\prime}=\frac{d u}{d x} \mathrm{e}^{\mathrm{x}} \\
& \mathrm{v}=\sin \mathrm{x} \Rightarrow \mathrm{v}^{\prime}=\frac{d v}{d x} \cos \mathrm{x} \\
& \text { i.e. } \mathrm{y}^{\prime}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x})+\sin \mathrm{x}\left(\mathrm{e}^{\mathrm{x}}\right) \\
& =\mathrm{e}^{\mathrm{x}}[\sin \mathrm{x}+\cos \mathrm{x}]
\end{aligned}
$
Question 8 .
$
\mathrm{y}=\frac{\tan x}{x}
$
Solution
$
\begin{aligned}
y & =\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
u & =\tan x \therefore u^{\prime}=\sec ^2 x \\
v & =x \Rightarrow v^{\prime}=1 \\
y & =\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2}=\frac{x \sec ^2 x-\tan x(1)}{x^2} \\
& =\frac{x \sec ^2 x-\tan x}{x^2}
\end{aligned}
$
Question 9.
$
\mathrm{y}=\frac{\sin x}{1+\cos x}
$
Solution:
$
\begin{aligned}
& \mathrm{y}=\frac{\sin x}{1+\cos x}=\frac{u}{v} \text { (say) } \\
& \mathrm{u}=\sin \mathrm{x}=1+\cos \mathrm{x} \\
& \mathrm{u}^{\prime}=\cos \mathrm{x} \mathrm{v}^{\prime}=-\sin \mathrm{x}
\end{aligned}
$

$
\begin{aligned}
& \mathrm{y}=\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& \text { (i.e.,) } \frac{d y}{d x}=\frac{(1+\cos x) \cos x-\sin (-\sin x)}{(1+\cos x)^2} \\
&=\frac{\cos x+\cos ^2 x+\sin ^2 x}{(1+\cos x)^2}=\frac{1+\cos x}{(1+\cos x)^2} \\
&=\frac{1}{1+\cos x}
\end{aligned}
$
Question 10.
$
\mathrm{y}=\frac{x}{\sin x+\cos x}
$
Solution:
$
\begin{aligned}
y=\frac{u}{v}(s a y) & \\
u & =x \text { and } v=\sin x+\cos x \\
u^{\prime} & =1 \text { and } v^{\prime}=\cos x-\sin x \\
y & =\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
\frac{d y}{d x} & =\frac{(\sin x+\cos x)(1)-x(\cos x-\sin x)}{(\sin x+\cos x)^2} \\
& =\frac{\sin x+\cos x-x \cos x+x \sin x}{(\sin x+\cos x)^2} \\
& =\frac{(1+x) \sin x+(1-x) \cos x}{(\sin x+\cos x)^2}
\end{aligned}
$
Question 11.
$
\mathrm{y}=\frac{\tan x-1}{\sec x}
$

Solution:
Here
$
\begin{aligned}
y & =\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
u & =\tan x-1 \Rightarrow u^{\prime}=\sec ^2 x \\
v & =\sec x \Rightarrow v^{\prime}=\sec x \tan x \\
\frac{d y}{d x} & =\frac{v u^{\prime}-u v^{\prime}}{v^2}=\frac{\sec x\left(\sec ^2 x\right)-(\tan x-1)(\sec x \tan x)}{\sec ^2 x} \\
& =\frac{\sec ^3 x-\sec x \tan ^2 x+\sec x \tan x}{\sec ^2 x} \\
& =\frac{\sec x\left[\left(\sec ^2 x-\tan ^2 x\right)+\tan x\right]}{\sec ^2 x} \\
& =\frac{\sec x(1+\tan x)}{\sec { }^2 x}=\frac{1+\tan x}{\sec x} \\
& =\frac{1+\frac{\sin x}{\cos x}}{\sec x}=\frac{\cos x+\sin x}{\cos x \times \frac{1}{\cos x}} \\
& =\cos x+\sin x
\end{aligned}
$

Question 12 .
$
\mathrm{y}=\frac{\sin x}{x^2}
$
Solution:
$
y=\frac{\sin x}{x^2}=\frac{u}{v} \text { (say) }
$
$
\begin{aligned}
& \text { Here } \\
& u=\sin x \text { and } v=x^2 \\
& \Rightarrow \quad u^{\prime}=\cos x \text { and } v^{\prime}=2 x \\
& y=\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& \frac{d y}{d x}=\frac{x^2(\cos x)-\sin x(2 x)}{\left(x^2\right)^2} \\
& =\frac{x[x \cos x-2 \sin x]}{x^4} \\
& =\frac{x \cos x-2 \sin x}{x^3} \\
&
\end{aligned}
$
Question 13 .
$
y=\tan \theta(\sin \theta+\cos \theta)
$

Solution:
Here
$
\begin{aligned}
y & =\tan \theta(\sin \theta+\cos \theta) \\
y & =u v \Rightarrow y^{\prime}=u v^{\prime}+v u^{\prime} \\
u & =\tan \theta ; v=\sin \theta+\cos \theta \\
\frac{d u}{d \theta} & =\sec ^2 \theta ; \frac{d v}{d \theta}=\cos \theta-\sin \theta
\end{aligned}
$
Now
$
\begin{aligned}
y & =v u \Rightarrow y^{\prime}=u v^{\prime}+v u^{\prime} \\
\frac{d y}{d \theta} & =\tan \theta(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) \sec ^2 \theta \\
& =\frac{\sin \theta}{\cos \theta}(\cos \theta-\sin \theta)+\frac{1}{\cos ^2 \theta}(\sin \theta+\cos \theta) \\
& =\sin \theta-\frac{\sin ^2 \theta}{\cos \theta}+\frac{\sin \theta}{\cos ^2 \theta}+\frac{1}{\cos \theta} \\
& =\sin \theta+\frac{\sin \theta}{\cos ^2 \theta}+\frac{1-\sin ^2 \theta}{\cos \theta} \\
& =\sin \theta+\frac{\sin \theta}{\cos 2}+\frac{\cos ^2 \theta}{\cos \theta} \\
& =(\sin \theta+\cos \theta)+\frac{\sin \theta}{\cos 2} \\
& =(\sin \theta+\cos \theta)+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta} \\
& =(\sin \theta+\cos \theta)+\tan \theta \sec \theta
\end{aligned}
$

Question 14 .
$
y=\operatorname{cosex} x \cdot \cot x
$
Solution:
$
\begin{aligned}
& \mathrm{y}=\mathrm{u} \mathrm{v}^{\prime} \Rightarrow \mathrm{y}^{\prime}=\mathrm{uv}^{\prime}+\mathrm{vu}^{\prime} \\
& \mathrm{u}=\operatorname{cosec} \mathrm{x} \Rightarrow \mathrm{u}^{\prime}=-\operatorname{cosec} \mathrm{x} \cot \mathrm{x} \\
& \mathrm{v}=\cot \mathrm{x} \Rightarrow \mathrm{v}^{\prime}=-\operatorname{cosec}^2 \mathrm{x} \\
& (\operatorname{cosec} \mathrm{x})\left(-\operatorname{cosec}^2 \mathrm{x}\right)+\cot \mathrm{x}(-\operatorname{cosec} \mathrm{x} \cot \mathrm{x}) \\
& =\operatorname{cosec}{ }^3 \mathrm{x}-\operatorname{cosec} \mathrm{x} \cot ^2 \mathrm{x} \\
& =-\operatorname{cosec} \mathrm{x}\left(\operatorname{cosec}^2 \mathrm{x}+\cot ^2 \mathrm{x}\right) \\
& =-\frac{1}{\sin x}\left(\frac{1+\cos ^2 x}{\sin ^2 x}\right)=-\frac{\left(1+\cos ^2 x\right)}{\sin ^3 x}
\end{aligned}
$
Question 15 .
$
\mathrm{y}=\mathrm{x} \sin \mathrm{x} \cos \mathrm{x}
$

Solution:
$
\begin{aligned}
& y=u v w \Rightarrow y^{\prime}=u v w\left(\frac{1}{u} u^{\prime}+\frac{1}{v} v^{\prime}+\frac{1}{w} w^{\prime}\right) \\
& \text { Here } \quad u=x \quad \cdot \quad v=\sin x \quad w=\cos x \\
& u^{\prime}=1 \quad v^{\prime}=\cos x \quad w^{\prime}=-\sin x \\
& \frac{d y}{d x}=x \sin x \cos x\left[\frac{1}{x}(1)+\frac{1}{\sin x}(\cos x)+\frac{1}{\cos x}(-\sin x)\right] \\
& =\sin x \cos x+x \cos ^2 x-x \sin ^2 x \\
& \left.=\sin x \cos x+x\left[\left(\cos ^2 x\right)-\sin ^2 x\right)\right] \\
& =x \cos 2 x+\sin x \cos x \\
&
\end{aligned}
$

Question 16.
$y=e^{-x} \cdot \log x$
Solution:
$
\begin{aligned}
& \mathrm{y}=\mathrm{e}^{-\mathrm{x}} \log \mathrm{x}=\mathrm{uv} \text { (say) } \\
& \text { Here } \mathrm{u}=\mathrm{e}^{-\mathrm{x}} \text { and } \mathrm{v}^{\prime}=\log \mathrm{x} \\
& \Rightarrow \mathrm{u}^{\prime}=-\mathrm{e}^{-\mathrm{x}} \text { and } \mathrm{v}^{\prime}=\frac{1}{x} \\
& \text { Now } \mathrm{y}=\mathrm{uv} \Rightarrow \mathrm{y}^{\prime}=\mathrm{uv}^{\prime}+\mathrm{vu} \\
& \text { (i.e.) } \frac{d y}{d x}=\mathrm{e}^{-\mathrm{x}}\left(\frac{1}{x}\right)+\log \mathrm{x}\left(-\mathrm{e}^{-\mathrm{x}}\right) \\
& =\mathrm{e}^{-\mathrm{x}}\left(\frac{1}{x}-\log \mathrm{x}\right)
\end{aligned}
$
Question 17.
$
y=\left(x^2+5\right) \log (1+x) e^{-3 x}
$
Solution:
Here
$
\begin{aligned}
& v=\log (1+x) \Rightarrow v^{\prime}=\frac{1}{1+x} \\
& w=e^{-3 x} \Rightarrow w^{\prime}=e^{-3 x}(-3)=-3 e^{-3 x}
\end{aligned}
$
$
\frac{d y}{d x}=\left(x^2+5\right) \log (1+x) e^{-3 x}\left[\frac{1}{x^2+5}(2 x)+\frac{1}{\log (1+x)} \frac{1}{(1+x)}+\frac{1}{e^{-3 x}}\left(-3 e^{-3 x}\right)\right]
$
$
\begin{aligned}
& =\left(x^2+5\right) \log (1+x) e^{-3 x}\left[\frac{2 x}{x^2+5}+\frac{1}{(1+x) \log (1+x)}-3\right] \\
& =e^{-3 x}\left[2 x \log (1+x)+\frac{x^2+5}{1+x}-3\left(x^2+5\right) \log (1+x)\right]
\end{aligned}
$
Question 18 .
$
\mathrm{y}=\sin \mathrm{x}^0
$

Solution:
$
\begin{aligned}
y & =\sin x\left[\left(\frac{\pi}{180}\right)\right] \\
y & =\sin x\left[\frac{\pi}{180}\right] \\
\frac{d y}{d x} & =\frac{\pi}{180} \cos x=\frac{\pi}{180} \times \cos \frac{\pi}{180} x
\end{aligned}
$
Question 19.
$
\mathrm{y}=\log _{10} \mathrm{x}
$
Solution:
$
\begin{aligned}
y & =\log _{10} x=\log _e x \cdot \log _{10} \\
\frac{d y}{d x} & =\log _{10} e\left[\frac{1}{x}\right] \\
& =\frac{\log _{10} e}{x}
\end{aligned}
$
Question 20.
Draw the function $f^{\prime}(x)$ if $f(x)=2 x^2-5 x+3$
Solution:
$
f(x)=2 x^2-5 x+3
$
$f^{\prime}(x)=4 x-5$ which is a linear function

$\text { (i.e.) } y=4 x-5$