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

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Find the derivation of following functions
Question 1.
$
\begin{aligned}
& 3 \sin \mathrm{x}+4 \cos \mathrm{x}-\mathrm{e}^{\mathrm{x}} \\
& \text { Solution: } \\
& \mathrm{y}=3 \sin \mathrm{x}+4 \cos \mathrm{x}-\mathrm{e}^{\mathrm{x}} \\
& \frac{d y}{d x}=3(\cos \mathrm{x})+4(-\sin \mathrm{x})-\left(\mathrm{e}^{\mathrm{x}}\right) \\
& =3 \cos \mathrm{x}-4 \sin \mathrm{x}-\mathrm{e}^{\mathrm{x}}
\end{aligned}
$
Question 2.
$
\sin 5+\log _{10} x+2 \sec x
$
Solution:
$
\begin{aligned}
& \mathrm{y}=\sin 5+\log _{10} \mathrm{x}+2 \sec \mathrm{x} \\
& \frac{d y}{d x}=0+\left(\frac{1}{x}\right) \log _{10} \mathrm{e}+2[\sec \mathrm{x}+\tan \mathrm{x}]=\frac{\log _{10} e}{x}+2 \sec \mathrm{x} \tan \mathrm{x}
\end{aligned}
$
Question 3.
$6 \sin x \log _{10} x+e$

Solution:
Let
$
\begin{aligned}
& y=6 \sin x \log _{10} x+e \\
& u=\sin x \Rightarrow u^{\prime}=\cos x \\
& v=\log _{10} x \Rightarrow v^{\prime}=\frac{1}{x} \log _{10} e
\end{aligned}
$
Now

$
\begin{aligned}
y & =6 u v+e \\
y^{\prime} & =6\left[u v^{\prime}+v u^{\prime}\right]+0 \\
\frac{d y}{d x} & =6\left\{\sin x \frac{1}{x} \log _{10} e+\log _{10} x(\cos x)\right\} \\
& =6\left[\frac{\sin x}{x} \log _{10} e+\cos x \log _{10} x\right] \\
& =6\left[\frac{\sin x}{x} \log _{10} e+\cos x \log _{10} e \cdot \log _e x\right] \\
& =6 \log _{10} e\left(\frac{\sin x}{x}+\cos x \cdot \log _e x\right)
\end{aligned}
$
Question 4.

$
\left(x^4-6 x^3+7 x^2+4 x+2\right)\left(x^3-1\right)
$
Solution:
Let $u=x^4-6 x^3+7 x^2+4 x+2$ and $v=x^3-1$
$
\begin{aligned}
& \mathrm{u}^{\prime}=4 \mathrm{x}^3-6\left(3 \mathrm{x}^2\right)+7(2 \mathrm{x})+4(1)+0 \\
& =4 \mathrm{x}^3-18 \mathrm{x}^2+14 \mathrm{x}+4 \\
& \mathrm{v}^{\prime}=3 \mathrm{x}^3 \\
& \mathrm{y}=\mathrm{uv}^{\prime}+\mathrm{vu}^{\prime} \\
& \text { i.e. } \frac{d y}{d x}=\left(\mathrm{x}^4-6 \mathrm{x}^3+7 \mathrm{x}^2+4 \mathrm{x}+2\right)\left(3 \mathrm{x}^2\right)+\left(\mathrm{x}^3-1\right)\left(4 \mathrm{x}^3-18 \mathrm{x}^2+14 \mathrm{x}+4\right) \\
& =3 \mathrm{x}^6-18 \mathrm{x}^5+21 \mathrm{x}^4+12 \mathrm{x}^3+6 \mathrm{x}^2+4 \mathrm{x}^6-18 \mathrm{x}^5+14 \mathrm{x}^4+4 \mathrm{x}^3-4 \mathrm{x}^3+18 \mathrm{x}^2-14 \mathrm{x}-4 \\
& =7 \mathrm{x}^6-36 \mathrm{x}^5+35 \mathrm{x}^4+12 \mathrm{x}^3+24 \mathrm{x}^2-14 \mathrm{x}-4
\end{aligned}
$
Question 5.
$
\left(3 \mathrm{x}^2+1\right)^2
$
Solution:
$
y=\left(3 x^2+1\right)^2=\left(3 x^2+1\right)\left(3 x^2+1\right)
$
Let $u=3 x^2+1$ and $v=3 x^2+1$
$
\begin{aligned}
& \therefore \mathrm{u}^{\prime}=3(2 \mathrm{x})=6 \mathrm{x} \text { and } \mathrm{v}^{\prime}=6 \mathrm{x} \\
& \mathrm{y}^{\prime}=\mathrm{uv} \mathrm{v}^{\prime}+\mathrm{vu} \\
& \text { (i.e.,) } \frac{d y}{d x}=\left(3 \mathrm{x}^2+1\right)(6 \mathrm{x})+\left(3 \mathrm{x}^2+1\right) 6 \mathrm{x}=12 \mathrm{x}\left(3 \mathrm{x}^2+1\right)
\end{aligned}
$
Question 6.
$
(3 \sec x-4 \operatorname{cosec} x)(2 \sin x+5 \cos x)
$

Solution:
$
\begin{aligned}
& y=(3 \sec x-4 \operatorname{cosec} x)(2 \sin x+5 \cos x) \\
& \text { Let } u=3 \sec x-4 \operatorname{cosec} x \text { and } v=2 \sin x+5 \cos x \\
& u^{\prime}=3(\sec x \tan x)-4(-\operatorname{cosec} x \cot x) ; v^{\prime}=2(\cos x)+5(-\sin x) \\
& \left.u^{\prime}=3 \sec x \tan x+4 \operatorname{cosec} x \cot x\right) ; v^{\prime}=2 \cos x-5 \sin x . \\
& \therefore y^{\prime}=u v^{\prime}+u^{\prime}
\end{aligned}
$

$
\begin{aligned}
& \text { So } \frac{d y}{d x}=(3 \sec \mathrm{x}-4 \operatorname{cosec} \mathrm{x})(2 \cos \mathrm{x}-5 \sin \mathrm{x})+(2 \sin \mathrm{x}+5 \cos \mathrm{x})(3 \sec \mathrm{x} \tan \mathrm{x}+4 \operatorname{cosec} \mathrm{x} \cot \mathrm{x})=6 \sec \\
& \mathrm{x} \cos \mathrm{x}-15 \sec \mathrm{x} \sin \mathrm{x}-8 \operatorname{cosec} \mathrm{x} \cos \mathrm{x}+20 \operatorname{cosec} \mathrm{x} \sin \mathrm{x}+6 \sin \mathrm{x} \sec \mathrm{x} \tan \mathrm{x}+8 \sin \mathrm{x} \operatorname{cosec} \mathrm{x} \cot \mathrm{x}+15 \\
& \cos \mathrm{x} \sec \mathrm{t} \tan \mathrm{x}+20 \cos \mathrm{x} \operatorname{cosec} \mathrm{x} \cot \mathrm{x} \\
& =6 \frac{1}{\cos x} \cos \mathrm{x}-15 \frac{1}{\cos x} \sin \mathrm{x}-8 \frac{1}{\sin x} \cos \mathrm{x}+20 \frac{1}{\sin x} \sin \mathrm{x}+6 \sin \mathrm{x} \frac{1}{\cos x} \tan \mathrm{x}+8 \sin \mathrm{x} \frac{1}{\sin x} \cot \mathrm{x}+15 \\
& \cos \mathrm{x} \frac{1}{\cos x} \tan \mathrm{x}+20 \cos \mathrm{x} \frac{1}{\sin x} \cot \mathrm{x} \\
& =6-15 \tan \mathrm{x}-8 \cot \mathrm{x}+20+6 \tan ^2 \mathrm{x}+8 \cot \mathrm{x}+15 \tan \mathrm{x}+20 \cot ^2 \mathrm{x} \\
& =26+6 \tan ^2 \mathrm{x}+20 \cot ^2 \mathrm{x}
\end{aligned}
$
Question 7.
$
\mathrm{x}^2 \mathrm{e}^x \sin x
$
Solution:
$
y=x^2 e^x \sin x
$
Let $\mathrm{u}=\mathrm{x}^2, \mathrm{v}=\mathrm{e}^{\mathrm{x}}$ and $\mathrm{w}=\sin \mathrm{x}$
$
\begin{aligned}
& u^{\prime}=2 x, v^{\prime}=e^x \text { and } w^{\prime}=\cos x \\
& y^{\prime}=u v w^{\prime}+v w u^{\prime}+u w v^{\prime} \\
& =\left(x^2 e^x\right) \cos x+\left(e^x \sin x\right)(2 x)+\left(x^2 \sin x\right) e^x \\
& =x^2 e^x \cos x+2 x e^x \sin x+x^2 e^x \sin x \\
& =x e^x\{x \cos x+2 \sin x+x \sin x\}
\end{aligned}
$
Question 8 .
$
\frac{\cos x+\log x}{x^2+e^x}
$
Solution:
$
\mathrm{y}=\frac{\cos x+\log x}{x^2+e^x}
$
Let $u=\cos x+\log x$ and $v=x^2+e^x$
$
\therefore \mathrm{u}^{\prime}=-\sin \mathrm{x}+\frac{1}{x}, \mathrm{v}^{\prime}=2 \mathrm{x}+\mathrm{e}^{\mathrm{x}}
$

$\begin{aligned}
& \therefore y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& \text { (i.e.,) } \quad \frac{d y}{d x}=\frac{\left(x^2+e^x\right)\left(-\sin x+\frac{1}{x}\right)-(\cos x+\log x)\left(2 x+e^x\right)}{\left(x^2+e^x\right)^2} \\
& =\frac{-x^2 \sin x+x-e^x \sin x+\frac{e^x}{x}-2 x \cos x-2 x \log x-e^x \cos x-e^x \log x}{\left(x^2+e^x\right)^2} \\
& =\frac{e^x\left\{\frac{1}{x}-\sin x-\cos x-\log x\right\}-2 x(\log x+\cos x)+x-x^2 \sin x}{\left(x^2+e^x\right)^2} \\
&
\end{aligned}$

Question 9.
$\frac{\tan x+1}{\tan x-1}$
Solution:
$
y=\frac{\tan x+1}{\tan x-1}
$
Let $u=\tan x+1$ and $v=\tan x-1 \therefore u^{\prime}=\sec ^2 x$ and $v^{\prime}=\sec ^2 x$
$
\begin{aligned}
& y=\frac{u}{v} \quad \therefore y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \\
& \text { (i.e.) } \frac{d y}{d x}=\frac{(\tan x-1) \sec ^2 x-(\tan x+1) \sec ^2 x}{(\tan x-1)^2} \\
& =\frac{\sec ^2 x[\tan x-1-\tan x-1]}{(\tan x-1)^2} \\
& =\frac{\sec ^2 x(-2)}{(\tan x-1)^2}=\frac{-2 \sec ^2 x}{(\tan x-1)^2} \\
&
\end{aligned}
$
Question 10 .
$\frac{\sin x+x \cos x}{x \sin x-\cos x}$
Solution:
$
\mathrm{y}=\frac{\sin x+x \cos x}{x \sin x-\cos x}
$
Let $\mathrm{u}=\sin \mathrm{x}+\mathrm{x} \cos \mathrm{x}$ and $\mathrm{v}=\mathrm{x} \sin \mathrm{x}-\cos \mathrm{x}$
$\mathrm{u}^{\prime}=\cos \mathrm{x}+\mathrm{x}(-\sin \mathrm{x})+\cos \mathrm{x}(1)$
$=\cos x-x \sin x+\cos x=2 \cos x-x \sin x$
$v^{\prime}=x(\cos \mathrm{x})+\sin x(1)-(-\sin x)$
$=x \cos x+\sin x+\sin x=2 \sin x+x \cos x$
$\mathrm{y}=\frac{u}{v} \therefore \mathrm{y}^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2}$
(i.e.,) $\frac{d y}{d x}=\frac{(x \sin x-\cos x)(2 \cos x-x \sin x)-(\sin x+\cos x)(2 \sin x+x \cos x)}{(x \sin x-\cos x)^2}$
$=\frac{-x^2 \sin ^2 x-x \sin x \cos x-2 \cos ^2 x+x \sin \cos x-x^2 \cos ^2 x-2 \sin ^2 x}{(x \sin x-\cos x)^2}$
$=\frac{-x^2\left(\sin ^2 x+\cos ^2 x\right)-2\left(\sin ^2 x+\cos ^2 x\right)}{(x \sin x-\cos x)^2}$
$=\frac{-x^2-2}{(x \sin x-\cos x)^2}=\frac{-\left(x^2+2\right)}{(x \sin x-\cos x)^2}$