Exercise 5.5 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Ex 5.5 Question 1 :

Differentiate the function with respect to $x$.
$\cos x \cdot \cos 2 x \cdot \cos 3 x$

Answer:
Let $y=\cos x \cdot \cos 2 x \cdot \cos 3 x$
Taking logarithm on both the sides, we obtain
$
\begin{aligned}
& \log y=\log (\cos x \cdot \cos 2 x \cdot \cos 3 x) \\
& \Rightarrow \log y=\log (\cos x)+\log (\cos 2 x)+\log (\cos 3 x)
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\frac{1}{\cos 2 x} \cdot \frac{d}{d x}(\cos 2 x)+\frac{1}{\cos 3 x} \cdot \frac{d}{d x}(\cos 3 x) \\
& \Rightarrow \frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)\right] \\
& \therefore \frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[\tan x+2 \tan 2 x+3 \tan 3 x]
\end{aligned}
$

Ex 5.5 Question 2:

Differentiate the function with respect to $x$.
$
\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}
$

Answer :
Let $y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
Taking logarithm on both the sides, we obtain
$
\begin{aligned}
& \log y=\log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \\
& \Rightarrow \log y=\frac{1}{2} \log \left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right] \\
& \Rightarrow \log y=\frac{1}{2}[\log \{(x-1)(x-2)\}-\log \{(x-3)(x-4)(x-5)\}] \\
& \Rightarrow \log y=\frac{1}{2}[\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5)]
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{l}
\frac{1}{x-1} \cdot \frac{d}{d x}(x-1)+\frac{1}{x-2} \cdot \frac{d}{d x}(x-2)-\frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \\
-\frac{1}{x-4} \cdot \frac{d}{d x}(x-4)-\frac{1}{x-5} \cdot \frac{d}{d x}(x-5)
\end{array}\right]
$

$\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\frac{y}{2}\left(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right) \\
& \therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right] \\
&
\end{aligned}$

Ex 5.5 Question 3 : 3

Differentiate the function with respect to $x$.
$
(\log x)^{\cos x}
$

Answer :
Let $y=(\log x)^{\cos x}$
Taking logarithm on both the sides, we obtain
$
\log y=\cos x \cdot \log (\log x)
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x) \times \log (\log x)+\cos x \times \frac{d}{d x}[\log (\log x)] \\
& \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=-\sin x \log (\log x)+\cos x \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{d y}{d x}=y\left[-\sin x \log (\log x)+\frac{\cos x}{\log x} \times \frac{1}{x}\right] \\
& \therefore \frac{d y}{d x}=(\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right]
\end{aligned}
$

Ex 5.5 Question 4:

Differentiate the function with respect to $x$.
$
x^x-2^{\sin x}
$

Answer:
Let $y=x^x-2^{\sin x}$
Also, let $x^x=u$ and $2^{\sin x}=v$
$
\begin{aligned}
& \therefore y=u-v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x} \\
& u=x^X
\end{aligned}
$

Taking logarithm on both the sides, we obtain
$
\log u=x \log x
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\left[\frac{d}{d x}(x) \times \log x+x \times \frac{d}{d x}(\log x)\right] \\
& \Rightarrow \frac{d u}{d x}=u\left[1 \times \log x+x \times \frac{1}{x}\right] \\
& \Rightarrow \frac{d u}{d x}=x^x(\log x+1) \\
& \Rightarrow \frac{d u}{d x}=x^x(1+\log x) \\
& v=2^{\sin x}
\end{aligned}
$

Taking logarithm on both the sides with respect to $x$, we obtain
$
\log v=\sin x \cdot \log 2
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \cdot \frac{d v}{d x}=\log 2 \cdot \frac{d}{d x}(\sin x) \\
& \Rightarrow \frac{d v}{d x}=v \log 2 \cos x \\
& \Rightarrow \frac{d v}{d x}=2^{\sin x} \cos x \log 2 \\
& \therefore \frac{d y}{d x}=x^x(1+\log x)-2^{\sin x} \cos x \log 2
\end{aligned}
$

Ex 5.5 Question 5 :

Differentiate the function with respect to $x$.
$
(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4
$

Answer :

 Let $y=(x+3)^2(x+4)^3(x+5)^4$ $\qquad$
Taking logs on both sides, we have
$
\begin{aligned}
& \Rightarrow \log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5) \\
& \Rightarrow \frac{d}{d x} \log y=2 \frac{d}{d x} \log (x+3)+3 \frac{d}{d x} \log (x+4)+4 \frac{d}{d x} \log (x+5) \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=2 \frac{1}{x+3} \frac{d}{d x}(x+3)+3 \frac{1}{x+4} \frac{d}{d x}(x+4)+4 \frac{1}{x+5} \frac{d}{d x}(x+5) \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}
\end{aligned}
$

$\begin{aligned}
& \Rightarrow \frac{d y}{d x}=y\left(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right) \\
& \Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4\left(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right) \\
& \Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4\left(\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right) \\
& \Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4\left(\frac{2 x^2+18 x+40+3 x^2+24 x+45+4 x^2+28 x+4}{(x+3)(x+4)(x+5)}\right) \\
& \Rightarrow \frac{d y}{d x}=(x+3)(x+4)^2(x+5)^3\left(9 x^2+70 x+133\right)
\end{aligned}$

Ex 5.5 Question 6.

$\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}$

Answer.

Let $y=\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}$
Putting $\quad\left(x+\frac{1}{x}\right)^x=u$ and $x^{\left(1+\frac{1}{x}\right)}=v$
$
\begin{aligned}
& y=u+v \\
& \therefore \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \ldots \ldots \ldots \text { (i) }
\end{aligned}
$

Now $u=\left(x+\frac{1}{x}\right)^x$
$
\begin{aligned}
& \Rightarrow \log u=\log \left(x+\frac{1}{x}\right)^x=x \log \left(x+\frac{1}{x}\right) \\
& \Rightarrow \frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{\left(x+\frac{1}{x}\right)} \frac{d}{d x}\left(x+\frac{1}{x}\right)+\log \left(x+\frac{1}{x}\right) \cdot 1
\end{aligned}
$

$\begin{aligned}
& \Rightarrow \frac{1}{u} \cdot \frac{d u}{d x}=x \cdot \frac{1}{\left(x+\frac{1}{x}\right)}\left(1-\frac{1}{x^2}\right)+\log \left(x+\frac{1}{x}\right) \cdot 1 \\
& \Rightarrow \frac{d u}{d x}=u\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right] \\
& =\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right] \ldots \ldots \ldots .(\mathrm{ii}) \\
& \text { Again } v=x^{\left(1+\frac{1}{x}\right)}
\end{aligned}$

$
\begin{aligned}
& \Rightarrow \log v=\log x^{\left(1+\frac{1}{x}\right)}=\left(1+\frac{1}{x}\right) \log x \\
& \Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}+\log x \cdot\left(\frac{-1}{x^2}\right) \\
& \Rightarrow \frac{d v}{d x}=v\left[\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}+\log x \cdot\left(\frac{-1}{x^2}\right)\right] \\
& \Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{1}{x}\left(1+\frac{1}{x}\right)-\frac{1}{x^2} \log x\right]
\end{aligned}
$

Putting the values from eq. (ii) and (iii) in eq. (i),
$
\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{1}{x}\left(1+\frac{1}{x}\right)-\frac{1}{x^2} \log x\right]
$

Ex 5.5 Question 7:

Differentiate the function with respect to $x$.
$
(\log x)^x+x^{\log x}
$

Answer:
Let $y=(\log x)^x+x^{\log x}$
Also, let $u=(\log x)^x$ and $v=x^{\log x}$
$
\begin{aligned}
& \therefore y=u+v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\
& u=(\log x)^x \\
& \Rightarrow \log u=\log \left[(\log x)^x\right] \\
& \Rightarrow \log u=x \log (\log x)
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\log x)+x \cdot \frac{d}{d x}[\log (\log x)] \\
& \Rightarrow \frac{d u}{d x}=u\left[1 \times \log (\log x)+x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)\right] \\
& \Rightarrow \frac{d u}{d x}=(\log x)^x\left[\log (\log x)+\frac{x}{\log x} \cdot \frac{1}{x}\right]
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \frac{d u}{d x}=(\log x)^x\left[\log (\log x)+\frac{1}{\log x}\right] \\
& \Rightarrow \frac{d u}{d x}=(\log x)^x\left[\frac{\log (\log x) \cdot \log x+1}{\log x}\right] \\
& \Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)] \\
& v=x^{\log x} \\
& \Rightarrow \log v=\log \left(x^{\log x}\right) \\
& \Rightarrow \log v=\log x \log x=(\log x)^2
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}\left[(\log x)^2\right] \\
& \Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x} \\
& \Rightarrow \frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x} \\
& \Rightarrow \frac{d v}{d x}=2 x^{\log x-1} \cdot \log x
\end{aligned}
$

Therefore, from (1), (2), and (3), we obtain
$
\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \cdot \log x
$

Ex 5.5 Question 8:

Differentiate the function with respect to $x$.
$
(\sin x)^x+\sin ^{-1} \sqrt{x}
$

Answer :
Let $y=(\sin x)^x+\sin ^{-1} \sqrt{x}$
Also, let $u=(\sin x)^x$ and $v=\sin ^{-1} \sqrt{x}$
$
\begin{aligned}
& \therefore y=u+v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\
& u=(\sin x)^x \\
& \Rightarrow \log u=\log (\sin x)^x \\
& \Rightarrow \log u=x \log (\sin x)
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \Rightarrow \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\sin x)+x \times \frac{d}{d x}[\log (\sin x)] \\
& \Rightarrow \frac{d u}{d x}=u\left[1 \cdot \log (\sin x)+x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right] \\
& \Rightarrow \frac{d u}{d x}=(\sin x)^x\left[\log (\sin x)+\frac{x}{\sin x} \cdot \cos x\right] \\
& \Rightarrow \frac{d u}{d x}=(\sin x)^x(x \cot x+\log \sin x) \\
& v=\sin ^{-1} \sqrt{x}
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{d x}(\sqrt{x}) \\
& \Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}} \\
& \Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^2}}
\end{aligned}
$

Therefore, from (1), (2), and (3), we obtain
$
\frac{d y}{d x}=(\sin x)^x(x \cot x+\log \sin x)+\frac{1}{2 \sqrt{x-x^2}}
$

Ex 5.5 Question 9:

Differentiate the function with respect to $x$.
$
x^{\sin x}+(\sin x)^{\cos x}
$

Answer :
Let $y=x^{\sin x}+(\sin x)^{\cos x}$
Also, let $u=x^{\sin x}$ and $v=(\sin x)^{\cos x}$
$
\begin{aligned}
& \therefore y=u+v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\
& u=x^{\sin x} \\
& \Rightarrow \log u=\log \left(x^{\sin x}\right) \\
& \Rightarrow \log u=\sin x \log x
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{d u}{d x}=u\left[\cos x \log x+\sin x \cdot \frac{1}{x}\right] \\
& \Rightarrow \frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]
\end{aligned}
$

$
\begin{aligned}
& v=(\sin x)^{\cos x} \\
& \Rightarrow \log v=\log (\sin x)^{\cos x} \\
& \Rightarrow \log v=\cos x \log (\sin x)
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}(\cos x) \times \log (\sin x)+\cos x \times \frac{d}{d x}[\log (\sin x)] \\
& \Rightarrow \frac{d v}{d x}=v\left[-\sin x \cdot \log (\sin x)+\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right] \\
& \Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}\left[-\sin x \log \sin x+\frac{\cos x}{\sin x} \cos x\right] \\
& \Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[-\sin x \log \sin x+\cot x \cos x] \\
& \Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log \sin x]
\end{aligned}
$

From (1), (2), and (3), we obtain
$
\frac{d y}{d x}=x^{\sin x}\left(\cos x \log x+\frac{\sin x}{x}\right)+(\sin x)^{\cos x}[\cos x \cot x-\sin x \log \sin x]
$

Ex 5.5 Question 10:

Differentiate the function with respect to $x$.
$
x^{x \cos x}+\frac{x^2+1}{x^2-1}
$

Answer :
Let $y=x^{x \cos x}+\frac{x^2+1}{x^2-1}$
Also, let $u=x^{x \cos x}$ and $v=\frac{x^2+1}{x^2-1}$
$
\begin{aligned}
& \therefore y=u+v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}
\end{aligned}
$
$
\begin{aligned}
& u=x^{x \cos x} \\
& \Rightarrow \log u=\log \left(x^{x \cos x}\right) \\
& \Rightarrow \log u=x \cos x \log x
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain

$\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \cdot \cos x \cdot \log x+x \cdot \frac{d}{d x}(\cos x) \cdot \log x+x \cos x \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{d u}{d x}=u\left[1 \cdot \cos x \cdot \log x+x \cdot(-\sin x) \log x+x \cos x \cdot \frac{1}{x}\right] \\
& \Rightarrow \frac{d u}{d x}=x^{\operatorname{rocos} x}(\cos x \log x-x \sin x \log x+\cos x) \\
& \Rightarrow \frac{d u}{d x}=x^{\operatorname{rocos} x}[\cos x(1+\log x)-x \sin x \log x] \\
& v=\frac{x^2+1}{x^2-1} \\
& \Rightarrow \log v=\log \left(x^2+1\right)-\log \left(x^2-1\right)
\end{aligned}$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \frac{d v}{d x}=\frac{2 x}{x^2+1}-\frac{2 x}{x^2-1} \\
& \Rightarrow \frac{d v}{d x}=v\left[\frac{2 x\left(x^2-1\right)-2 x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}\right] \\
& \Rightarrow \frac{d v}{d x}=\frac{x^2+1}{x^2-1} \times\left[\frac{-4 x}{\left(x^2+1\right)\left(x^2-1\right)}\right] \\
& \Rightarrow \frac{d v}{d x}=\frac{-4 x}{\left(x^2-1\right)^2}
\end{aligned}
$

From (1), (2), and (3), we obtain
$
\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \sin x \log x]-\frac{4 x}{\left(x^2-1\right)^2}
$

Ex 5.5 Question 11:

Differentiate the function with respect to $x$.
$
(x \cos x)^x+(x \sin x)^{\frac{1}{x}}
$

Answer:
Let $y=(x \cos x)^x+(x \sin x)^{\frac{1}{x}}$
Also, let $u=(x \cos x)^x$ and $v=(x \sin x)^{\frac{1}{x}}$
$
\begin{aligned}
& \therefore y=u+v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}
\end{aligned}
$
$
\begin{aligned}
& u=(x \cos x)^x \\
& \Rightarrow \log u=\log (x \cos x)^x \\
& \Rightarrow \log u=x \log (x \cos x) \\
& \Rightarrow \log u=x[\log x+\log \cos x] \\
& \Rightarrow \log u=x \log x+x \log \cos x
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain

$\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x \log x)+\frac{d}{d x}(x \log \cos x) \\
& \Rightarrow \frac{d u}{d x}=u\left[\left\{\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)\right\}+\left\{\log \cos x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos x)\right\}\right] \\
& \Rightarrow \frac{d u}{d x}=(x \cos x)^x\left[\left(\log x \cdot 1+x \cdot \frac{1}{x}\right)+\left\{\log \cos x \cdot 1+x \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)\right\}\right] \\
& \Rightarrow \frac{d u}{d x}=(x \cos x)^x\left[(\log x+1)+\left\{\log \cos x+\frac{x}{\cos x} \cdot(-\sin x)\right\}\right] \\
& \Rightarrow \frac{d u}{d x}=(x \cos x)^x[(1+\log x)+(\log \cos x-x \tan x)] \\
& \Rightarrow \frac{d u}{d x}=(x \cos x)^x[1-x \tan x+(\log x+\log \cos x)] \\
& \Rightarrow \frac{d u}{d x}=(x \cos x)^x[1-x \tan x+\log (x \cos x)]
\end{aligned}$

$
\begin{aligned}
& v=(x \sin x)^{\frac{1}{x}} \\
& \Rightarrow \log v=\log (x \sin x)^{\frac{1}{x}} \\
& \Rightarrow \log v=\frac{1}{x} \log (x \sin x) \\
& \Rightarrow \log v=\frac{1}{x}(\log x+\log \sin x) \\
& \Rightarrow \log v=\frac{1}{x} \log x+\frac{1}{x} \log \sin x
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}\left(\frac{1}{x} \log x\right)+\frac{d}{d x}\left[\frac{1}{x} \log (\sin x)\right] \\
& \Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{1}{x} \cdot \frac{d}{d x}(\log x)\right]+\left[\log (\sin x) \cdot \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{1}{x} \cdot \frac{d}{d x}\{\log (\sin x)\}\right] \\
& \Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot\left(-\frac{1}{x^2}\right)+\frac{1}{x} \cdot \frac{1}{x}\right]+\left[\log (\sin x) \cdot\left(-\frac{1}{x^2}\right)+\frac{1}{x} \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right] \\
& \Rightarrow \frac{1}{v} \frac{d v}{d x}=\frac{1}{x^2}(1-\log x)+\left[-\frac{\log (\sin x)}{x^2}+\frac{1}{x \sin x} \cdot \cos x\right] \\
& \Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x}{x^2}+\frac{-\log (\sin x)+x \cot x}{x^2}\right]
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x-\log (\sin x)+x \cot x}{x^2}\right] \\
& \Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log (x \sin x)+x \cot x}{x^2}\right]
\end{aligned}
$

From (1), (2), and (3), we obtain
$
\frac{d y}{d x}=(x \cos x)^x[1-x \tan x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{x \cot x+1-\log (x \sin x)}{x^2}\right]
$

Ex 5.5 Question 12:

Find $\frac{d y}{d x}$ of function.
$
x^y+y^x=1
$

Answer :
The given function is $x^y+y^x=1$
Let $x^y=u$ and $y^x=v$
Then, the function becomes $u+v=1$
$
\begin{aligned}
& \therefore \frac{d u}{d x}+\frac{d v}{d x}=0 \\
& u=x^y \\
& \Rightarrow \log u=\log \left(x^y\right) \\
& \Rightarrow \log u=y \log x
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{d u}{d x}=u\left[\log x \frac{d y}{d x}+y \cdot \frac{1}{x}\right] \\
& \Rightarrow \frac{d u}{d x}=x^y\left(\log x \frac{d y}{d x}+\frac{y}{x}\right)
\end{aligned}
$

$
\begin{aligned}
& v=y^x \\
& \Rightarrow \log v=\log \left(y^x\right) \\
& \Rightarrow \log v=x \log y
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \cdot \frac{d v}{d x}=\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y) \\
& \Rightarrow \frac{d v}{d x}=v\left(\log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right) \\
& \Rightarrow \frac{d v}{d x}=y^x\left(\log y+\frac{x}{y} \frac{d y}{d x}\right)
\end{aligned}
$

From (1), (2), and (3), we obtain
$
\begin{aligned}
& x^y\left(\log x \frac{d y}{d x}+\frac{y}{x}\right)+y^x\left(\log y+\frac{x}{y} \frac{d y}{d x}\right)=0 \\
& \Rightarrow\left(x^y \log x+x y^{x-1}\right) \frac{d y}{d x}=-\left(y x^{y-1}+y^x \log y\right) \\
& \therefore \frac{d y}{d x}=-\frac{y y^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}
\end{aligned}
$

Ex 5.5 Question 13 :

Find $\frac{d y}{d x}$ of function.
$
y^x=x^y
$

Answer :
The given function is $y^x=x^y$
Taking logarithm on both the sides, we obtain
$
x \log y=y \log x
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)=\log x \cdot \frac{d}{d x}(y)+y \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+y \cdot \frac{1}{x} \\
& \Rightarrow \log y+\frac{x}{y} \frac{d y}{d x}=\log x \frac{d y}{d x}+\frac{y}{x} \\
& \Rightarrow\left(\frac{x}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}-\log y \\
& \Rightarrow\left(\frac{x-y \log x}{y}\right) \frac{d y}{d x}=\frac{y-x \log y}{x} \\
& \therefore \frac{d y}{d x}=\frac{y}{x}\left(\frac{y-x \log y}{x-y \log x}\right)
\end{aligned}
$

Ex 5.5 Question 14 :

Find $\frac{d y}{d x}$ of function.
$
(\cos x)^y=(\cos y)^x
$

Answer:
The given function is $(\cos x)^y=(\cos y)^x$
Taking logarithm on both the sides, we obtain
$
y \log \cos x=x \log \cos y
$

Differentiating both sides, we obtain
$
\begin{aligned}
& \log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log \cos x)=\log \cos y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos y) \\
& \Rightarrow \log \cos x \frac{d y}{d x}+y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)=\log \cos y \cdot 1+x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y) \\
& \Rightarrow \log \cos x \frac{d y}{d x}+\frac{y}{\cos x} \cdot(-\sin x)=\log \cos y+\frac{x}{\cos y}(-\sin y) \cdot \frac{d y}{d x} \\
& \Rightarrow \log \cos x \frac{d y}{d x}-y \tan x=\log \cos y-x \tan y \frac{d y}{d x} \\
& \Rightarrow(\log \cos x+x \tan y) \frac{d y}{d x}=y \tan x+\log \cos y \\
& \therefore \frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}
\end{aligned}
$

Ex 5.5 Question 15:

Find $\frac{d y}{d x}$ of function.
$
x y=e^{(x-y)}
$

Answer :
The given function is $x y=e^{(x-y)}$
Taking logarithm on both the sides, we obtain
$
\begin{aligned}
& \log (x y)=\log \left(e^{x-y}\right) \\
& \Rightarrow \log x+\log y=(x-y) \log e \\
& \Rightarrow \log x+\log y=(x-y) \times 1 \\
& \Rightarrow \log x+\log y=x-y
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}(\log x)+\frac{d}{d x}(\log y)=\frac{d}{d x}(x)-\frac{d y}{d x} \\
& \Rightarrow \frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x} \\
& \Rightarrow\left(1+\frac{1}{y}\right) \frac{d y}{d x}=1-\frac{1}{x} \\
& \Rightarrow\left(\frac{y+1}{y}\right) \frac{d y}{d x}=\frac{x-1}{x} \\
& \therefore \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}
\end{aligned}
$

Ex 5.5 Question 16:

Find the derivative of the function given by $f(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$ and hence find $f^{\prime}(1)$.

Answer:
The given relationship is $f(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$
Taking logarithm on both the sides, we obtain
$\log f(x)=\log (1+x)+\log \left(1+x^2\right)+\log \left(1+x^4\right)+\log \left(1+x^8\right)$
Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]=\frac{d}{d x} \log (1+x)+\frac{d}{d x} \log \left(1+x^2\right)+\frac{d}{d x} \log \left(1+x^4\right)+\frac{d}{d x} \log \left(1+x^8\right) \\
& \Rightarrow \frac{1}{f(x)} \cdot f^{\prime}(x)=\frac{1}{1+x} \cdot \frac{d}{d x}(1+x)+\frac{1}{1+x^2} \cdot \frac{d}{d x}\left(1+x^2\right)+\frac{1}{1+x^4} \cdot \frac{d}{d x}\left(1+x^4\right)+\frac{1}{1+x^8} \cdot \frac{d}{d x}\left(1+x^8\right) \\
& \Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{1+x}+\frac{1}{1+x^2} \cdot 2 x+\frac{1}{1+x^4} \cdot 4 x^3+\frac{1}{1+x^8} \cdot 8 x^7\right] \\
& \therefore f^{\prime}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}\right]
\end{aligned}
$

Hence, $f^{\prime}(1)=(1+1)\left(1+1^2\right)\left(1+1^4\right)\left(1+1^8\right)\left[\frac{1}{1+1}+\frac{2 \times 1}{1+1^2}+\frac{4 \times 1^3}{1+1^4}+\frac{8 \times 1^7}{1+1^8}\right]$
$
\begin{aligned}
& =2 \times 2 \times 2 \times 2\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right] \\
& =16 \times\left(\frac{1+2+4+8}{2}\right) \\
& =16 \times \frac{15}{2}=120
\end{aligned}
$

Ex 5.5 Question 17:

Differentiate $\left(x^5-5 x+8\right)\left(x^3+7 x+9\right)$ in three ways mentioned below
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial.
(iii By logarithmic differentiation.
Do they all give the same answer?

Answer :
Let $y=\left(x^5-5 x+8\right)\left(x^3+7 x+9\right)$
(i)

Let $x^2-5 x+8=u$ and $x^3+7 x+9=v$
$
\begin{aligned}
& \therefore y=u v \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x} \quad \text { (By using product rule) } \\
& \Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^2-5 x+8\right) \cdot\left(x^3+7 x+9\right)+\left(x^2-5 x+8\right) \cdot \frac{d}{d x}\left(x^3+7 x+9\right) \\
& \Rightarrow \frac{d y}{d x}=(2 x-5)\left(x^3+7 x+9\right)+\left(x^2-5 x+8\right)\left(3 x^2+7\right) \\
& \Rightarrow \frac{d y}{d x}=2 x\left(x^3+7 x+9\right)-5\left(x^3+7 x+9\right)+x^2\left(3 x^2+7\right)-5 x\left(3 x^2+7\right)+8\left(3 x^2+7\right) \\
& \Rightarrow \frac{d y}{d x}=\left(2 x^4+14 x^2+18 x\right)-5 x^3-35 x-45+\left(3 x^4+7 x^2\right)-15 x^3-35 x+24 x^2+56 \\
& \therefore \frac{d y}{d x}=5 x^4-20 x^3+45 x^2-52 x+11
\end{aligned}
$

(ii)
$
\begin{aligned}
y= & \left(x^2-5 x+8\right)\left(x^3+7 x+9\right) \\
= & x^2\left(x^3+7 x+9\right)-5 x\left(x^3+7 x+9\right)+8\left(x^3+7 x+9\right) \\
= & x^5+7 x^3+9 x^2-5 x^4-35 x^2-45 x+8 x^3+56 x+72 \\
= & x^5-5 x^4+15 x^3-26 x^2+11 x+72 \\
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(x^5-5 x^4+15 x^3-26 x^2+11 x+72\right) \\
& =\frac{d}{d x}\left(x^5\right)-5 \frac{d}{d x}\left(x^4\right)+15 \frac{d}{d x}\left(x^3\right)-26 \frac{d}{d x}\left(x^2\right)+11 \frac{d}{d x}(x)+\frac{d}{d x}( \\
& =5 x^4-5 \times 4 x^3+15 \times 3 x^2-26 \times 2 x+11 \times 1+0 \\
& =5 x^4-20 x^3+45 x^2-52 x+11
\end{aligned}
$

(iii) $y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)$

Taking logarithm on both the sides, we obtain
$\log y=\log \left(x^2-5 x+8\right)+\log \left(x^3+7 x+9\right)$
Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x} \log \left(x^2-5 x+8\right)+\frac{d}{d x} \log \left(x^3+7 x+9\right) \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{x^2-5 x+8} \cdot \frac{d}{d x}\left(x^2-5 x+8\right)+\frac{1}{x^3+7 x+9} \cdot \frac{d}{d x}\left(x^3+7 x+9\right) \\
& \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x^2-5 x+8} \times(2 x-5)+\frac{1}{x^3+7 x+9} \times\left(3 x^2+7\right)\right] \\
& \Rightarrow \frac{d y}{d x}=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)\left[\frac{2 x-5}{x^2-5 x+8}+\frac{3 x^2+7}{x^3+7 x+9}\right] \\
& \Rightarrow \frac{d y}{d x}=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)\left[\frac{(2 x-5)\left(x^3+7 x+9\right)+\left(3 x^2+7\right)\left(x^2-5 x+8\right)}{\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)}\right] \\
& \Rightarrow \frac{d y}{d x}=2 x\left(x^3+7 x+9\right)-5\left(x^3+7 x+9\right)+3 x^2\left(x^2-5 x+8\right)+7\left(x^2-5 x+8\right) \\
& \Rightarrow \frac{d y}{d x}=\left(2 x^4+14 x^2+18 x\right)-5 x^3-35 x-45+\left(3 x^4-15 x^3+24 x^2\right)+\left(7 x^2-35 x+56\right) \\
& \Rightarrow \frac{d y}{d x}=5 x^4-20 x^3+45 x^2-52 x+11
\end{aligned}
$

From the above three observations, it can be concluded that all the results of $\frac{d y}{d x}$ are same.

Ex 5.5 Quetion 18:

If $u, v$ and $w$ are functions of $x$, then show that
$
\frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x}, w+u \cdot v \cdot \frac{d w}{d x}
$
in two ways-first by repeated application of product rule, second by logarithmic differentiation.

Answer :
Let $y=u, v . w=u .(v . w)$
By applying product rule, we obtain
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d u}{d x} \cdot(v \cdot w)+u \cdot \frac{d}{d x}(v \cdot w) \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x} v \cdot w+u\left[\frac{d v}{d x} \cdot w+v \cdot \frac{d w}{d x}\right] \quad \text { (Again applying product rule) } \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}
\end{aligned}
$

By taking logarithm on both sides of the equation $y=u . v . w$, we obtain
$
\log y=\log u+\log v+\log w
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\log u)+\frac{d}{d x}(\log v)+\frac{d}{d x}(\log w) \\
& \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x} \\
& \Rightarrow \frac{d y}{d x}=y\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right) \\
& \Rightarrow \frac{d y}{d x}=u \cdot v \cdot w \cdot\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right) \\
& \therefore \frac{d y}{d x}=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}
\end{aligned}
$