Question 1

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

Accepted Answer

Let $y=\cos x \cdot \cos 2 x \cdot \cos 3 x$

Taking logarithm on both the sides, we obtain

$\log y=\log (\cos x \cdot \cos 2 x \cdot \cos 3 x)$

$\Rightarrow \log y=\log (\cos x)+\log (\cos 2 x)+\log (\cos 3 x)$

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

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\frac{1}{\cos 2 x} \cdot \frac{d}{d x}(\cos 2 x)+\frac{1}{\cos 3 x} \cdot \frac{d}{d x}(\cos 3 x)$

$\Rightarrow \frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)ight]$

$	herefore \frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[	an x+2 	an 2 x+3 	an 3 x]$

Question 2

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

Accepted Answer

Let $y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Taking logarithm on both the sides, we obtain

$\log y=\log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

$\Rightarrow \log y=\frac{1}{2} \log \left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}ight]$

$\Rightarrow \log y=\frac{1}{2}[\log \{(x-1)(x-2)\}-\log \{(x-3)(x-4)(x-5)\}]$

$\Rightarrow \log y=\frac{1}{2}[\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5)]$

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

$\frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{l}\frac{1}{x-1} \cdot \frac{d}{d x}(x-1)+\frac{1}{x-2} \cdot \frac{d}{d x}(x-2)-\frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \ -\frac{1}{x-4} \cdot \frac{d}{d x}(x-4)-\frac{1}{x-5} \cdot \frac{d}{d x}(x-5)\end{array}ight]$

$\Rightarrow \frac{d y}{d x}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}ight]$

$	herefore \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}ight]$

Question 3

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

Accepted 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

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x) \cdot \log (\log x)+\cos x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=-\sin x \log (\log x)+\cos x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d}{d x}=y\left[-\sin x \log (\log x)+\frac{\cos x}{\log x} \cdot \frac{1}{x}ight]$

$	herefore \frac{d y}{d x}=(\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)ight]$

Question 4

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

Accepted Answer

Let $y=x^{x}-2^{\sin x}$

Also, let $x^{x}=u$ and $2^{\sin x}=v$

$	herefore y=u-v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$

$u=x^{x}$

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

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\left[\frac{d}{d x}(x) 	imes \log x+x 	imes \frac{d}{d x}(\log x)ight]$

$\Rightarrow \frac{d u}{d x}=u\left[1 	imes \log x+x 	imes \frac{1}{x}ight]$

$\Rightarrow \frac{d u}{d x}=x^{x}(\log x+1)$

$\Rightarrow \frac{d u}{d x}=x^{x}(1+\log x)$

$v=2^{\sin x}$

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

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

$\frac{1}{v} \cdot \frac{d v}{d x}=\log 2 \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow \frac{d v}{d x}=v \log 2 \cos x$

$\Rightarrow \frac{d v}{d x}=2^{\sin x} \cos x \log 2$

$	herefore \frac{d y}{d x}=x^{x}(1+\log x)-2^{\sin x} \cos x \log 2$

Question 5

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

Accepted Answer

Let $y=(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Taking logarithm on both the sides, we obtain

$\log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4}$

$\Rightarrow \log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5)$

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

$\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)$

$\Rightarrow \frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}ight]$

$\Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}ight]$

$\Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot\left[\begin{array}{l}2(x+4)(x+5)+3(x+3)(x+5) \ \frac{+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\end{array}ight]$

$\Rightarrow \frac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3} \cdot\left[\begin{array}{l}2\left(x^{2}+9 x+20ight)+3\left(x^{2}+8 x+15ight) \ +4\left(x^{2}+7 x+12ight)\end{array}ight]$

$	herefore \frac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3}\left(9 x^{2}+70 x+133ight)$

Question 6

Differentiate the function with respect to $x:\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$

Accepted Answer

Let $y=\left(x+\frac{1}{x}ight)^{x}+x^{\left(1+\frac{1}{x}ight)}$

Also, let $u=\left(x+\frac{1}{x}ight)^{x}$ and $v=x^{\left(1+\frac{1}{x}ight)}$

$$

\begin{align*}

& 	herefore y=u+v \

& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{align*}

$$

Then, $u=\left(x+\frac{1}{x}ight)^{x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x+\frac{1}{x}ight)^{x}$

$\Rightarrow \log u=x \log \left(x+\frac{1}{x}ight)$

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) 	imes \log \left(x+\frac{1}{x}ight)+x 	imes \frac{d}{d x}\left[\log \left(x+\frac{1}{x}ight)ight]$

$\Rightarrow \frac{1}{u} \cdot \frac{d u}{d x}=1 	imes \log \left(x+\frac{1}{x}ight)+x 	imes \frac{1}{\left(x+\frac{1}{x}ight)} \cdot \frac{d}{d x}\left(x+\frac{1}{x}ight)$

$\Rightarrow \frac{d u}{d x}=u\left[\log \left(x+\frac{1}{x}ight)+\frac{x}{\left(x+\frac{1}{x}ight)} 	imes\left(1-\frac{1}{x^{2}}ight)ight]$

$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}ight)^{x}\left[\log \left(x+\frac{1}{x}ight)+\frac{\left(x-\frac{1}{x}ight)}{\left(x+\frac{1}{x}ight)}ight]$

$\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}ight)^{x}\left[\log \left(x+\frac{1}{x}ight)+\frac{x^{2}-1}{x^{2}+1}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}ight)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}ight)ight] 	ag{2}

\end{equation*}

$$

Now, $v=x^{\left(1+\frac{1}{x}ight)}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left[x^{\left(1+\frac{1}{x}ight)}ight]$

$\Rightarrow \log v=\left(1+\frac{1}{x}ight) \log x$

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

$\frac{1}{v} \cdot \frac{d v}{d x}=\left[\frac{d}{d x}\left(1+\frac{1}{x}ight)ight] 	imes \log x+\left(1+\frac{1}{x}ight) \cdot \frac{d}{d x} \log x$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=\left(-\frac{1}{x^{2}}ight) \log x+\left(1+\frac{1}{x}ight) \cdot \frac{1}{x}$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=-\frac{\log x}{x^{2}}+\frac{1}{x}+\frac{1}{x^{2}}$

$\Rightarrow \frac{d v}{d x}=v\left[\frac{-\log x+x+1}{x^{2}}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}ight)}\left[\frac{-\log x+x+1}{x^{2}}ight] 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=\left(x+\frac{1}{x}ight)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}ight)ight]+x^{\left(1+\frac{1}{x}ight)}\left[\frac{x+1-\log x}{x^{2}}ight]$

Question 7

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

Accepted Answer

Let $y=(\log x)^{x}+x^{\log x}$

Also, let $u=(\log x)^{x}$ and $v=x^{\log x}$

$	herefore y=u+v$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{equation*}

$$

Then, $u=(\log x)^{x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left[(\log x)^{x}ight]$

$\Rightarrow \log u=x \log (\log x)$

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) 	imes \log (\log x)+x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{d u}{d x}=u\left[1 	imes \log (\log x)+x \cdot \frac{1}{(\log x)} \cdot \frac{d}{d x}(\log x)ight]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\log (\log x)+\frac{x}{(\log x)} \cdot \frac{1}{x}ight]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\log (\log x)+\frac{1}{(\log x)}ight]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\frac{\log (\log x) \cdot \log x+1}{\log x}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)] 	ag{2}

\end{equation*}

$$

$v=x^{\log x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left(x^{\log x}ight)$

$\Rightarrow \log v=\log x \log x=(\log x)^{2}$

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

$\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}\left[(\log x)^{2}ight]$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x}$

$\Rightarrow \frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x}$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=2 x^{\log x-1} \cdot \log x 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \cdot \log x$

Question 8

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

Accepted Answer

Let $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}$

Also, let $u=(\sin x)^{x}$ and $v=\sin ^{-1} \sqrt{x}$

$	herefore y=u+v$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{equation*}

$$

Then, $u=(\sin x)^{x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log (\sin x)^{x}$

$\Rightarrow \log u=x \log (\sin x)$

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) 	imes \log (\sin x)+x \cdot \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d u}{d x}=u\left[1 	imes \log (\sin x)+x \cdot \frac{1}{(\sin x)} \cdot \frac{d}{d x}(\sin x)ight]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}\left[\log (\sin x)+\frac{x}{(\sin x)} \cdot \cos xight]$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}=(\sin x)^{x}[x \cot x+\log \sin x] 	ag{2}

\end{equation*}

$$

$v=\sin ^{-1} \sqrt{x}$

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

$\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x})$

$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}} 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$

Question 9

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

Accepted Answer

Let $y=x^{\sin x}+(\sin x)^{\cos x}$

Also, let $u=x^{\sin x}$ and $v=(\sin x)^{\cos x}$

$	herefore y=u+v$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{equation*}

$$

Then, $u=x^{\sin x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x^{\sin x}ight)$

$\Rightarrow \log u=\sin x \log x$

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[\cos x \log x+\sin x \cdot \frac{1}{x}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}ight] 	ag{2}

\end{equation*}

$$

$v=(\sin x)^{\cos x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log (\sin x)^{\cos x}$

$\Rightarrow \log v=\cos x \log (\sin x)$

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

$\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}(\cos x) 	imes \log (\sin x)+\cos x 	imes \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d v}{d x}=v\left[-\sin x \cdot \log (\sin x)+\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)ight]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}\left[-\sin x \log (\sin x)+\frac{\cos x}{\sin x} \cos xight]$

$\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[-\sin x \log (\sin x)+\cot x \cos x]$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)] 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}ight]+(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]$

Question 10

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

Accepted Answer

Let $y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$

Also, let $u=x^{x \cos x}$ and $v=\frac{x^{2}+1}{x^{2}-1}$

$	herefore y=u+v$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{equation*}

$$

Then, $u=x^{x \cos x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x^{x \cos x}ight)$

$\Rightarrow \log u=x \cos x \log x$

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \cdot \cos x \cdot \log x+x \cdot \frac{d}{d x}(\cos x) \cdot \log x+x \cos x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[1 \cdot \cos x \cdot \log x+x \cdot(-\sin x) \log x+x \cos x \cdot \frac{1}{x}ight]$

$\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x \log x-x . \sin x \log x+\cos x]$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \sin x \log x] 	ag{2}

\end{equation*}

$$

$v=\frac{x^{2}+1}{x^{2}-1}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left(x^{2}+1ight)-\log \left(x^{2}-1ight)$

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

$\frac{1}{v} \frac{d v}{d x}=\frac{2 x}{x^{2}+1}-\frac{2 x}{x^{2}-1}$

$\Rightarrow \frac{d v}{d x}=v\left[\frac{2 x\left(x^{2}-1ight)-2 x\left(x^{2}+1ight)}{\left(x^{2}+1ight)\left(x^{2}-1ight)}ight]$

$\Rightarrow \frac{d v}{d x}=\frac{x^{2}+1}{x^{2}-1} 	imes\left[\frac{-4 x}{\left(x^{2}+1ight)\left(x^{2}-1ight)}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=\frac{-4 x}{\left(x^{2}-1ight)^{2}} 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \sin x \log x]-\frac{4 x}{\left(x^{2}-1ight)^{2}}$

Question 11

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

Accepted Answer

Let $y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

Also, let $u=(x \cos x)^{x}$ and $v=(x \sin x)^{\frac{1}{x}}$

$	herefore y=u+v$

$$

\begin{equation*}

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} 	ag{1}

\end{equation*}

$$

Then, $u=(x \cos x)^{x}$

Taking logarithm on both the sides, we obtain

$$

\begin{aligned}

& \Rightarrow \log u=(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{align*}

& \frac{1}{u} \cdot \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)ight\}+\left\{\log \cos x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos x)ight\}ight] \

& \Rightarrow \frac{d u}{d x}=(x \cos x)^{x}\left[\left(\log x \cdot 1+x \cdot \frac{1}{x}ight)+\left\{\log \cos x \cdot 1+x \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)ight\}ight] \

& \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)ight\}ight] \

& \Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[(1+\log x)+(\log \cos x-x 	an x)] \

& \Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[(1-x 	an x)+(\log x+\log \cos x)] \

& \Rightarrow \frac{d u}{d x}=(x \cos x)^{x}[1-x 	an x+\log (x \cos x)]  	ag{2}\

& v=(x \sin x)^{\frac{1}{x}}

\end{align*}

$$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log (x \sin x)^{\frac{1}{x}}$

$\Rightarrow \log v=\frac{1}{x} \log (x \sin x)$

$\Rightarrow \log v=\frac{1}{x}(\log x+\log \sin x)$

$\Rightarrow \log v=\frac{1}{r} \log x+\frac{1}{r} \log \sin x$

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

$\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}\left(\frac{1}{x} \log xight)+\frac{d}{d x}\left[\frac{1}{x} \log (\sin x)ight]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot \frac{d}{d x}\left(\frac{1}{x}ight)+\frac{1}{x} \cdot \frac{d}{d x}(\log x)ight]+\left[\log (\sin x) \cdot \frac{d}{d x}\left(\frac{1}{x}ight)+\frac{1}{x} \cdot \frac{d}{d x}\{\log (\sin x)\}ight]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot\left(-\frac{1}{x^{2}}ight)+\frac{1}{x} \cdot \frac{1}{x}ight]+\left[\log (\sin x) \cdot\left(-\frac{1}{x^{2}}ight)+\frac{1}{x} \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)ight]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\frac{1}{x^{2}}(1-\log x)+\left[-\frac{\log (\sin x)}{x^{2}}+\frac{1}{x \sin x} \cdot \cos xight]$

$\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x}{x^{2}}+\frac{-\log (\sin x)+x \cot x}{x^{2}}ight]$

$\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x-\log (\sin x)+x \cot x}{x^{2}}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log (x \sin x)+x \cot x}{x^{2}}ight] 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$\frac{d y}{d x}=(x \cos x)^{x}[1-x 	an x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{x \cot x+1-\log (x \sin x)}{x^{2}}ight]$

Question 12

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

Accepted Answer

The given function is $x^{y}+y^{x}=1$

Let, $x^{y}=u$ and $y^{x}=v$

$	herefore u+v=1$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}+\frac{d v}{d x}=0 	ag{1}

\end{equation*}

$$

Then, $u=x^{y}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log u=\log \left(x^{y}ight)$

$\Rightarrow \log u=y \log x$

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

$\frac{1}{u} \cdot \frac{d u}{d x}=\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[\log x \frac{d y}{d x}+y \cdot \frac{1}{x}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d u}{d x}=x^{y}\left[\log x \frac{d y}{d x}+\frac{y}{x}ight] 	ag{2}

\end{equation*}

$$

Now, $v=y^{x}$

Taking logarithm on both the sides, we obtain

$\Rightarrow \log v=\log \left(y^{x}ight)$

$\Rightarrow \log v=x \log y$

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

$\frac{1}{v} \frac{d v}{d x}=\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)$

$\Rightarrow \frac{d v}{d x}=v\left[\log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}ight]$

$$

\begin{equation*}

\Rightarrow \frac{d v}{d x}=y^{x}\left[\log y+\frac{x}{y} \cdot \frac{d y}{d x}ight] 	ag{3}

\end{equation*}

$$

Therefore, from (1), (2) and (3);

$x^{y}\left[\log x \frac{d y}{d x}+\frac{y}{x}ight]+y^{x}\left[\log y+\frac{x}{y} \cdot \frac{d y}{d x}ight]=0$

$\Rightarrow\left(x^{y} \log x+x y^{x-1}ight) \frac{d y}{d x}=-\left(y x^{y-1}+y^{x} \log yight)$

$	herefore \frac{d y}{d x}=\frac{-\left(y x^{y-1}+y^{x} \log yight)}{\left(x^{y} \log x+x y^{x-1}ight)}$

Question 13

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

Accepted 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

$\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)=\log x \cdot \frac{d}{d x}(y)+y \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+y \cdot \frac{1}{x}$

$\Rightarrow \log y+\frac{x}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+\frac{y}{x}$

$\Rightarrow\left(\frac{x}{y}-\log xight) \frac{d y}{d x}=\frac{y}{x}-\log y$

$\Rightarrow\left(\frac{x-y \log x}{y}ight) \frac{d y}{d x}=\frac{y-x \log y}{x}$

$	herefore \frac{d y}{d x}=\frac{y}{x}\left(\frac{y-x \log y}{x-y \log x}ight)$

Question 14

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

Accepted 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 with respect to $x$, we obtain

$\log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log \cos x)=\log \cos y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos y)$

$\Rightarrow \log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)=\log \cos y \cdot 1+x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y)$

$\Rightarrow \log \cos x \cdot \frac{d y}{d x}+\frac{y}{\cos x} \cdot(-\sin x)=\log \cos y+\frac{x}{\cos y} \cdot(-\sin y) \cdot \frac{d y}{d x}$

$\Rightarrow \log \cos x \cdot \frac{d y}{d x}-y 	an x=\log \cos y-x 	an y \frac{d y}{d x}$

$\Rightarrow(\log \cos x+x 	an y) \frac{d y}{d x}=y 	an x+\log \cos y$

$	herefore \frac{d y}{d x}=\frac{y 	an x+\log \cos y}{x 	an y+\log \cos x}$

Question 15

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

Accepted Answer

The given function is $x y=e^{(x-y)}$

Taking logarithm on both the sides, we obtain

$\log (x y)=\log \left(e^{x-y}ight)$

$\Rightarrow \log x+\log y=(x-y) \log e$

$\Rightarrow \log x+\log y=(x-y) 	imes 1$

$\Rightarrow \log x+\log y=(x-y)$

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}ight) \frac{d y}{d x}=1-\frac{1}{x} \

& \Rightarrow\left(\frac{y+1}{y}ight) \frac{d y}{d x}=\frac{x-1}{x} \

& 	herefore \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}

\end{aligned}

$$

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 $f^{\prime}(1)$.

Accepted Answer

The given function is $f(x)=(1+x)\left(1+x^{2}ight)\left(1+x^{4}ight)\left(1+x^{8}ight)$

Taking logarithm on both the sides, we obtain

$$

\log f(x)=\log (1+x)+\log \left(1+x^{2}ight)+\log \left(1+x^{4}ight)+\log \left(1+x^{8}ight)

$$

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}ight) \

& +\frac{d}{d x} \log \left(1+x^{4}ight)+\frac{d}{d x} \log \left(1+x^{8}ight) \

\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}ight) \

& +\frac{1}{1+x^{4}} \cdot \frac{d}{d x}\left(1+x^{4}ight)+\frac{1}{1+x^{8}} \cdot \frac{d}{d x}\left(1+x^{8}ight) \

\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}ight] \

herefore & f^{\prime}(x)=(1+x)\left(1+x^{2}ight)\left(1+x^{4}ight)\left(1+x^{8}ight)\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}}ight]

\end{aligned}

$$

Hence,

$$

\begin{aligned}

f^{\prime}(1) & =(1+1)\left(1+1^{2}ight)\left(1+1^{4}ight)\left(1+1^{8}ight)\left[\frac{1}{1+1}+\frac{2(1)}{1+1^{2}}+\frac{4(1)^{3}}{1+1^{4}}+\frac{8(1)^{7}}{1+1^{8}}ight] \

& =2 	imes 2 	imes 2 	imes 2\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}ight] \

& =16\left(\frac{15}{2}ight) \

& =120

\end{aligned}

$$

Question 17

Differentiate $\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$ 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?

Accepted Answer

Let $y=\left(x^{2}-5 x+8ight)\left(x^{3}+7 x+9ight)$

(i) By using product rule

Let $u=\left(x^{2}-5 x+8ight)$ and $v=x^{3}+7 x+9$

$$

\begin{aligned}

& 	herefore y=u v \

& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x} \

& \Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^{2}-5 x+8ight) \cdot\left(x^{3}+7 x+9ight)+\left(x^{2}-5 x+8ight) \cdot \frac{d}{d x}\left(x^{3}+7 x+9ight) \

& \Rightarrow \frac{d y}{d x}=(2 x-5)\left(x^{3}+7 x+9ight)+\left(x^{2}-5 x+8ight)\left(3 x^{2}+7ight) \

& \Rightarrow \frac{d y}{d x}=2 x\left(x^{3}+7 x+9ight)-5\left(x^{3}+7 x+9ight)+x^{2}\left(3 x^{2}+7ight) \

& -5 x\left(3 x^{2}+7ight)+8\left(3 x^{2}+7ight) \

& \Rightarrow \frac{d y}{d x}=\left(2 x^{4}+14 x^{2}+18 xight)-5 x^{3}-35 x-45+\left(3 x^{4}+7 x^{2}ight) \

& \Rightarrow \frac{d y}{d x}=-15 x^{3}-35 x+24 x^{2}+56 \

& 	herefore \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11

\end{aligned}

$$

(ii) By expanding the product to obtain a single polynomial.

$$

\begin{aligned}

y & =\left(x^{2}-5 x+8ight)\left(x^{3}+7 x+9ight) \

& =x^{2}\left(x^{3}+7 x+9ight)-5 x\left(x^{3}+7 x+9ight)+8\left(x^{3}+7 x+9ight) \

& =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

\end{aligned}

$$

Therefore,

$$

\begin{align*}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72ight) \

& =\frac{d}{d x}\left(x^{5}ight)-5 \frac{d}{d x}\left(x^{4}ight)+15 \frac{d}{d x}\left(x^{3}ight)-26 \frac{d}{d x}\left(x^{2}ight)+11 \frac{d}{d x}(x)+\frac{d}{d x}(  	ag{72}\

& =5 x^{4}-5\left(4 x^{3}ight)+15\left(3 x^{2}ight)-26(2 x)+11(1)+0 \

& =5 x^{4}-20 x^{3}+45 x^{2}-52 x+11

\end{align*}

$$

(iii) By logarithmic differentiation.

$$

y=\left(x^{2}-5 x+8ight)\left(x^{3}+7 x+9ight)

$$

Taking logarithm on both the sides, we obtain

$$

\log y=\log \left(x^{2}-5 x+8ight)+\log \left(x^{3}+7 x+9ight)

$$

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 \left(x^{2}-5 x+8ight)+\frac{d}{d x} \log \left(x^{3}+7 x+9ight) \

& \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{x^{2}-5 x+8} \cdot \frac{d}{d x}\left(x^{2}-5 x+8ight)+\frac{1}{x^{3}+7 x+9} \cdot \frac{d}{d x}\left(x^{3}+7 x+9ight) \

& \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x^{2}-5 x+8} \cdot(2 x-5)+\frac{1}{x^{3}+7 x+9} \cdot\left(3 x^{2}+7ight)ight] \

& \Rightarrow \frac{d y}{d x}=\left(x^{2}-5 x+8ight)\left(x^{3}+7 x+9ight)\left[\frac{2 x-5}{x^{2}-5 x+8}+\frac{3 x^{2}+7}{x^{3}+7 x+9}ight] \

& \Rightarrow \frac{d y}{d x}=\left(x^{2}-5 x+8ight)\left(x^{3}+7 x+9ight)\left[\frac{(2 x-5)\left(x^{3}+7 x+9ight)+\left(3 x^{2}+7ight)\left(x^{2}-5 x+8ight)}{\left(x^{2}-5 x+8ight)\left(x^{3}+7 x+9ight)}ight] \

& \Rightarrow \frac{d y}{d x}=2 x\left(x^{3}+7 x+9ight)-5\left(x^{3}+7 x+9ight)+3 x^{2}\left(x^{2}-5 x+8ight)+7\left(x^{2}-5 x+8ight) \

& \Rightarrow \frac{d y}{d x}=2 x^{4}+14 x^{2}+18 x-5 x^{3}-35 x-45+3 x^{5}-15 x^{3}+24 x^{2}+7 x^{2}-35 x+56 \

& \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.

Question 18

If $u, v$ and $w$ are functions of $x$, then show that

$\frac{d}{d x}(u . v . w)=\frac{d u}{d x} v . w+u . \frac{d v}{d x} . w+u . v . \frac{d w}{d x}$

in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Accepted Answer

Let $y=u \cdot v \cdot w=u \cdot(v \cdot w)$

By applying product rule, we get

$$

\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} \cdot(v \cdot w)+u\left[\frac{d v}{d x} \cdot w+v \cdot \frac{d w}{d x}ight] \quad 	ext { (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}

$$

Taking logarithm on both the 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}ight) \

& \Rightarrow \frac{d y}{d x}=u \cdot v \cdot w\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}ight) \

& 	herefore \frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}

\end{aligned}

$$

\section*{EXERCISE 5.6}