Solution

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

$ \begin{aligned} & \dfrac{1}{y} \dfrac{d y}{d x}=\dfrac{1}{\cos x} \cdot \dfrac{d}{d x}(\cos x)+\dfrac{1}{\cos 2 x} \cdot \dfrac{d}{d x}(\cos 2 x)+\dfrac{1}{\cos 3 x} \cdot \dfrac{d}{d x}(\cos 3 x) \\ & \Rightarrow \dfrac{d y}{d x}=y[-\dfrac{\sin x}{\cos x}-\dfrac{\sin 2 x}{\cos 2 x} \cdot \dfrac{d}{d x}(2 x)-\dfrac{\sin 3 x}{\cos 3 x} \cdot \dfrac{d}{d x}(3 x)] \\ & \therefore \dfrac{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} $

Solution

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

Taking logarithm on both the sides, we obtain

$ \begin{aligned} & \log y=\log \sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \\ & \Rightarrow \log y=\dfrac{1}{2} \log [\dfrac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}] \\ & \Rightarrow \log y=\dfrac{1}{2}[\log {(x-1)(x-2)}-\log {(x-3)(x-4)(x-5)}] \\ & \Rightarrow \log y=\dfrac{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

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

Solution

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} & \dfrac{1}{y} \cdot \dfrac{d y}{d x}= \log (\log x) \dfrac{d}{d x}(\cos x) +\cos x \times \dfrac{d}{d x}[\log (\log x)] \\ & \Rightarrow \dfrac{1}{y} \cdot \dfrac{d y}{d x}=-\sin x \log (\log x)+\cos x \times \dfrac{1}{\log x} \cdot \dfrac{d}{d x}(\log x) \\ & \Rightarrow \dfrac{d y}{d x}=y[-\sin x \log (\log x)+\dfrac{\cos x}{\log x} \times \dfrac{1}{x}] \\ & \therefore \dfrac{d y}{d x}=(\log x)^{\cos x}[\dfrac{\cos x}{x \log x}-\sin x \log (\log x)] \end{aligned} $

Solution

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

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

$\therefore y=u-v$

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

$u=x^{x}$

Taking logarithm on both the sides, we obtain

$\log u=x \log x$

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

$\dfrac{1}{u} \dfrac{d u}{d x}=[\log x \times \dfrac{d}{d x}(x) +x \times \dfrac{d}{d x}(\log x)]$

$\Rightarrow \dfrac{d u}{d x}=u[1 \times \log x+x \times \dfrac{1}{x}]$

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

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

$v=2^{\sin x}$

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

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

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

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

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

Solution

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

Taking logarithm on both the sides, we obtain

$ \begin{aligned} & \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) \end{aligned} $

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

$ \begin{aligned} & \dfrac{1}{y} \cdot \dfrac{d y}{d x}=2 \cdot \dfrac{1}{x+3} \cdot \dfrac{d}{d x}(x+3)+3 \cdot \dfrac{1}{x+4} \cdot \dfrac{d}{d x}(x+4)+4 \cdot \dfrac{1}{x+5} \cdot \dfrac{d}{d x}(x+5) \\ & \Rightarrow \dfrac{d y}{d x}=y[\dfrac{2}{x+3}+\dfrac{3}{x+4}+\dfrac{4}{x+5}] \\ & \Rightarrow \dfrac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot[\dfrac{2}{x+3}+\dfrac{3}{x+4}+\dfrac{4}{x+5}] \\ & \Rightarrow \dfrac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot[\dfrac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}] \\ & \Rightarrow \dfrac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3} \cdot[2(x^{2}+9 x+20)+3(x^{2}+8 x+15)+4(x^{2}+7 x+12)] \\ & \therefore \dfrac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3}(9 x^{2}+70 x+133) \end{aligned} $

Solution

Let $y=(x+\dfrac{1}{x})^{x}+x^{(1+\dfrac{1}{x})}$

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

$\therefore y=u+v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x}+\dfrac{d v}{d x} \qquad …(1)$

Then, $u=(x+\dfrac{1}{x})^{x}$

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

$\Rightarrow \log u=x \log (x+\dfrac{1}{x})$

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

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

$\Rightarrow \dfrac{1}{u} \dfrac{d u}{d x}=1 \times \log (x+\dfrac{1}{x})+x \times \dfrac{1}{(x+\dfrac{1}{x})} \cdot \dfrac{d}{d x}(x+\dfrac{1}{x})$

$\Rightarrow \dfrac{d u}{d x}=u[\log (x+\dfrac{1}{x})+\dfrac{x}{(x+\dfrac{1}{x})} \times(1-\dfrac{1}{x^{2}})]$

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

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

$\Rightarrow \dfrac{d u}{d x}=(x+\dfrac{1}{x})^{x}[\dfrac{x^{2}-1}{x^{2}+1}+\log (x+\dfrac{1}{x})] \qquad …(2)$

$v=x^{(1+\dfrac{1}{x})}$

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

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

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

$\dfrac{1}{v} \cdot \dfrac{d v}{d x}=[\dfrac{d}{d x}(1+\dfrac{1}{x})] \times \log x+(1+\dfrac{1}{x}) \cdot \dfrac{d}{d x} \log x$

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

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

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

$\Rightarrow \dfrac{d v}{d x}=x^{(1+\dfrac{1}{x})}(\dfrac{x+1-\log x}{x^{2}}) \qquad …(3)$

Therefore, from (1), (2), and (3), we obtain

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

Solution

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

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

$\therefore y=u+v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x}+\dfrac{d v}{d x} \qquad …(1)$

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

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

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

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

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

$\Rightarrow \dfrac{d u}{d x}=u[1 \times \log (\log x)+x \cdot \dfrac{1}{\log x} \cdot \dfrac{d}{d x}(\log x)]$

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

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

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

$\Rightarrow \dfrac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)] \qquad …(2)$

$v=x^{\log x}$

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

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

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

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

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

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

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

$\Rightarrow \dfrac{d v}{d x}=2 x^{\log x-1} \cdot \log x \qquad …(3)$

Therefore, from (1), (2), and (3), we obtain

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

Solution

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

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

$\therefore y=u+v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x}+\dfrac{d v}{d x} \qquad …(1)$

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

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

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

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

$\Rightarrow \dfrac{1}{u} \dfrac{d u}{d x}= \log (\sin x) \times \dfrac{d}{d x}(x) +x \times \dfrac{d}{d x}[\log (\sin x)]$

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

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

$\Rightarrow \dfrac{d u}{d x}=(\sin x)^{x}(x \cot x+\log \sin x) \qquad …(2)$

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

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

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

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

$\Rightarrow \dfrac{d v}{d x}=\dfrac{1}{2 \sqrt{x-x^{2}}} \qquad …(3)$

Therefore, from (1), (2), and (3), we obtain

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

Solution

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

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

$\therefore y=u+v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x}+\dfrac{d v}{d x} \qquad …(1)$

$u=x^{\sin x}$

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

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

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

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

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

$\Rightarrow \dfrac{d u}{d x}=x^{\sin x}[\cos x \log x+\dfrac{\sin x}{x}] \qquad …(2)$

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

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

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

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

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

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

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

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

$\Rightarrow \dfrac{d v}{d x}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log \sin x] \qquad …(3)$

From (1), (2), and (3), we obtain

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

Solution

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

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

$\therefore y=u+v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x}+\dfrac{d v}{d x} \qquad …(1)$

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

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

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

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

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

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

$\Rightarrow \dfrac{d u}{d x}=x^{x \cos x}(\cos x \log x-x \sin x \log x+\cos x)$

$\Rightarrow \dfrac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \sin x \log x] \qquad …(2)$

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

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

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

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

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

$\Rightarrow \dfrac{d v}{d x}=\dfrac{x^{2}+1}{x^{2}-1} \times[\dfrac{-4 x}{(x^{2}+1)(x^{2}-1)}]$

$\Rightarrow \dfrac{d v}{d x}=\dfrac{-4 x}{(x^{2}-1)^{2}} \qquad …(3)$

From (1), (2), and (3), we obtain

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

Solution

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

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

$\therefore y=u+v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x}+\dfrac{d v}{d x} \qquad …(1)$

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

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

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

$\Rightarrow \log u=x[\log x+\log \cos x]$

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

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

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

$\Rightarrow \dfrac{d u}{d x}=u[{\log x \cdot \dfrac{d}{d x}(x)+x \cdot \dfrac{d}{d x}(\log x)}+{\log \cos x \cdot \dfrac{d}{d x}(x)+x \cdot \dfrac{d}{d x}(\log \cos x)}]$

$\Rightarrow \dfrac{d u}{d x}=(x \cos x)^{x}[(\log x \cdot 1+x \cdot \dfrac{1}{x})+{\log \cos x \cdot 1+x \cdot \dfrac{1}{\cos x} \cdot \dfrac{d}{d x}(\cos x)}]$

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

$\Rightarrow \dfrac{d u}{d x}=(x \cos x)^{x}[(1+\log x)+(\log \cos x-x \tan x)]$

$\Rightarrow \dfrac{d u}{d x}=(x \cos x)^{x}[1-x \tan x+(\log x+\log \cos x)]$

$\Rightarrow \dfrac{d u}{d x}=(x \cos x)^{x}[1-x \tan x+\log (x \cos x)] \qquad …(2)$

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

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

$$ \begin{align*} & \dfrac{1}{v} \dfrac{d v}{d x}=\dfrac{d}{d x}(\dfrac{1}{x} \log x)+\dfrac{d}{d x}[\dfrac{1}{x} \log (\sin x)] \\ & \Rightarrow \dfrac{1}{v} \dfrac{d v}{d x}=[\log x \cdot \dfrac{d}{d x}(\dfrac{1}{x})+\dfrac{1}{x} \cdot \dfrac{d}{d x}(\log x)]+[\log (\sin x) \cdot \dfrac{d}{d x}(\dfrac{1}{x})+\dfrac{1}{x} \cdot \dfrac{d}{d x}{\log (\sin x)}] \\ & \Rightarrow \dfrac{1}{v} \dfrac{d v}{d x}=[\log x \cdot(-\dfrac{1}{x^{2}})+\dfrac{1}{x}+\dfrac{1}{x}]+[\log (\sin x) \cdot(-\dfrac{1}{x^{2}})+\dfrac{1}{x} \cdot \dfrac{1}{\sin x} \cdot \dfrac{d}{d x}(\sin x)] \\ & \Rightarrow \dfrac{1}{v} \dfrac{d v}{d x}=\dfrac{1}{x^{2}}(1-\log x)+[-\dfrac{\log (\sin x)}{x^{2}}+\dfrac{1}{x \sin x} \cdot \cos x] \\ & \Rightarrow \dfrac{d v}{d x}=(x \sin x)^{\dfrac{1}{x}}[\dfrac{1-\log x}{x^{2}}+\dfrac{-\log (\sin x)+x \cot x}{x^{2}}] \\ & \Rightarrow \dfrac{d v}{d x}=(x \sin x)^{\dfrac{1}{x}}[\dfrac{1-\log x-\log (\sin x)+x \cot x}{x^{2}}] \\ & \Rightarrow \dfrac{d v}{d x}=(x \sin x)^{\dfrac{1}{x}}[\dfrac{1-\log (x \sin x)+x \cot x}{x^{2}}] \qquad …(3) \end{align*} $$

From (1), (2), and (3), we obtain

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

Solution

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

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

Then, the function becomes $u+v=1$

$\therefore \dfrac{d u}{d x}+\dfrac{d v}{d x}=0 \qquad …(1)$

$u=x^{y}$

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

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

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

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

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

$\Rightarrow \dfrac{d u}{d x}=x^{y}(\log x \dfrac{d y}{d x}+\dfrac{y}{x}) \qquad …(2)$

$v=y^{x}$

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

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

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

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

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

$\Rightarrow \dfrac{d v}{d x}=y^{x}(\log y+\dfrac{x}{y} \dfrac{d y}{d x}) \qquad …(3)$

From (1), (2), and (3), we obtain

$ \begin{aligned} & x^{y}(\log x \dfrac{d y}{d x}+\dfrac{y}{x})+y^{x}(\log y+\dfrac{x}{y} \dfrac{d y}{d x})=0 \\ & \Rightarrow(x^{y} \log x+x y^{x-1}) \dfrac{d y}{d x}=-(y x^{y-1}+y^{x} \log y) \\ & \therefore \dfrac{d y}{d x}=-\dfrac{y x^{y-1}+y^{x} \log y}{x^{y} \log x+x y^{x-1}} \end{aligned} $

Solution

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

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

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

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

$\Rightarrow \log \cos x \dfrac{d y}{d x}-y \tan x=\log \cos y-x \tan y \dfrac{d y}{d x}$

$\Rightarrow(\log \cos x+x \tan y) \dfrac{d y}{d x}=y \tan x+\log \cos y$

$\therefore \dfrac{d y}{d x}=\dfrac{y \tan x+\log \cos y}{x \tan y+\log \cos x}$

Solution

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 \dfrac{d}{d x}(x)+x \cdot \dfrac{d}{d x}(\log y)=\log x \cdot \dfrac{d}{d x}(y)+y \cdot \dfrac{d}{d x}(\log x) \\ & \Rightarrow \log y \cdot 1+x \cdot \dfrac{1}{y} \cdot \dfrac{d y}{d x}=\log x \cdot \dfrac{d y}{d x}+y \cdot \dfrac{1}{x} \\ & \Rightarrow \log y+\dfrac{x}{y} \dfrac{d y}{d x}=\log x \dfrac{d y}{d x}+\dfrac{y}{x} \\ & \Rightarrow(\dfrac{x}{y}-\log x) \dfrac{d y}{d x}=\dfrac{y}{x}-\log y \\ & \Rightarrow(\dfrac{x-y \log x}{y}) \dfrac{d y}{d x}=\dfrac{y-x \log y}{x} \\ & \therefore \dfrac{d y}{d x}=\dfrac{y}{x}(\dfrac{y-x \log y}{x-y \log x}) \end{aligned} $

Solution

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

Taking logarithm on both the sides, we obtain

$ \begin{aligned} & \log (x y)=\log (e^{x-y}) \\ & \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

$\dfrac{d}{d x}(\log x)+\dfrac{d}{d x}(\log y)=\dfrac{d}{d x}(x)-\dfrac{d y}{d x}$

$\Rightarrow \dfrac{1}{x}+\dfrac{1}{y} \dfrac{d y}{d x}=1-\dfrac{d y}{d x}$

$\Rightarrow(1+\dfrac{1}{y}) \dfrac{d y}{d x}=1-\dfrac{1}{x}$

$\Rightarrow(\dfrac{y+1}{y}) \dfrac{d y}{d x}=\dfrac{x-1}{x}$

$\therefore \dfrac{d y}{d x}=\dfrac{y(x-1)}{x(y+1)}$

Solution

The given relationship is $f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$

Taking logarithm on both the sides, we obtain

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

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

$ \begin{aligned} & \dfrac{1}{f(x)} \cdot \dfrac{d}{d x}[f(x)]=\dfrac{d}{d x} \log (1+x)+\dfrac{d}{d x} \log (1+x^{2})+\dfrac{d}{d x} \log (1+x^{4})+\dfrac{d}{d x} \log (1+x^{8}) \\ & \Rightarrow \dfrac{1}{f(x)} \cdot f^{\prime}(x)=\dfrac{1}{1+x} \cdot \dfrac{d}{d x}(1+x)+\dfrac{1}{1+x^{2}} \cdot \dfrac{d}{d x}(1+x^{2})+\dfrac{1}{1+x^{4}} \cdot \dfrac{d}{d x}(1+x^{4})+\dfrac{1}{1+x^{8}} \cdot \dfrac{d}{d x}(1+x^{8}) \\ & \Rightarrow f^{\prime}(x)=f(x)[\dfrac{1}{1+x}+\dfrac{1}{1+x^{2}} \cdot 2 x+\dfrac{1}{1+x^{4}} \cdot 4 x^{3}+\dfrac{1}{1+x^{8}} \cdot 8 x^{7}] \\ & \begin{matrix} \therefore f^{\prime}(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})[\dfrac{1}{1+x}+\dfrac{2 x}{1+x^{2}}+\dfrac{4 x^{3}}{1+x^{4}}+\dfrac{8 x^{7}}{1+x^{8}}] \\ \text{ Hence, } f^{\prime}(1)=(1+1)(1+1^{2})(1+1^{4})(1+1^{8})[\dfrac{1}{1+1}+\dfrac{2 \times 1}{1+1^{2}}+\dfrac{4 \times 1^{3}}{1+1^{4}}+\dfrac{8 \times 1^{7}}{1+1^{8}}] \\ =2 \times 2 \times 2 \times 2[\dfrac{1}{2}+\dfrac{2}{2}+\dfrac{4}{2}+\dfrac{8}{2}] \\ =16 \times(\dfrac{1+2+4+8}{2}) \\ =16 \times \dfrac{15}{2}=120 \end{matrix} \end{aligned} $

Solution

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

Let $x^{2}-5 x+8=u$ and $x^{3}+7 x+9=v$

$\therefore y=u v$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x} \cdot v+u \cdot \dfrac{d v}{d x} \quad$ (By using product rule)

$\Rightarrow \dfrac{d y}{d x}=(x^{3}+7 x+9)\dfrac{d}{d x}(x^{2}-5 x+8) +(x^{2}-5 x+8) \cdot \dfrac{d}{d x}(x^{3}+7 x+9)$

$\Rightarrow \dfrac{d y}{d x}=(2 x-5)(x^{3}+7 x+9)+(x^{2}-5 x+8)(3 x^{2}+7)$

$\Rightarrow \dfrac{d y}{d x}=2 x(x^{3}+7 x+9)-5(x^{3}+7 x+9)+x^{2}(3 x^{2}+7)-5 x(3 x^{2}+7)+8(3 x^{2}+7)$

$\Rightarrow \dfrac{d y}{d x}=(2 x^{4}+14 x^{2}+18 x)-5 x^{3}-35 x-45+(3 x^{4}+7 x^{2})-15 x^{3}-35 x+24 x^{2}+56$

$\therefore \dfrac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$

(ii) $y=(x^{2}-5 x+8)(x^{3}+7 x+9)$

$ \begin{aligned} & =x^{2}(x^{3}+7 x+9)-5 x(x^{3}+7 x+9)+8(x^{3}+7 x+9) \\ & =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 \dfrac{d y}{d x}=\dfrac{d}{d x}(x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72) \\ & =\dfrac{d}{d x}(x^{5})-5 \dfrac{d}{d x}(x^{4})+15 \dfrac{d}{d x}(x^{3})-26 \dfrac{d}{d x}(x^{2})+11 \dfrac{d}{d x}(x)+\dfrac{d}{d x}(72) \\ & =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) $y=(x^{2}-5 x+8)(x^{3}+7 x+9)$

Taking logarithm on both the sides, we obtain

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

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

$ \begin{aligned} & \dfrac{1}{y} \dfrac{d y}{d x}=\dfrac{d}{d x} \log (x^{2}-5 x+8)+\dfrac{d}{d x} \log (x^{3}+7 x+9) \\ & \Rightarrow \dfrac{1}{y} \dfrac{d y}{d x}=\dfrac{1}{x^{2}-5 x+8} \cdot \dfrac{d}{d x}(x^{2}-5 x+8)+\dfrac{1}{x^{3}+7 x+9} \cdot \dfrac{d}{d x}(x^{3}+7 x+9) \\ & \Rightarrow \dfrac{d y}{d x}=y[\dfrac{1}{x^{2}-5 x+8} \times(2 x-5)+\dfrac{1}{x^{3}+7 x+9} \times(3 x^{2}+7)] \\ & \Rightarrow \dfrac{d y}{d x}=(x^{2}-5 x+8)(x^{3}+7 x+9)[\dfrac{2 x-5}{x^{2}-5 x+8}+\dfrac{3 x^{2}+7}{x^{3}+7 x+9}] \\ & \Rightarrow \dfrac{d y}{d x}=(x^{2}-5 x+8)(x^{3}+7 x+9)[\dfrac{(2 x-5)(x^{3}+7 x+9)+(3 x^{2}+7)(x^{2}-5 x+8)}{(x^{2}-5 x+8)(x^{3}+7 x+9)}] \\ & \Rightarrow \dfrac{d y}{d x}=2 x(x^{3}+7 x+9)-5(x^{3}+7 x+9)+3 x^{2}(x^{2}-5 x+8)+7(x^{2}-5 x+8) \\ & \Rightarrow \dfrac{d y}{d x}=(2 x^{4}+14 x^{2}+18 x)-5 x^{3}-35 x-45+(3 x^{4}-15 x^{3}+24 x^{2})+(7 x^{2}-35 x+56) \\ & \Rightarrow \dfrac{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 $\dfrac{d y}{d x}$ are same.

Solution

Let $y=u . v . w=u .(v . w)$

By applying product rule, we obtain $\dfrac{d y}{d x}=\dfrac{d u}{d x} \cdot(v \cdot w)+u \cdot \dfrac{d}{d x}(v \cdot w)$

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x} v \cdot w+u[\dfrac{d v}{d x} \cdot w+v \cdot \dfrac{d w}{d x}] \quad$ (Again applying product rule)

$\Rightarrow \dfrac{d y}{d x}=\dfrac{d u}{d x} \cdot v \cdot w+u \cdot \dfrac{d v}{d x} \cdot w+u \cdot v \cdot \dfrac{d w}{d x}$

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

$\dfrac{1}{y} \cdot \dfrac{d y}{d x}=\dfrac{d}{d x}(\log u)+\dfrac{d}{d x}(\log v)+\dfrac{d}{d x}(\log w)$

$\Rightarrow \dfrac{1}{y} \cdot \dfrac{d y}{d x}=\dfrac{1}{u} \dfrac{d u}{d x}+\dfrac{1}{v} \dfrac{d v}{d x}+\dfrac{1}{w} \dfrac{d w}{d x}$

$\Rightarrow \dfrac{d y}{d x}=y(\dfrac{1}{u} \dfrac{d u}{d x}+\dfrac{1}{v} \dfrac{d v}{d x}+\dfrac{1}{w} \dfrac{d w}{d x})$

$\Rightarrow \dfrac{d y}{d x}=u . v . w \cdot(\dfrac{1}{u} \dfrac{d u}{d x}+\dfrac{1}{v} \dfrac{d v}{d x}+\dfrac{1}{w} \dfrac{d w}{d x})$

$\therefore \dfrac{d y}{d x}=\dfrac{d u}{d x} \cdot v \cdot w+u \cdot \dfrac{d v}{d x} \cdot w+u \cdot v \cdot \dfrac{d w}{d x}$