Question

${ \displaystyle{{\boxed{ \bold \red{{\frac{d}{dy} \bigg( \frac{ \sqrt[In2]{ {x}^{2} {e}^{x} }\sin^{\pi}x }{Inx \: \int^{x}_{0} { \tan}^{ - 1}y \: dy } \bigg)}}}}}}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\dfrac{ \sqrt[In2]{ {x}^{2} {e}^{x} }\sin^{\pi}x }{Inx \: \int^{x}_{0} { \tan}^{ - 1}y \: dy }$

Let assume that

$\rm :\longmapsto\:f(x) = \dfrac{ \sqrt[In2]{ {x}^{2} {e}^{x} }\sin^{\pi}x }{Inx \: \int^{x}_{0} { \tan}^{ - 1}y \: dy }$

Now, Consider

$\rm :\longmapsto\:\displaystyle \int \: {tan}^{ - 1}y \: dy$

can be rewritten as

$\rm \: = \: \displaystyle \int \:1. \: {tan}^{ - 1}y \: dy$

$\rm \: = \: {tan}^{ - 1}y\displaystyle \int \: dy - \displaystyle \int \: \bigg[\dfrac{d}{dy} {tan}^{ - 1}y\displaystyle \int \: dy\bigg]dy$

$\rm \: = \: {tan}^{ - 1}y \: (y) - \displaystyle \int \: \dfrac{1}{ {y}^{2} + 1} \times y \: dy$

$\rm \: = \: {tan}^{ - 1}y \: (y) - \displaystyle \int \: \dfrac{y}{ {y}^{2} + 1} \: dy$

$\rm \: = \: {tan}^{ - 1}y \: (y) - \dfrac{1}{2} \displaystyle \int \: \dfrac{2y}{ {y}^{2} + 1} \: dy$

$\rm \: = \: {tan}^{ - 1}y \: (y) - \dfrac{1}{2}log | {y}^{2} + 1|$

Therefore,

$\rm :\longmapsto\:\displaystyle \int_0^x \: {tan}^{ - 1}y \: dy = x{tan}^{ - 1}x - \frac{1}{2}log | {x}^{2} + 1 | - 0$

$\rm :\longmapsto\:\displaystyle \int_0^x \: {tan}^{ - 1}y \: dy = x{tan}^{ - 1}x - \frac{1}{2}log | {x}^{2} + 1 |$

So,

$\rm :\longmapsto\:f(x) = \dfrac{ \sqrt[In2]{ {x}^{2} {e}^{x} }\sin^{\pi}x }{Inx \: [x{tan}^{ - 1}x - \frac{1}{2}log | {x}^{2} + 1 |]}$

On differentiating both sides w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{d}{dy} f(x) =\dfrac{d}{dy}\bigg[ \dfrac{ \sqrt[In2]{ {x}^{2} {e}^{x} }\sin^{\pi}x }{Inx \: [x{tan}^{ - 1}x - \frac{1}{2}log | {x}^{2} + 1 |]}\bigg]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }}$

So,

$\rm :\longmapsto\:\dfrac{d}{dy}f(x) = 0$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dy}\:\dfrac{ \sqrt[In2]{ {x}^{2} {e}^{x} }\sin^{\pi}x }{Inx \: \int^{x}_{0} { \tan}^{ - 1}y \: dy } \: = \: 0 \: }}}$

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More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

$\red{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}}$