Question

${ \boxed{ \bold \red{ \frac{d}{dx} \bigg( \frac{ \sqrt[In2]{x {}^{2} e {}^{x} \sin x} }{Inx \: \tan^{ - 1} y} \bigg)}}}$

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Answer 1

Answer:

${ \boxed{ \bold \red{ \frac{d}{dx} \bigg( \frac{ \sqrt[In2]{x {}^{2} e {}^{x} \sin x} }{Inx \: \tan^{ - 1} y} \bigg)}}}$

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Answer 2

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

Given function is

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

Let assume that

$\rm :\longmapsto\:y = \dfrac{ \sqrt[In2]{x {}^{2} e {}^{x} \sin x} }{Inx \: \tan^{ - 1} y}$

can be rewritten as

$\rm :\longmapsto\:y = \dfrac{ {\bigg( {x}^{2} {e}^{x}sinx \bigg) }^{\dfrac{1}{ln2} } }{Inx \: \tan^{ - 1} y}$

Now, taking ln (log to the base e) on both sides, we get

$\rm :\longmapsto\:ln(y) =ln \dfrac{ {\bigg( {x}^{2} {e}^{x}sinx \bigg) }^{\dfrac{1}{ln2} } }{Inx \: \tan^{ - 1} y}$

We know,

$\boxed{ \tt{ \: log {x}^{y} = ylogx \: }} \\ \\ \boxed{ \tt{ \: logxy = logx + logy \: }} \\ \\ \boxed{ \tt{ \: log \frac{x}{y} = logx - logy \: }}$

So, using this, we get

$\rm \: lny = \dfrac{1}{ln2}\bigg[2ln(x) + xln(e) + ln(sinx)\bigg] - ln(ln(x) - ln( {tan}^{ - 1}x)$

$\rm \: lny = \dfrac{1}{ln2}\bigg[2ln(x) + x + ln(sinx)\bigg] - ln(ln(x) - ln( {tan}^{ - 1}x)$

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

$\rm \:\dfrac{d}{dx}lny =\dfrac{d}{dx}\bigg\{\dfrac{1}{ln2}\bigg[2ln(x) + x + ln(sinx)\bigg] - ln(ln(x) - ln( {tan}^{ - 1}x) \bigg\}$

We know,

$\\ \boxed{ \tt{ \: \dfrac{d}{dx}ln(x) = \frac{1}{x} \: }} \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}sinx = cosx \: }} \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}x = 1 \: }} \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx} {tan}^{ - 1}x = \frac{1}{1 + {x}^{2} } \: }} \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}f(g(x)) = f'(g(x))g'(x) \: }}$

So, using these, we get

$\rm \:\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{ln2}\bigg[\dfrac{2}{x} + 1 + \dfrac{1}{sinx}(cosx) \bigg] - \dfrac{1}{x \: ln(x)} - \dfrac{1}{ {tan}^{ - 1}x} \times \dfrac{1}{1 + {x}^{2} }$

$\rm \:\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{ln2}\bigg[\dfrac{2}{x}+1 + cotx\bigg] - \dfrac{1}{x \: ln(x)}-\dfrac{1}{(1 + {x}^{2}){tan}^{ - 1}x}$

$\rm \:\dfrac{dy}{dx}=y\bigg\{\dfrac{1}{ln2}\bigg[\dfrac{2}{x}+1 + cotx\bigg] - \dfrac{1}{x \: ln(x)}-\dfrac{1}{(1 + {x}^{2}){tan}^{ - 1}x}\bigg\}$

$\rm \:\dfrac{dy}{dx}=\dfrac{ \sqrt[In2]{x {}^{2} e {}^{x} \sin x} }{Inx \: \tan^{ - 1} y}\bigg\{\dfrac{1}{ln2}\bigg[\dfrac{2}{x}+1 + cotx\bigg] - \dfrac{1}{x \: ln(x)}-\dfrac{1}{(1 + {x}^{2}){tan}^{ - 1}x}\bigg\}$

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More to know :-

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