Question

If
$y = {sin}^{ - 1} \frac{ {2}^{x + 1} \times {5}^{x} }{1 + {100}^{x} } \: find \: \frac{dy}{dx}$

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Answer 1

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

Given function is

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{ {2}^{x + 1} \times {5}^{x} }{1 + {100}^{x} } \bigg]$

can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{2 \times {2}^{x} \times {5}^{x} }{1 + {( {10}^{2}) }^{x} } \bigg]$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{2 \times {(2 \times 5)}^{x} }{1 + {10}^{2x} } \bigg]$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{2 \times {10}^{x} }{1 + {10}^{2x} } \bigg]$

We know,

$\boxed{ \tt{ \: {sin}^{ - 1}\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg] = 2 {tan}^{ - 1}x \: }}$

So, using this, we get

$\rm :\longmapsto\:y \: = \: 2 \: {tan}^{ - 1} {10}^{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \: \dfrac{d}{dx}[2 \: {tan}^{ - 1} {10}^{x}]$

$\rm :\longmapsto\:\dfrac{dy}{dx}\: = 2 \: \dfrac{d}{dx}\: {tan}^{ - 1} {10}^{x}$

We know that,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{1 + {( {10}^{x})}^{2} }\dfrac{d}{dx} {10}^{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{1 + {100}^{x} } \times {10}^{x} \times log_{e}(10)$

$\purple{\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{ {10}^{x} \: log_{e}(10) }{1 + {100}^{x} } \: }}}$

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Additional Information :-

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