Question

sin^-1(2^x)=?differentiate



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Answer 1

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

Given function is

$\rm :\longmapsto\: {sin}^{ - 1} {2}^{x}$

Let assume that

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

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

$\rm :\longmapsto\: \dfrac{d}{dx} y =\dfrac{d}{dx} {sin}^{ - 1} {2}^{x}$

We know,

$\boxed{\tt{ \dfrac{d}{dx}[f(g(x)] = f'[g(x)]\dfrac{d}{dx}g(x) \: }}$

and

$\boxed{\tt{ \dfrac{d}{dx} {sin}^{ - 1}x = \frac{1}{ \sqrt{1 - {x}^{2} } }}}$

So, using this, we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{1 - {4}^{x}} } \: \times {2}^{x} log_{e}(2)$

$\purple{\rm\implies \:\boxed{\tt{ \: \: \: \: \: \: \dfrac{d}{dx} {sin}^{ - 1}{2}^{x} = \frac{{2}^{x} log_{e}(2) }{ \sqrt{1 - {4}^{x}} } \: \: \: \: \: }}}$

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Additional Information

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