Question

Differentiate the above with clear steps

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Differentiate the following function with respect to x

$\rm :\longmapsto\:y = {sin}^{ - 1}(logx)$

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

Given function is

$\rm :\longmapsto\:y = {sin}^{ - 1}(logx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {sin}^{ - 1}(logx)$

We know,

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

So, using this result, we get

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

We know that,

$\boxed{ \bf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

So, using this result, we get

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

So,

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{x \sqrt{1 - {(logx)}^{2} } }$

Additional Information :-

$\boxed{ \bf{ \: \dfrac{d}{dx}{cos}^{ - 1}x = - \: \frac{1}{ \sqrt{1 - {x}^{2} } } }}$

$\boxed{ \bf{ \: \dfrac{d}{dx}{tan}^{ - 1}x = \frac{1}{1 + {x}^{2} }}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}{cot}^{ - 1}x = - \: \frac{1}{1 + {x}^{2} }}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}{sec}^{ - 1}x = \frac{1}{ x\sqrt{ {x}^{2} - 1 } } }}$

$\boxed{ \bf{ \: \dfrac{d}{dx}{cosec}^{ - 1}x = - \: \frac{1}{ x\sqrt{ {x}^{2} - 1 } } }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x}}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga}}$