Question

Differentiate the above. Write clear steps.

Answer 1

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

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{dy}{dx}$

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

Given that

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

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

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

We know,

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

So, using this, we get

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

We know,

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

So, using this, we have

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

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \: \dfrac{{e}^{ {sin}^{ - 1} {x}^{2} }}{ \sqrt{1 - {{x}^{4}}} } \: \dfrac{d}{dx} \: {{x}^{2} }$

We know,

$\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

So, using this result, we get

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

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \: \dfrac{2x \: \: {e}^{ {sin}^{ - 1} {x}^{2} }}{ \sqrt{1 - {{x}^{4}}} }$

Hence,

$\bf :\longmapsto\:\dfrac{dy}{dx} \: = \: \dfrac{2x \: \: {e}^{ {sin}^{ - 1} {x}^{2} }}{ \sqrt{1 - {{x}^{4}}} }$

Additional Information :-

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

$\boxed{ \rm{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}$

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

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

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

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosx = - \: sinx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = - \: {cosec}^{2}x}}$