Question

find dy/dx if y=sin^-1(x^2√1-x^2 -x√1-x^4

Answer 1

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

Given that

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg( {x}^{2} \sqrt{1 - {x}^{2}} - x \sqrt{1 - {x}^{4}}\bigg)$

can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg( {x}^{2} \sqrt{1 - {x}^{2}} - x \sqrt{1 - {( {x}^{2})}^{2}}\bigg)$

We know,

$\boxed{ \sf{ \: {sin}^{x} - {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2}} - y \sqrt{1 - {x}^{2}}}}$

So, using this identity, we get

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

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

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

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

$\: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \dfrac{d}{dx} {sin}^{ - 1}x \: = \: \dfrac{1}{ \sqrt{1 - {x}^{2} } }\bigg \}}$

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

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

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

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

Additional Information :-

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1}x = - \: \dfrac{1}{ x\sqrt{{x}^{2} - 1} }}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {sec}^{ - 1}x = \: \dfrac{1}{ x\sqrt{{x}^{2} - 1} }}$

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

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx}logx = \dfrac{1}{x}}$