Question

If y = sin-? [x√(1-x) - √x√1–x^2], find dy/dx​

Answer 1

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

Given that,

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

can be rewritten as

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

We know,

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

Using this, we get

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

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

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

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

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

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

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

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

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

Additional information :-

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx}k = 0}$

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

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

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

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

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