Question

Find dy/dx ;y = (sin^-1x)^x



plz answer this

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Answer 1

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

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

On taking log on both sides

$\rm :\longmapsto\:logy = log {( {sin}^{ - 1} x)}^{x}$

$\rm :\longmapsto\:logy = x \: log {( {sin}^{ - 1} x)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log( {x}^{y} ) = y \: logx \bigg \}}$

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

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

We know,

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

$\red{\rm :\longmapsto\:\dfrac{d}{dx} u.v = u \dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u}$

Using these Identities, we get

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

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

We know,

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

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

Using all, we get now

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

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

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

Shortcut

$\rm :\longmapsto\:\dfrac{d}{dx} {f}^{g}$

where f and g both are function of x or either function of x or y.

$\rm \: \: = \: {f}^{g}\bigg(\dfrac{g}{f} \dfrac{d}{dx}f \: + \: logf \: \dfrac{d}{dx}g\bigg)$

For example,

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{x}$

$\rm \: \: = \: {x}^{x}\bigg(\dfrac{x}{x} \dfrac{d}{dx}x + log(x)\dfrac{d}{dx}x \bigg)$

$\rm \: \: = \: {x}^{x}(1 + logx)$