Question

if y=sin-¹[5x+12√1-x²/13] , dy/dx =...?​

Answer 1

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

Given expression is

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{5x + 12 \sqrt{1 - {x}^{2} } }{13} \bigg]$

Let we use method of Substitution to evaluate this to simplest form.

Let Substitute

$\red{\rm :\longmapsto\:x = sin \theta \: }$

So, above expression can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{5sin\theta + 12 \sqrt{1 - {sin}^{2} \theta } }{13} \bigg]$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{5sin\theta + 12 \sqrt{ {cos}^{2} \theta } }{13} \bigg]$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{5sin\theta + 12cos\theta }{13} \bigg]$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg[\dfrac{5}{13}sin\theta + \dfrac{12}{13}cos\theta \bigg]$

Let further assume that

$\red{\rm :\longmapsto\:\dfrac{5}{13} = cos \alpha }$

$\red{\rm\implies \:\dfrac{12}{13} = sin \alpha }$

So,

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(sin\theta cos \alpha + sin \alpha cos\theta \bigg)$

$\rm :\longmapsto\:y = {sin}^{ - 1}\bigg(sin(\theta + \alpha ) \bigg)$

$\rm :\longmapsto\:y = \theta + \alpha$

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

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}[ {sin}^{ - 1}x + \alpha ]$

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

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MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$