Question

यदि ${\displaystyle y={\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2},0<x<1}$  है तो ${\displaystyle \frac{dy}{dx}}$ ज्ञात कीजिए।

Answer 1

∴ y = sin^-1 x + sin^-1 ${\displaystyle \sqrt{1-x^2}}$

x = sin θ रखने पर,

y = ${\displaystyle \theta +{\sin }^{-1} \cos \theta }$

${\displaystyle =\theta +{\sin }^{-1}\left[\sin (\frac{\pi }{2}-\theta )\right]}$

${\displaystyle =\theta +\frac{\pi }{2}-\theta }$

=${\displaystyle \frac{\pi }{2}}$

${\displaystyle \therefore \frac{dy}{dx}=0}$

Answer 2

यहाँ, ${\displaystyle y={\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2}}$

माना,  ${\displaystyle u={\sin }^{-1}x \text{और} v={\sin }^{-1}\sqrt{1-x^2}}$

${\displaystyle \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}}$

अब ${\displaystyle v={\sin }^{-1}\sqrt{1-x^2}}$

x = cos θ

∴ ${\displaystyle v={\sin }^{-1}\sqrt{1-{\cos }^2\theta }={\sin }^{-1}\sqrt{{\sin }^2\theta }}$

${\displaystyle ={\sin }^{-1}\left(\sin \theta \right)=\theta ={\cos }^{-1}x}$

∴${\displaystyle \frac{dv}{dx}=-\frac{1}{\sqrt{1-x^2}}}$

${\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}}$

${\displaystyle =\frac{1}{\sqrt{1-x^2}}+-\frac{1}{\sqrt{1-x^2}}=0}$