Question

The derivative of ${\sin }^{-1}x$ with respect to ${\cos }^{-1}\sqrt{1-x^2}$ is?

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. ${\cos }^{-1}x$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

Assume that $\theta ={\sin }^{-1}x$

$\therefore \sin \theta =x⟹\cos \theta =\sqrt{1-x^2}$

differentiating $\sin \theta$ wrt $x$,

$\cos \theta \frac{d\theta }{dx}=1$

$\therefore \frac{d\theta }{dx}=\frac{1}{\cos \theta }$

$\therefore \frac{d\theta }{dx}=\frac{1}{\sqrt{1-x^2}}$

Assume that $\gamma ={\cos }^{-1}\sqrt{1-x^2}⟹\sin \gamma =x$

Also, differentiating $\cos \gamma$ wrt $x$,

$-\sin \gamma \frac{d\gamma }{dx}=\frac{1}{2}\frac{1}{\sqrt{1-x^2}}(-2x)$

$\therefore -x\frac{d\gamma }{dx}=\frac{1}{\sqrt{1-x^2}}(-x)$

$\therefore \frac{d\gamma }{dx}=\frac{1}{\sqrt{1-x^2}}$

Now, we require

$\frac{d{\sin }^{-1}x}{d{\cos }^{-1}\sqrt{1-x^2}}=\frac{d\theta }{d\gamma }$

$=\frac{d\theta }{dx}\times \frac{dx}{d\gamma }$

$=\frac{1}{\sqrt{1-x^2}}\times \frac{\sqrt{1-x^2}}{1}$

$=1$

This is the required answer.