Question

The derivative of ${\sin }^{-1}x$ w.r.t. ${\cos }^{-1}\sqrt{1-x^2}$ is ($-1\le x\le 1$)

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

Answer 1

The correct option is B $1$

$\frac{du}{dx}=\frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$
$\frac{dv}{dx}=\frac{d}{dx}\left({\cos }^{-1}\left(\sqrt{1-x^2}\right)\right)=-\frac{\frac{-2x}{1\sqrt{1-x^2}}}{\sqrt{1-{(1-x^2)}^2}}$
$=+\frac{x}{\sqrt{1-x^2}}\times \sqrt{x^2}=\frac{1}{\sqrt{1-x^2}}$

$\frac{du}{dv}=1$
Therefore derivative of ${\sin }^{-1}x$ w.r.t ${\cos }^{-1}\sqrt{1-x^2}$ is $1$