Question

If $y={\sin }^{-1}\left(\sqrt{1-x^2}\right)+4{\sec }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)$, then $\frac{dy}{dx}$ equals

• A. $0$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. $\frac{2}{\sqrt{1-x^2}}$
• D. $\frac{3}{\sqrt{1-x^2}}$

Answer 1

The correct option is D $\frac{3}{\sqrt{1-x^2}}$

$y={\sin }^{-1}\left(\sqrt{1-x^2}\right)+4{\sec }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)={\sin }^{-1}\left(\sqrt{1-x^2}\right)+4{\cos }^{-1}\left(\sqrt{1-x^2}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{-2x}{2\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}}-\frac{\frac{-8x}{2\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}}=\frac{3}{\sqrt{1-x^2}}$