Question

Find the value $f^{\prime }(x)$, if $f(x)={\sin }^{-1}(1-x^2)$

• (A) $\frac{1}{\sqrt{1-x^2}}$
• (B) $\frac{2}{\sqrt{2-x^2}}$
• (C) $\frac{-2}{\sqrt{2-x^2}}$
• (D) $\frac{-1}{\sqrt{1-x^2}}$

Answer 1

The Correct Option is C

Explanation:
Concept:$\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$$\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$Let $u=1-x^2$$\frac{du}{dx}=-2x$$y={\sin }^{-1}(1-x^2)={\sin }^{-1}u$$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$$\frac{dy}{dx}=\frac{d}{du}{\sin }^{-1}u\times (-2x)$$\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\times (-2x)$$\frac{dy}{dx}=\frac{-2x}{\sqrt{1-{(1-x^2)}^2}}$$\frac{dy}{dx}=\frac{-2x}{\sqrt{2x^2-x^4}}$$\frac{dy}{dx}=\frac{-2}{\sqrt{2-x^2}}$Hence, the correct option is (C).