Question

If $y = \sin^{-1} x$, then $(1 - x^2) \frac{d^2y}{dx^2}$ is equal to :

• A. $x \frac{dy}{dx}$
• B. $-x \frac{dy}{dx}$
• C. $x^2 \frac{dy}{dx}$
• D. $-x^2 \frac{dy}{dx}$

Hint:

For inverse trigonometric functions, use the chain rule and remember the identity $\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}$.

Answer 1

The Correct Option is D

Let $y = \sin^{-1} x$. Differentiating with respect to $x$ yields: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \] Differentiating a second time with respect to $x$ produces: \[ \frac{d^2y}{dx^2} = \frac{-x}{(1 - x^2)^{3/2}} \] Multiplying by $(1 - x^2)$ yields the result: \[ (1 - x^2) \frac{d^2y}{dx^2} = -x^2 \frac{dy}{dx} \] Therefore, the result is $-x^2 \frac{dy}{dx}$.