Question

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

• A. \( \sec y \)
• B. \( \sec y \tan y \)
• C. \( \sec^2 y \tan y \)
• D. \( \tan^2 y \sec y \)

Hint:

For derivatives of inverse trigonometric functions, use chain rule and trigonometric identities to simplify higher derivatives.

Answer 1

The Correct Option is C

Step 1: First derivative of \( y = \sin^{-1} x \)
Let \( y = \sin^{-1} x \). Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}. \] Step 2: Second derivative of \( y \)
Differentiate \( \frac{dy}{dx} \) with respect to \( x \): \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{1}{\sqrt{1 - x^2}} \right). \] Using the chain rule: \[ \frac{d^2 y}{dx^2} = \frac{0 - \left( \frac{1}{2} \right)(1 - x^2)^{-3/2}(-2x)}{\sqrt{1 - x^2}^2}. \] Simplify: \[ \frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^{3/2}}. \] Step 3: Express in terms of \( y \)
Since \( y = \sin^{-1} x \), we know: \[ \cos y = \sqrt{1 - x^2}, \quad {and} \quad \sec y = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}. \] Also, \( \tan y = \frac{\sin y}{\cos y} = x \sec y \). Substituting back: \[ \frac{d^2 y}{dx^2} = \sec^2 y \tan y. \] Conclusion: The second derivative is \( \sec^2 y \tan y \).