Question

If \( f(x) = \sin^{-1}(\cos x) \), then \( \frac{d^2 y}{dx^2} \) at \( x = \frac{\pi}{4} \) is:

• A. \( -\frac{1}{4} \)
• B. \( -1 \)
• C. 1
• D. \( \frac{1}{2} \)
• E. 0

Hint:

The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \), so make sure to apply the chain rule properly.

Answer 1

Answer: (E)

We need to compute the second derivative \( \frac{d^2y}{dx^2} \) at \( x = \frac{\pi}{4} \). Start by differentiating \( f(x) = \sin^{-1}(\cos x) \). 
Using the chain rule: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot (-\sin x) \] Since \( 1 - (\cos x)^2 = \sin^2 x \), we get: \[ \frac{dy}{dx} = -\frac{\sin x}{\sin x} = -1 \] Now, compute the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-1) = 0 \] Thus, \( \frac{d^2y}{dx^2} = 0 \) at \( x = \frac{\pi}{4} \).