Question

If \( f(x)=\sqrt{\cos ^{-1} \sqrt{1-x^{2}}} \), then \( f^{\prime}\left(\frac{1}{2}\right)= \)

• A. \( \sqrt{\frac{2}{\pi}} \)
• B. \( \sqrt{\frac{\pi}{2}} \)
• C. \( \frac{2}{\sqrt{\pi}} \)
• D. \( \frac{\sqrt{\pi}}{2} \)

Hint:

Using substitution (like \( x = \sin \theta \)) often simplifies inverse trigonometric functions significantly before differentiating. Always define the domain to ensure \( \cos^{-1}(\cos \theta) = \theta \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We are given a composite function involving square roots and inverse trigonometric functions. The goal is to find the derivative of this function at a specific point \( x = 1/2 \). To simplify the differentiation, we can use trigonometric substitution.
Step 2: Key Formula or Approach:

1. Substitution: Let \( x = \sin \theta \). Then \( \sqrt{1-x^2} = \cos \theta \) (for \( x \in [0, 1] \)). 2. Inverse Identity: \( \cos^{-1}(\cos \theta) = \theta \). 3. Chain Rule: \( \frac{d}{dx} (\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \). 4. Derivative of inverse sine: \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \).
Step 3: Detailed Explanation:

Given \( f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}} \). Let \( x = \sin \theta \), where \( \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \). Since we evaluate at \( x = 1/2 \), we can assume \( x \) is in the principal domain. Then \( \sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \cos \theta \). Now substitute this back into the function: \[ f(x) = \sqrt{\cos^{-1}(\cos \theta)} = \sqrt{\theta} \] Since \( x = \sin \theta \implies \theta = \sin^{-1} x \), we have: \[ f(x) = \sqrt{\sin^{-1} x} \] Now, differentiate \( f(x) \) with respect to \( x \) using the chain rule: \[ f'(x) = \frac{d}{dx} \left( (\sin^{-1} x)^{1/2} \right) \] \[ f'(x) = \frac{1}{2}(\sin^{-1} x)^{-1/2} \cdot \frac{d}{dx}(\sin^{-1} x) \] \[ f'(x) = \frac{1}{2\sqrt{\sin^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \] Evaluate \( f'(x) \) at \( x = \frac{1}{2} \): 1. Calculate \( \sin^{-1}(1/2) \): \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] 2. Calculate \( \sqrt{1 - (1/2)^2} \): \[ \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] Substitute these values into the derivative expression: \[ f'\left(\frac{1}{2}\right) = \frac{1}{2\sqrt{\frac{\pi}{6}}} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \] \[ = \frac{1}{2\frac{\sqrt{\pi}}{\sqrt{6}}} \cdot \frac{2}{\sqrt{3}} \] \[ = \frac{\sqrt{6}}{2\sqrt{\pi}} \cdot \frac{2}{\sqrt{3}} \] Cancel the 2s and combine square roots: \[ = \frac{\sqrt{6}}{\sqrt{3}\sqrt{\pi}} = \frac{\sqrt{2}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{\pi}} = \frac{\sqrt{2}}{\sqrt{\pi}} = \sqrt{\frac{2}{\pi}} \]
Step 4: Final Answer:

The value of \( f'(\frac{1}{2}) \) is \( \sqrt{\frac{2}{\pi}} \).