Question

If \( f(x) = \sin^{-1} \left( \sqrt{\frac{1 - x}{2}} \right) \), then \[ f'(x) = \]

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

Hint:

When differentiating inverse trigonometric functions, apply the chain rule and simplify the resulting expressions.

Answer 1

The Correct Option is A

Step 1: Differentiate the inverse sine function.
We are given \( f(x) = \sin^{-1} \left( \sqrt{\frac{1 - x}{2}} \right) \). To differentiate this, we first use the chain rule. The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1 - u^2}} \), and we differentiate the inner function \( \sqrt{\frac{1 - x}{2}} \). After applying the chain rule, we find that: \[ f'(x) = \frac{-1}{2\sqrt{1 - x^2}}. \]
Step 2: Conclusion.
Thus, the correct answer is \( \frac{-1}{2\sqrt{1 - x^2}} \), corresponding to option (A).