Question

The derivative of $ \frac{d}{dx} \left( x \sqrt{a^2} - x^2 + a^2 \sin^{-1} \left( \frac{x}{a} \right) \right) \text{ is equal to} $

• A. \( \sqrt{a^2 - x^2} \)
• B. \( 2\sqrt{a^2 - x^2} \)
• C. \( \frac{1}{\sqrt{a^2 - x^2}} \)
• D. None of these

Hint:

Use the product rule and the chain rule when differentiating composite functions, and remember the standard derivative of \( \sin^{-1}(x) \).

Answer 1

The Correct Option is B

The given function is: \[ f(x) = x \sqrt{a^2 - x^2} + a^2 \sin^{-1} \left( \frac{x}{a} \right) \] To differentiate this, apply the product rule for the first term and use the derivative formula for \( \sin^{-1}(x) \): - The derivative of \( x \sqrt{a^2 - x^2} \) is: \[ \frac{d}{dx} \left( x \sqrt{a^2 - x^2} \right) = \sqrt{a^2 - x^2} - \frac{x^2}{\sqrt{a^2 - x^2}} \] - The derivative of \( a^2 \sin^{-1} \left( \frac{x}{a} \right) \) is: \[ \frac{d}{dx} \left( a^2 \sin^{-1} \left( \frac{x}{a} \right) \right) = \frac{1}{\sqrt{a^2 - x^2}} \] Thus, combining these results: \[ f'(x) = 2\sqrt{a^2 - x^2} \]
Step 4: Conclusion. The derivative is \( 2\sqrt{a^2 - x^2} \), which corresponds to option (b).