Question

If \( y = \tan^{-1} \left( \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} \right) \), then \( \frac{dy}{dx} \) is

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

Hint:

When differentiating inverse trigonometric functions, apply the standard derivative formulas and simplify the expression carefully.

Answer 1

The Correct Option is C

Step 1: Simplify the given expression for \( y \).
The given expression for \( y \) is: \[ y = \tan^{-1} \left( \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} \right) \]
Step 2: Apply the derivative formula.
The derivative of \( \tan^{-1}(z) \) is \( \frac{d}{dx} \left( \tan^{-1}(z) \right) = \frac{1}{1 + z^2} \cdot \frac{dz}{dx} \). In our case, the expression for \( z \) involves \( x \), so we differentiate it to obtain: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]
Step 3: Conclusion.
The correct answer is \( \frac{1}{\sqrt{1 - x^2}} \).