Question

If $y + \sin^{-1}(1 - x^{2}) = e^{x}$, then $\frac{dy}{dx} =$

• A. $e^{x} - \frac{2}{\sqrt{2-x^{2}}}$
• B. $e^{x} - \frac{2}{\sqrt{2+x^{2}}}$
• C. $e^{x} + \frac{2}{\sqrt{2-x^{2}}}$
• D. $e^{x} + \frac{2}{\sqrt{2+x^{2}}}$

Hint:

Watch your signs! The derivative of the inner function $(-2x)$ cancels the negative sign from the subtraction.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We isolate \( y \) or differentiate implicitly. The derivative of \( \sin^{-1} u \) requires the chain rule: \( \frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \)[cite: 1].
Step 2: Key Formula or Approach:
1. \( \frac{d}{dx}(e^x) = e^x \)[cite: 1].
2. \( \frac{d}{dx}(1 - x^2) = -2x \)[cite: 1].
Step 3: Detailed Explanation:
Differentiate the entire equation: \[ \frac{dy}{dx} + \frac{1}{\sqrt{1 - (1 - x^2)^2}} \cdot \frac{d}{dx}(1 - x^2) = e^x \] \[ \frac{dy}{dx} + \frac{1}{\sqrt{1 - (1 - 2x^2 + x^4)}} \cdot (-2x) = e^x \] \[ \frac{dy}{dx} - \frac{2x}{\sqrt{2x^2 - x^4}} = e^x \] Rearranging for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = e^x + \frac{2x}{x\sqrt{2 - x^2}} = e^x + \frac{2}{\sqrt{2 - x^2}} \]
Step 4: Final Answer:
The derivative is \( \frac{dy}{dx} = e^x + \frac{2}{\sqrt{2-x^2}} \).