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: Concept
Isolate $y$ as $y = e^x - \sin^{-1}(1-x^2)$ and differentiate using the chain rule.

Step 2: Meaning
The derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$.

Step 3: Analysis
$\frac{dy}{dx} = e^x - \frac{1}{\sqrt{1-(1-x^2)^2}} \cdot (-2x) = e^x + \frac{2x}{\sqrt{1-(1 - 2x^2 + x^4)}} = e^x + \frac{2x}{\sqrt{2x^2 - x^4}}$.

Step 4: Conclusion
Factoring $x^2$ out of the root: $\frac{2x}{x\sqrt{2-x^2}} = \frac{2}{\sqrt{2-x^2}}$. Thus, $\frac{dy}{dx} = e^x + \frac{2}{\sqrt{2-x^{2}}}$.
Final Answer: (C)