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:

For \(\sin^{-1}u\), always apply chain rule carefully: derivative is \(\frac{u'}{\sqrt{1-u^2}}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This problem requires the differentiation of an implicit function containing an inverse trigonometric term and an exponential term.
The derivative of \( \sin^{-1} u \) is \( \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \).
Step 2: Key Formula or Approach:
1. Rewrite as \( y = e^x - \sin^{-1}(1-x^2) \).
2. Apply chain rule for the inverse sine term.
Step 3: Detailed Explanation:

Differentiate the equation with respect to \( x \):
\[ \frac{dy}{dx} + \frac{d}{dx}(\sin^{-1}(1-x^2)) = \frac{d}{dx}(e^x) \]

Evaluate the inverse sine derivative:
Let \( u = 1 - x^2 \). Then \( \frac{du}{dx} = -2x \).
The derivative is \( \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} = \frac{-2x}{\sqrt{1 - (1 - x^2)^2}} \).

Simplify the denominator:
\( 1 - (1 - x^2)^2 = 1 - (1 - 2x^2 + x^4) = 2x^2 - x^4 = x^2(2 - x^2) \).
The square root is \( |x|\sqrt{2 - x^2} \).

So, \( \frac{dy}{dx} + \frac{-2x}{x\sqrt{2-x^2}} = e^x \) (assuming \( x>0 \)).
\[ \frac{dy}{dx} - \frac{2}{\sqrt{2-x^2}} = e^x \]

Move the term to the other side:
\[ \frac{dy}{dx} = e^x + \frac{2}{\sqrt{2-x^2}} \]

Step 4: Final Answer:
The derivative is \( e^x + \frac{2}{\sqrt{2-x^2}} \).