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

Concept: Use: \[ \frac{d}{dx}\left(\sin^{-1}u\right)=\frac{u'}{\sqrt{1-u^2}} \]

Step 1: Given: \[ y+\sin^{-1}(1-x^2)=e^x \]

Step 2: Differentiate both sides. \[ \frac{dy}{dx}+\frac{d}{dx}\left[\sin^{-1}(1-x^2)\right]=e^x \] Let: \[ u=1-x^2 \] Then: \[ u'=-2x \]

Step 3: Differentiate inverse sine term. \[ \frac{d}{dx}\left[\sin^{-1}(1-x^2)\right] = \frac{-2x}{\sqrt{1-(1-x^2)^2}} \]

Step 4: Simplify denominator. \[ 1-(1-x^2)^2 = 1-(1-2x^2+x^4) \] \[ =2x^2-x^4 \] \[ =x^2(2-x^2) \] So: \[ \sqrt{x^2(2-x^2)}=x\sqrt{2-x^2} \] Thus: \[ \frac{-2x}{x\sqrt{2-x^2}} = -\frac{2}{\sqrt{2-x^2}} \]

Step 5: Substitute in the differentiated equation. \[ \frac{dy}{dx}-\frac{2}{\sqrt{2-x^2}}=e^x \] \[ \frac{dy}{dx}=e^x+\frac{2}{\sqrt{2-x^2}} \] Therefore, \[ \boxed{e^x+\frac{2}{\sqrt{2-x^2}}} \]