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Mathematics · JEE | CONTINUITY AND DIFFERENTIABILITY

हिन्दी

Differentiate \( y = e^{\sin^{-1} x} \) with respect to \( x \).

  1. A
    \( \frac{1}{\sqrt{1 - x^2}} \)
  2. B
    \( \frac{-e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \)
  3. C
    \( \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \)
  4. D
    \( e^{\sin^{-1} x} \sqrt{1 - x^2} \)
View Answer & Solution

Correct answer · Option C

\( \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \)

Explanation

Applying the chain rule, \( \frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}} \).
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