Question

If \( x = e^{y^2} \), prove that \[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}. \]

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is it{not} the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

Logarithmic differentiation is useful for functions involving exponents and logarithms.

Answer 1

The Correct Option is D

1. Given the equation: \[ x = e^{y^2}. \] Take the natural logarithm of both sides: \[ \log x = y^2. \] 2. Differentiate both sides with respect to \( x \): \[ \frac{1}{x} \cdot \frac{dx}{dx} = 2y \cdot \frac{dy}{dx}. \] 3. Rearrange to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2y} \cdot \frac{1}{x}. \] 4. Substitute \( y^2 = \log x \) (which implies \( y = \sqrt{\log x} \)) into the expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2 \sqrt{\log x}} \cdot \frac{1}{x}. \] 5. Express \( \frac{dy}{dx} \) solely in terms of \( \log x \): \[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}. \]
Proved.