Question

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

Hint:

Logarithmic differentiation is useful for functions involving exponents and logarithms.

Answer 1

1. Start with the given equation: \[ x = e^{y^2}. \] Take the natural logarithm on 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 for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2y} \cdot \frac{1}{x}. \] 4. Substitute \( y^2 = \log x \) into \( y = \sqrt{\log x} \): \[ \frac{dy}{dx} = \frac{1}{2 \sqrt{\log x}} \cdot \frac{1}{x}. \] 5. Express \( \frac{dy}{dx} \) in terms of \( \log x \): \[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}. \]
Proved.