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. Given equation: \[ x = e^{y^2}. \] Taking the natural logarithm of both sides yields: \[ \log x = y^2. \] 2. Differentiating both sides with respect to \( x \): \[ \frac{1}{x} \cdot \frac{dx}{dx} = 2y \cdot \frac{dy}{dx}. \] 3. Rearranging to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{2y} \cdot \frac{1}{x}. \] 4. Substituting \( y^2 = \log x \), 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. Expressing \( \frac{dy}{dx} \) in terms of \( \log x \): \[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}. \]
Proved.