Question

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

Hint:

For equations involving logarithms and exponentials, taking the natural logarithm can simplify differentiation.

Answer 1

Step 1: Take the natural logarithm on both sides.
Given: \[ x = e^{\frac{x}{y}} \] Taking \( \log \) on both sides: \[ \log x = xy. \] Step 2: Differentiate both sides using implicit differentiation.
Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} (\log x) = \frac{d}{dx} (xy). \] Using derivative rules: \[ \frac{1}{x} \cdot \frac{dx}{dx} = x \frac{dy}{dx} + y \frac{dx}{dx}. \] Since \( \frac{dx}{dx} = 1 \), we get: \[ \frac{1}{x} = x \frac{dy}{dx} + y. \] Step 3: Solve for \( \frac{dy}{dx} \).
Rearrange the equation: \[ \frac{1}{x} - y = x \frac{dy}{dx}. \] Dividing by \( x \): \[ \frac{dy}{dx} = \frac{\frac{1}{x} - y}{x}. \] Rewriting in simplified form: \[ \frac{dy}{dx} = \frac{x - y}{x \log x}. \] Thus, the required result is proved.