Question

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

Hint:

For logarithmic differentiation, always apply the quotient rule carefully and simplify step-by-step.

Answer 1

Step 1: Rewrite the given equation
Given \( x = e^{x/y} \), take the natural logarithm on both sides: \[ \log x = \frac{x}{y}. \] 
Step 2: Express \( y \) in terms of \( x \)
Rearranging: \[ y = \frac{x}{\log x}. \] 
Step 3: Differentiate with respect to \( x \)
Using the quotient rule: 
\[ \frac{dy}{dx} = \frac{(\log x)(1) - x \cdot \frac{1}{x}}{(\log x)^2}. \] Simplify: \[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}. \] 
Step 4: Conclude the result
Thus, \( \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2} \).