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