Question

If \(x^y = e^{x-y}\), \(\frac{dy}{dx} = ?\)

• A. \(\frac{\log x}{(1 + \log x)^2}\)
• B. \(\frac{\log x^2}{1 + \log x}\)
• C. \(\left(\frac{\log x}{1 + \log x}\right)^2\)
• D. \(\frac{\log x}{1 + \log x}\)

Hint:

Standard calculus identity: $x^y = e^{x-y} \implies y = \frac{x}{1+\ln x} \implies \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Logarithmic differentiation simplifies implicit exponential-variable relations into explicit rational logarithmic functions.
Key Formula or Approach:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \]

Step 2: Detailed Explanation:
Given the equation:
\[ x^y = e^{x-y} \]
Taking natural logarithm (base \(e\)) on both sides:
\[ \ln(x^y) = \ln(e^{x-y}) \]
\[ y \ln x = (x - y) \ln e = x - y \]
Rearranging to express \(y\) explicitly in terms of \(x\):
\[ y \ln x + y = x \]
\[ y(1 + \ln x) = x \implies y = \frac{x}{1 + \ln x} \]
Differentiating with respect to \(x\) using the quotient rule:
\[ \frac{dy}{dx} = \frac{(1 + \ln x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1 + \ln x)}{(1 + \ln x)^2} \]
\[ \frac{dy}{dx} = \frac{(1 + \ln x)(1) - x\left(\frac{1}{x}\right)}{(1 + \ln x)^2} = \frac{1 + \ln x - 1}{(1 + \ln x)^2} = \frac{\ln x}{(1 + \ln x)^2} \]

Step 3: Final Answer:
Thus, the derivative is \(\frac{\log x}{(1 + \log x)^2}\), matching option (A).