If \( x^y=e^{x-y} \) then \( \displaystyle\frac{dy}{dx}= \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{\log x}{(1+\log x)^2} \)
Taking logarithms on both sides brings y down from the power:
\( \displaystyle y\log x=x-y \)
Collecting the y terms makes y the subject:
\( \displaystyle y(1+\log x)=x \)
\( \displaystyle y=\frac{x}{1+\log x} \)
By the quotient rule:
\( \displaystyle \frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x\cdot\frac{1}{x}}{(1+\log x)^2} \)
\( \displaystyle =\frac{1+\log x-1}{(1+\log x)^2} \)
\( \displaystyle =\frac{\log x}{(1+\log x)^2} \)
Why the other options are incorrect
Option (1) \( y\,x^{y-1}+e^{x-y} \) — Incorrect. The power rule cannot be applied to \( x^y \) when y is a function of x, and dy/dx has not been separated; this is not the derivative.
Option (2) \( \displaystyle\frac{2+\log x}{(1+\log x)^2} \) — Incorrect. This results from adding instead of subtracting in the quotient rule: \( (1+\log x)+x\cdot\frac{1}{x}=2+\log x \).
Option (3) \( \displaystyle\frac{x}{1+\log x} \) — Incorrect. This is y itself, not its derivative.
NCERT Exemplar:
Continuity and Differentiability
