If \(e^x + e^y = e^{x+y}\), then \(\displaystyle\frac{dy}{dx}=\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-e^{y-x}\)
Differentiating both sides with respect to x (implicit differentiation):
\( \displaystyle e^x + e^y\frac{dy}{dx} = e^{x+y}\left(1 + \frac{dy}{dx}\right) \)
\( \displaystyle \left(e^y - e^{x+y}\right)\frac{dy}{dx} = e^{x+y} - e^x \)
From the given relation, \(e^{x+y} - e^x = e^y\) and \(e^y - e^{x+y} = -e^x\). So
\( \displaystyle -e^x\frac{dy}{dx} = e^y \)
\( \displaystyle \frac{dy}{dx} = -\frac{e^y}{e^x} = -e^{y-x} \)
NCERT Exemplar:
Continuity and Differentiability
