MCQ
If $e^x+e^y=e^{x+y}$, then $\frac{d y}{d x}$ is
  • A
    $e^{y-x}$
  • B
    $e^{y+x}$
  • C
    $-e^{y-x}$
  • D
    $2 e^{x-y}$

Answer

We have, $e^x+e^y=e^{x+y}$
$
\Rightarrow e^{-y}+e^{-x}=1
$
Differentiating w.r.t. $x$, we get
$
-e^{-y} \frac{d y}{d x}-e^{-x}=0 \Rightarrow \frac{d y}{d x}=-e^{y-x}
$

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