Find the derivative of e^{x}+e^{y}=e^{x+y}. | vedclass

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Question

(C) Given equation: $e^{x} + e^{y} = e^{x+y}$.Divide both sides by $e^{x+y}$:$\frac{e^{x}}{e^{x+y}} + \frac{e^{y}}{e^{x+y}} = 1$$e^{-y} + e^{-x} = 1$D

Vedclass · Accepted Answer

$-e^{y-x}$

Vedclass · Answer

(C) Given equation: $e^{x} + e^{y} = e^{x+y}$.Divide both sides by $e^{x+y}$:$\frac{e^{x}}{e^{x+y}} + \frac{e^{y}}{e^{x+y}} = 1$$e^{-y} + e^{-x} = 1$Differentiating both sides with respect to $x$:$\frac{d}{dx}(e^{-y}) + \frac{d}{dx}(e^{-x}) = \frac{d}{dx}(1)$$-e^{-y} \frac{dy}{dx} - e^{-x} = 0$$-e^{-y} \frac{dy}{dx} = e^{-x}$$\frac{dy}{dx} = -\frac{e^{-x}}{e^{-y}}$$\frac{dy}{dx} = -e^{y-x}$

Find the derivative of $e^{x}+e^{y}=e^{x+y}$.

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