# How do you differentiate e^x+e^y=e^(x+y)?

Jun 8, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} \left({e}^{y} - 1\right)}{{e}^{y} \left(1 - {e}^{x}\right)}$

#### Explanation:

Differentiating ${e}^{x} + {e}^{y} = {e}^{x + y}$

${e}^{x} + {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + y} \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

or ${e}^{x} + {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + y} + {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}}$

or ${e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + y} - {e}^{x}$

or $\left({e}^{y} - {e}^{x + y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \left({e}^{x + y} - {e}^{x}\right)$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x + y} - {e}^{x}}{{e}^{y} - {e}^{x + y}} = \frac{{e}^{x} \left({e}^{y} - 1\right)}{{e}^{y} \left(1 - {e}^{x}\right)}$

Jul 13, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{y - x}$.

#### Explanation:

${e}^{x} + {e}^{y} = {e}^{x + y} = {e}^{x} \cdot {e}^{y}$.

$\therefore \frac{{e}^{x} + {e}^{y}}{{e}^{x} \cdot {e}^{y}} = 1$.

$\therefore {e}^{x} / \left({e}^{x} \cdot {e}^{y}\right) + {e}^{y} / \left({e}^{x} \cdot {e}^{y}\right) = 1$.

$\therefore {e}^{-} y + {e}^{-} x = 1$.

Diff.ing w.r.t. $x$, we get,

$\frac{d}{\mathrm{dx}} \left({e}^{-} y\right) + \frac{d}{\mathrm{dx}} \left({e}^{-} x\right) = \frac{d}{\mathrm{dx}} \left(1\right)$.

:. d/dy(e^-y)*d/dx(-y)+e^-x*d/dx(-x)=0......[because," the Chain Rule]".

$\therefore {e}^{-} y \left(- \frac{\mathrm{dy}}{\mathrm{dx}}\right) + {e}^{-} x \cdot - 1 = 0$.

$\therefore - \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{-} \frac{x}{e} ^ - y = {e}^{y - x}$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{y - x}$, as Respected Shwetank Mauria has