# What is the derivative of e^x/(1+e^y)?

Aug 30, 2015

$\frac{d}{\mathrm{dx}} \left({e}^{x} / \left(1 + {e}^{y}\right)\right) = \frac{{e}^{x} \left(1 + {e}^{y} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{1 + {e}^{y}} ^ 2$

$\frac{d}{\mathrm{dt}} \left({e}^{x} / \left(1 + {e}^{y}\right)\right) = \frac{{e}^{x} \left(\frac{\mathrm{dx}}{\mathrm{dt}} + \frac{\mathrm{dx}}{\mathrm{dt}} {e}^{y} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dt}}\right)}{1 + {e}^{y}} ^ 2$

#### Explanation:

Assuming that we are differentiation with respect to $x$ and that $y$ is some function of $x$, we can differentiate implicitly using the quotient rule:

$\frac{d}{\mathrm{dx}} \left({e}^{x} / \left(1 + {e}^{y}\right)\right) = \frac{{e}^{x} \left(1 + {e}^{y}\right) - {e}^{x} \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{1 + {e}^{y}} ^ 2$

If we are differentiating with respect to some $t$ and assuming that $x$ and $y$ are functions of $t$, we get:

$\frac{d}{\mathrm{dt}} \left({e}^{x} / \left(1 + {e}^{y}\right)\right) = \frac{{e}^{x} \frac{\mathrm{dx}}{\mathrm{dt}} \left(1 + {e}^{y}\right) - {e}^{x} \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dt}}\right)}{1 + {e}^{y}} ^ 2$

$= \frac{{e}^{x} \left[\frac{\mathrm{dx}}{\mathrm{dt}} \left(1 + {e}^{y}\right) - {e}^{x} \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dt}}\right)\right]}{1 + {e}^{y}} ^ 2$

$= \frac{{e}^{x} \left(\frac{\mathrm{dx}}{\mathrm{dt}} + \frac{\mathrm{dx}}{\mathrm{dt}} {e}^{y} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dt}}\right)}{1 + {e}^{y}} ^ 2$