# How do you differentiate f(x) =1/(e^(3-x)+1) using the quotient rule?

Feb 12, 2016

$f ' \left(x\right) = \frac{{e}^{x + 3}}{{e}^{x} + {e}^{3}} ^ 2$

#### Explanation:

According to the quotient rule,

$f ' \left(x\right) = \frac{\left({e}^{3 - x} + 1\right) \frac{d}{\mathrm{dx}} \left(1\right) - 1 \frac{d}{\mathrm{dx}} \left({e}^{3 - x} + 1\right)}{{e}^{3 - x} + 1} ^ 2$

Finding both of the internal derivatives, we see that

$\frac{d}{\mathrm{dx}} \left(1\right) = 0$

To find the second derivative, use the chain rule:

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x} \text{ "=>" } \frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Since $u = 3 - x$,

$\frac{d}{\mathrm{dx}} \left({e}^{3 - x} + 1\right) = {e}^{3 - x} \frac{d}{\mathrm{dx}} \left(3 - x\right) = {e}^{3 - x} \left(- 1\right) = - {e}^{3 - x}$

The $1$ will be neglected since it is a constant.

Plug these both back in:

$f ' \left(x\right) = \frac{\left({e}^{3 - x} + 1\right) \left(0\right) - 1 \left(- {e}^{3 - x}\right)}{{e}^{3 - x} + 1} ^ 2$

$f ' \left(x\right) = {e}^{3 - x} / {\left({e}^{3 - x} + 1\right)}^{2}$

We can try to simplify further:

$f ' \left(x\right) = \frac{{e}^{3 - x}}{{e}^{3} / {e}^{x} + {e}^{x} / {e}^{x}} ^ 2$

$f ' \left(x\right) = \frac{{e}^{3 - x}}{\frac{{\left({e}^{3} + {e}^{x}\right)}^{2}}{e} ^ \left(2 x\right)}$

$f ' \left(x\right) = \frac{{e}^{3 - x} {e}^{2 x}}{{e}^{x} + {e}^{3}} ^ 2$

$f ' \left(x\right) = \frac{{e}^{x + 3}}{{e}^{x} + {e}^{3}} ^ 2$