# How do you differentiate f(x)=A/(B+Ce^x)?

Apr 18, 2018

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

#### Explanation:

If we want to differentiate with the Quotient Rule:

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

$\frac{d}{\mathrm{dx}} A = 0$, the derivative of a constant is zero.

$\frac{d}{\mathrm{dx}} \left(B + C {e}^{x}\right) = C {e}^{x}$

Thus,

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

Apr 18, 2018

$f ' \left(x\right) = - \setminus \frac{A \cdot C \cdot {e}^{x}}{{\left(B + C \cdot {e}^{x}\right)}^{2}}$

#### Explanation:

Remember:
${\left(\setminus \frac{f \left(x\right)}{g \left(x\right)}\right)}^{'} = \setminus \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$
in this case $f \left(x\right) = A$ the derivative of a constant is zero
and $g \left(x\right) = \left(B + C \cdot {e}^{x}\right)$, it become:
${\left(\setminus \frac{A}{g \left(x\right)}\right)}^{'} = \setminus \frac{- A g ' \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$
${\left(\setminus \frac{A}{g \left(x\right)}\right)}^{'} = \setminus \frac{- A {\left(B + C \cdot {e}^{x}\right)}^{'}}{{\left(B + C \cdot {e}^{x}\right)}^{2}} =$
$= \setminus \frac{- A \cdot C \cdot {e}^{x}}{{\left(B + C \cdot {e}^{x}\right)}^{2}}$