# How do you find the derivative of  (7e^x)/(2e^x+1)?

Jun 4, 2016

$\frac{d}{d x} \left[\frac{7 {e}^{x}}{2 {e}^{x} + 1}\right] = \frac{7 {e}^{x} \cdot l n \left(e\right)}{2 {e}^{x} + 1} - \frac{14 {e}^{2 x} \cdot l n \left(e\right)}{2 {e}^{2} + 1} ^ 2$

#### Explanation:

d/(d x)[(7e^x)/(2e^x+1)]=?

$\frac{{\left(7 {e}^{x}\right)}^{'} \cdot \left(2 {e}^{x} + 1\right) - {\left(2 {e}^{x} + 1\right)}^{'} \cdot 7 {e}^{x}}{{\left(2 {e}^{x} + 1\right)}^{2}}$

$= \frac{7 {e}^{x} \cdot l n \left(e\right) \cdot \left(2 {e}^{x} + 1\right) - \left(2 \cdot {e}^{x} \cdot l n \left(e\right) \cdot 7 {e}^{x}\right)}{2 {e}^{x} + 1} ^ 2$

$\frac{d}{d x} \left[\frac{7 {e}^{x}}{2 {e}^{x} + 1}\right] = \frac{7 {e}^{x} \cdot l n \left(e\right)}{2 {e}^{x} + 1} - \frac{7 {e}^{x} \cdot 2 {e}^{x} \cdot l n \left(e\right)}{2 {e}^{2} + 1} ^ 2$

$\frac{d}{d x} \left[\frac{7 {e}^{x}}{2 {e}^{x} + 1}\right] = \frac{7 {e}^{x} \cdot l n \left(e\right)}{2 {e}^{x} + 1} - \frac{14 {e}^{2 x} \cdot l n \left(e\right)}{2 {e}^{x} + 1} ^ 2$