# What is the derivative of y=(e^(2x))/(e^(2x)+1)?

Apr 9, 2018

#### Explanation:

$y = {e}^{2 x} / \left({e}^{2 x} + 1\right)$

$= \frac{{e}^{2 x} + 1 - 1}{{e}^{2 x} + 1}$

$= \frac{{e}^{2 x} + 1}{{e}^{2 x} + 1} - \frac{1}{{e}^{2 x} + 1}$

$= 1 - \frac{1}{{e}^{2 x} + 1}$

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

$y ' = {\left({e}^{2 x} + 1\right)}^{-} 2 \times {e}^{2 x} \times 2$

$y ' = \frac{2 {e}^{2 x}}{{e}^{2 x} + 1} ^ 2$

Apr 9, 2018

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

#### Explanation:

We have,

$y = \frac{{e}^{2 x}}{{e}^{2 x} + 1}$

$\text{Using "color(green)"Quotient Rule}$ ,

(dy)/(dx)=((e^(2x)+1)d/(dx)(e^(2x))-e^(2x)d/(dx) (e^(2x)+1))/((e^(2x)+1)^2)

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

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

$= \frac{2 {e}^{2 x}}{{\left({e}^{2 x} + 1\right)}^{2}}$