How do you differentiate f(x)=(1+e^(2x))/(2-e^(2x))?

Feb 25, 2017

$f ' \left(x\right) = \frac{6 {e}^{2 x}}{2 - {e}^{2 x}} ^ 2$

Explanation:

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

$\text{Given "f(x)=(g(x))/(h(x))" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here } g \left(x\right) = 1 + {e}^{2 x} \Rightarrow g ' \left(x\right) = {e}^{2 x} . \frac{d}{\mathrm{dx}} \left(2 x\right) = 2 {e}^{2 x}$

$\text{and } h \left(x\right) = 2 - {e}^{2 x} \Rightarrow h ' \left(x\right) = - {e}^{2 x} . \frac{d}{\mathrm{dx}} \left(2 x\right) = - 2 {e}^{2 x}$

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

$\textcolor{w h i t e}{\times \times \times x} = \frac{4 {e}^{2 x} \cancel{- 2 {e}^{4 x}} + 2 {e}^{2 x} \cancel{+ 2 {e}^{4 x}}}{2 - {e}^{2 x}} ^ 2$

$\textcolor{w h i t e}{\times \times \times x} = \frac{6 {e}^{2 x}}{2 - {e}^{2 x}} ^ 2$