# How do you find the derivative of sqrt(e^(2x) +e^(-2x))?

Jul 22, 2016

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

#### Explanation:

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

We need to use the chain rule as we have $y \left(u \left(x\right)\right)$.

In this case $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$u = {e}^{2 x} + {e}^{- 2 x} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 {e}^{2 x} - 2 {e}^{- 2 x}$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2} {\left({e}^{2 x} + {e}^{- 2 x}\right)}^{- \frac{1}{2}}$ using the power rule

Hence:

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

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