# How do you find the derivative of  h(x) = ln(cosh(2x)) ?

Oct 23, 2016

$h ' \left(x\right) = 2 \tanh 2 x$

#### Explanation:

Using the chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

let $y = h \left(x\right) = \ln \left(\cosh \left(2 x\right)\right)$

& $u = \cosh 2 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 \sinh 2 u$

so $y = \ln u \implies \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$

substitute in the chai rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times 2 \sinh 2 x$

$\therefore h ' \left(x\right) = \frac{2 \sinh 2 x}{\cosh 2 x} = 2 \tanh 2 x$