# How do you find the derivative of y = e^cosh(2x)?

Jun 7, 2016

$\frac{d}{\mathrm{dx}} \left({e}^{\cosh \left(2 x\right)}\right) = {e}^{\cosh \left(2 x\right)} \sinh \left(2 x\right) 2$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({e}^{\cosh \left(2 x\right)}\right)$
Applying the chain rule, $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$L e t , \cosh \left(2 x\right) = u$
$= \frac{d}{\mathrm{du}} \left({e}^{u}\right) \frac{d}{\mathrm{dx}} \left(\cosh \left(2 x\right)\right)$
We know,
$\frac{d}{\mathrm{du}} \left({e}^{u}\right) = {e}^{u}$
$\frac{d}{\mathrm{dx}} \left(\cosh \left(2 x\right)\right) = \sinh \left(2 x\right) 2$
So,
$\frac{d}{\mathrm{dx}} \left(\cosh \left(2 x\right)\right) = \sinh \left(2 x\right) 2$

substituted back,$u = \cosh \left(2 x\right)$

we get,

${e}^{\cosh \left(2 x\right)} \sinh \left(2 x\right) 2$