# How do you differentiate g(x) =e^(1-x)coshx using the product rule?

Jan 8, 2016

$g ' \left(x\right) = {e}^{1 - x} \left(\sinh x - \cosh x\right)$

#### Explanation:

According to the product rule, which will need to be used here,

$g ' \left(x\right) = \cosh x \frac{d}{\mathrm{dx}} \left({e}^{1 - x}\right) + {e}^{1 - x} \frac{d}{\mathrm{dx}} \left(\cosh x\right)$

Find each derivative separately.

The first will require the chain rule: $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$

Thus,

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

The second requires the knowledge of hyperbolic trig functions. One of their basic characteristics is that hyperbolic sine and hyperbolic cosine are the derivatives of one another:

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

(Also, $\frac{d}{\mathrm{dx}} \left(\sinh x\right) = \cosh x$.)

This gives:

$g ' \left(x\right) = - {e}^{1 - x} \cosh x + {e}^{1 - x} \sinh x$

Factored:

$g ' \left(x\right) = {e}^{1 - x} \left(\sinh x - \cosh x\right)$