# What is the second derivative of  e^x(cosx-sinx) ?

Apr 3, 2017

And doing the same again we get:

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} {e}^{x} \left(\cos x - \sin x\right) = - 2 {e}^{x} \left(\cos x + \sin x\right)$

#### Explanation:

Before tackling the problem I will derive two useful derivative results using the product rule:

$\frac{d}{\mathrm{dx}} {e}^{x} \sin x = {e}^{x} \frac{d}{\mathrm{dx}} \sin x + \frac{d}{\mathrm{dx}} {e}^{x} \sin x$
$\text{ } = {e}^{x} \cos x + {e}^{x} \sin x$

And:

$\frac{d}{\mathrm{dx}} {e}^{x} \cos x = {e}^{x} \frac{d}{\mathrm{dx}} \cos x + \frac{d}{\mathrm{dx}} {e}^{x} \cos x$
$\text{ } = - {e}^{x} \sin x + {e}^{x} \cos x$
$\text{ } = {e}^{x} \cos x - {e}^{x} \sin x$

Let:

$y = {e}^{x} \left(\cos x - \sin x\right)$
$\setminus \setminus = {e}^{x} \cos x - {e}^{x} \sin x$

Then using the above derivative results we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({e}^{x} \cos x - {e}^{x} \sin x\right) - \left({e}^{x} \cos x + {e}^{x} \sin x\right)$
$\text{ } = {e}^{x} \cos x - {e}^{x} \sin x - {e}^{x} \cos x - {e}^{x} \sin x$
$\text{ } = - 2 {e}^{x} \sin x$

And doing the same again we get:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 2 \left({e}^{x} \cos x + {e}^{x} \sin x\right)$
$\text{ } = - 2 {e}^{x} \left(\cos x + \sin x\right)$