# How do you find the derivative of f(x) = -5 e^{x \cos x} using the chain rule?

Jun 4, 2017

$\frac{d}{\mathrm{dx}} f \left(x\right) = - 5 {e}^{x \cos x} \left(\cos x - x \sin x\right)$

#### Explanation:

Let's not worry about the $- 5$ for now. We'll tack that on at the end since it doesn't affect the derivative process at all.

In order to use chain rule, we need to treat $x \cos x$ like a single variable.

Let $u = x \cos x$. Then:

$\frac{d}{\mathrm{dx}} \left({e}^{x \cos x}\right) = \frac{d}{\mathrm{du}} {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

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

Now, we need product rule to differentiate:

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

You can simplify this if needed, but this is a fairly good stopping point. All we have to do is multiply by $- 5$ since the problem originally multiplied by $- 5$.

$\frac{d}{\mathrm{dx}} f \left(x\right) = - 5 {e}^{x \cos x} \left(\cos x - x \sin x\right)$