# How do you find the derivative of  cos^7(e^x) using the chain rule?

Nov 16, 2015

$- 7 {e}^{x} {\cos}^{6} \left({e}^{x}\right) \sin \left({e}^{x}\right)$

#### Explanation:

The overriding issue in this scenario is the fact that the $\cos$ function is to the $7 t h$ power.

With the chain rule, we can say that $f ' \left(x\right) = 7 {\cos}^{6} \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left[\cos \left({e}^{x}\right)\right]$.

We can use chain rule again to determine $\frac{d}{\mathrm{dx}} \left[\cos \left({e}^{x}\right)\right]$.
$\frac{d}{\mathrm{dx}} \left[\cos \left({e}^{x}\right)\right] = - \sin \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = - {e}^{x} \sin \left({e}^{x}\right)$

We can plug this back into our equation from earlier:
f'(x)=7cos^6(e^x)*-e^xsin(e^x)=color(blue)(-7e^xcos^6(e^x)sin(e^x)