# Question #e54b4

Dec 7, 2017

$\frac{d}{\mathrm{dx}} \left[5 {\cos}^{2} x\right] = - 10 \cos x \sin x$

#### Explanation:

The chain rule states that given the functions $y \left(u \left(x\right)\right)$ and $u \left(x\right)$, $\frac{d}{\mathrm{dx}} \left[y\right] = \frac{d}{\mathrm{du}} y \times \frac{d}{\mathrm{dx}} u$. Let's apply this to our problem.

First use the constant rule ($\frac{d}{\mathrm{dx}} \left[a u\right] = a \times \frac{d}{\mathrm{dx}} u$) to simplify the problem.

$\frac{d}{\mathrm{dx}} \left[5 {\cos}^{2} x\right] = 5 \frac{d}{\mathrm{dx}} \left[{\cos}^{2} x\right]$

Now it's time for the chain rule. Let $u = \cos x$ and let $y = {u}^{2}$.

$5 \frac{d}{\mathrm{dx}} \left[{\cos}^{2} x\right] = 5 \frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = 5 \frac{d}{\mathrm{dx}} \left[y\right]$

By the chain rule,

$\frac{d}{\mathrm{dx}} \left[y\right] = \frac{d}{\mathrm{du}} y \times \frac{d}{\mathrm{dx}} u = \frac{d}{\mathrm{du}} {u}^{2} \times \frac{d}{\mathrm{dx}} \cos x$

Solving these derivatives,

$\frac{d}{\mathrm{du}} {u}^{2} \times \frac{d}{\mathrm{dx}} \cos x = 2 u \times \left(- \sin x\right) = - 2 u \sin x = - 2 \cos x \sin x$

So since $\frac{d}{\mathrm{dx}} \left[y\right] = - 2 \cos x \sin x$, $5 \frac{d}{\mathrm{dx}} \left[y\right] = 5 \left(- 2 \cos x \sin x\right) = - 10 \cos x \sin x$
$\frac{d}{\mathrm{dx}} \left[5 {\cos}^{2} x\right] = - 10 \cos x \sin x$