# What is the derivative of f(x)=cos^2x*cos2x?

Mar 25, 2018

$f ' \left(x\right) = - 2 \sin \left(2 x\right) {\cos}^{2} \left(x\right) - 2 \cos \left(2 x\right) \sin \left(2 x\right)$

#### Explanation:

Recall that the derivative of two multiplied functions, $f \left(x\right) = a \left(x\right) b \left(x\right) ,$ is given by $f ' \left(x\right) = a \left(x\right) b ' \left(x\right) + b \left(x\right) a ' \left(x\right)$

Here, we see

$a \left(x\right) = {\cos}^{2} \left(x\right)$

$a ' \left(x\right) = 2 \cos \left(x\right) \cdot \frac{d}{\mathrm{dx}} \cos \left(x\right) = - 2 \cos x \sin x = - \sin \left(2 x\right)$ (From the identity $\sin \left(2 x\right) = 2 \sin x \cos x$)

$b \left(x\right) = \cos \left(2 x\right)$

$b ' \left(x\right) = - \sin \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x\right) = - 2 \sin \left(2 x\right)$

Thus,

$f ' \left(x\right) = - 2 \sin \left(2 x\right) {\cos}^{2} \left(x\right) - 2 \cos \left(2 x\right) \sin \left(2 x\right)$

Mar 25, 2018

$f ' \left(x\right) = - 8 {\cos}^{3} \left(x\right) \sin \left(x\right) + 2 \cos \left(x\right) \sin \left(x\right)$

#### Explanation:

We have $f \left(x\right) = {\cos}^{2} \left(x\right) \cos \left(2 x\right)$

Let's use the trigonometric identity:

${\cos}^{2} \left(x\right) = \frac{1 + \cos \left(2 x\right)}{2} \Rightarrow \cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right) - 1$

so we can express the trig functions in terms of a common angle $x$

$\Rightarrow f \left(x\right) = {\cos}^{2} \left(x\right) \left(2 {\cos}^{2} \left(x\right) - 1\right)$

$\Rightarrow f \left(x\right) = 2 {\cos}^{4} \left(x\right) - {\cos}^{2} \left(x\right)$

Now we can take the derivative using the chain rule:

$\Rightarrow f ' \left(x\right) = 8 {\cos}^{3} \left(x\right) \left(- \sin \left(x\right)\right) - 2 \cos \left(x\right) \left(- \sin \left(x\right)\right)$

$\Rightarrow f ' \left(x\right) = - 8 {\cos}^{3} \left(x\right) \sin \left(x\right) + 2 \cos \left(x\right) \sin \left(x\right)$