# What is the derivative of cos^4(x)-sin^4(x)?

Aug 30, 2016

$f ' \left(x\right) = - 2 \sin 2 x$

#### Explanation:

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

First let's do some simplification.

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

Since $\left({\cos}^{2} x + {\sin}^{2} x\right) = 1 \to f \left(x\right) = \left({\cos}^{2} x - {\sin}^{2} x\right)$

Applying the identity $\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x$

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

$f ' \left(x\right) = - 2 \sin \left(2 x\right)$ (Chain rule)