# How do you find the derivative of 2cos^2(x)?

Apr 1, 2018

$- 2 \sin 2 x$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{Given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cos x \times \frac{d}{\mathrm{dx}} \left(\cos x\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - 4 \sin x \cos x = - 2 \sin 2 x$