# How do you find the derivative of cos2x-5cos^2x?

May 4, 2018

$3 \sin 2 x$

#### Explanation:

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

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

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

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

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

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\cos 2 x - 5 {\cos}^{2} x\right)$

$= - 2 \sin 2 x + 5 \sin 2 x = 3 \sin 2 x$

May 4, 2018

$3 \sin 2 x$

#### Explanation:

Given: $\frac{d}{\mathrm{dx}} \left(\cos 2 x - 5 {\cos}^{2} x\right)$.

$= \frac{d}{\mathrm{dx}} \left(\cos 2 x\right) - \frac{d}{\mathrm{dx}} \left(5 {\cos}^{2} x\right)$

$= \frac{d}{\mathrm{dx}} \left(\cos 2 x\right) - 5 \frac{d}{\mathrm{dx}} \left({\cos}^{2} x\right)$

Using the chain rule,

$\frac{d}{\mathrm{dx}} \left(\cos 2 x\right)$

$= 2 \cdot - \sin 2 x$

$= - 2 \sin 2 x$

For the second part, We find: $\frac{d}{\mathrm{dx}} \left({\cos}^{2} x\right)$.

Let $u = \cos x , \frac{\mathrm{du}}{\mathrm{dx}} = - \sin x$, $\therefore f = {u}^{2} , \frac{\mathrm{df}}{\mathrm{du}} = 2 u$.

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

$= - 2 \cos x \sin x$

$= - \sin 2 x$ $\left(\because 2 \cos x \sin x = \sin 2 x\right)$

So, the whole differential becomes:

$= - 2 \sin 2 x - 5 \left(- \sin 2 x\right)$

$= - 2 \sin 2 x + 5 \sin 2 x$

$= 3 \sin 2 x$