# How doyou find the derivative of 2sinxcosx?

Apr 17, 2018

$\frac{d}{\mathrm{dx}} 2 \sin x \cos x = 2 \cos 2 x$

#### Explanation:

$2 \sin x \cos x = \sin 2 x$

So

$\frac{d}{\mathrm{dx}} 2 \sin x \cos x = \frac{d}{\mathrm{dx}} \sin 2 x = 2 \cos 2 x$

Apr 17, 2018

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

#### Explanation:

Using the linearity of the derivative and product rule this would be:

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

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

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

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

but you can also use the trigonometric identity:

$2 \sin x \cos x = \sin 2 x$

and then:

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