# How do you differentiate f(x)= 2x*sinx*cosx?

Aug 6, 2015

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

#### Explanation:

Use the product rule:

$f = g h k$ => $f ' = g ' h k + g h ' k + g h k '$

With:
$g = 2 x$ => $g ' = 2 x$
$h = \sin x$ => $h ' = \cos x$
$k = \cos x$ => $k ' = - \sin x$

We then have:
$f ' \left(x\right) = 2 \sin x \cos x + 2 x {\cos}^{2} x - 2 x {\sin}^{2} x$

Aug 6, 2015

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

#### Explanation:

$f ' \left(x\right) = \left(2 x\right) ' \cdot \left(\sin \left(x\right) \cdot \cos \left(x\right)\right) + 2 x \cdot \left(\sin \left(x\right) \cdot \cos \left(x\right)\right) '$

$\left(2 x\right) ' = 2$

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

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