# How do you find the derivative of h(x)= sin2xcos2x?

Jun 6, 2015

We must use the product rule here, which states that, be $y = f \left(x\right) g \left(x\right)$, then $y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Thus, let's just bear in mind the derivatives of these trigonometric functions:

• $y = \sin \left(u\right)$, then $y ' = u ' \cos u$
• $y = \cos \left(u\right)$, then $y ' = - u ' \sin u$

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

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