# How do you find the first derivative for f(x)= (sin x)(cos x)?

Apr 11, 2015

Using the multiplication rule for derivatives, which states that the derivative of a product $\left(f \cdot g\right) '$ equals the following:

$\left(f \cdot g\right) ' = f ' \cdot g + f \cdot g '$

Since the derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$, and the derivative of $\cos \left(x\right)$ is $- \sin \left(x\right)$, we have that

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

This would of course be a perfect answer, but I think that noticing that ${\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) = \cos \left(2 x\right)$ would be more elegant.