How do you differentiate f(x) = sin(2x)cos(2x) using the product rule?

Oct 29, 2015

See the explanation section below.

Explanation:

Differentiate $f \left(x\right) = \sin \left(2 x\right) \cos \left(2 x\right)$

using the product rule
I use the order: the derivative of a product of functions is the derivative of the first times the second, plus: the first times the derivative of the second.
$\frac{d}{\mathrm{dx}} \left(F S\right) = F ' S + F S '$

Note that we shall need the chain rule for the derivatives of $\sin \left(2 x\right)$ and $\cos \left(2 x\right)$

You may choose to write $F$, $S$, $F '$ and $S '$ before using the formula. I do not have that habit, so

$f ' \left(x\right) = \left[\cos \left(2 x\right) \left(2\right)\right] \cos \left(2 x\right) + \sin \left(2 x\right) \left[- \sin \left(2 x\right) \left(2\right)\right]$

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

Your teacher/textbook may well prefer to rewrite this answer using $\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$, to get

$= 2 \left[{\cos}^{2} \left(2 x\right) - {\sin}^{2} \left(2 x\right)\right]$

$= 2 \cos \left(4 x\right)$

Although if you're going to do that, I suggest

Rewriting the function

Use $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$ to rewrite $f \left(x\right)$ as

$f \left(x\right) = \sin \left(2 x\right) \cos \left(2 x\right) = \frac{1}{2} \sin \left(4 x\right)$.

Now we do not need the product rule, only the chain rule (which we needed in the other method also).