# What is the derivative of sin^2(x) * cos^2(x)?

Nov 7, 2016

By the chain and product rules, $\frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(x\right) \cdot {\cos}^{2} \left(x\right)\right) = - 2 \cos \left(x\right) \sin \left(x\right) \left({\sin}^{2} \left(x\right) - {\cos}^{2} \left(x\right)\right)$

#### Explanation:

In order to evaluate this derivative, we need to use both the product and chain rules.

Starting with ${\sin}^{2} \left(x\right)$ and using the chain rule to take its derivative, we have:

$2 \sin \left(x\right) \cdot \cos \left(x\right)$

Now, by the product rule, we multiply this by our second term, ${\cos}^{2} \left(x\right)$, so the left side of the derivative is

$2 \sin \left(x\right) \cdot \cos \left(x\right) \cdot {\cos}^{2} \left(x\right)$ or $2 {\cos}^{3} \left(x\right) \sin \left(x\right)$

Now for the right side, we use the chain rule to take the derivative of ${\cos}^{2} \left(x\right)$:

$2 \cos \left(x\right) \cdot \left(- \sin x\right)$ or $- 2 \cos \left(x\right) \sin \left(x\right)$

Similarly to with the left side, we now multiply this by our ${\sin}^{x}$ term, so the right side of the derivative is

$- 2 \cos \left(x\right) \sin \left(x\right) \cdot {\sin}^{2} \left(x\right)$ or $- 2 {\sin}^{3} \left(x\right) \cos \left(x\right)$

Continuing with the product rule, we add the left- and right-hand derivatives we calculated above together, so our final answer is:

2cos^3(x)sin(x)-2sin^3(x)cos(x))

This can be simplified in several ways, but one simplified version of the derivative may be:

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

Nov 19, 2016

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

#### Explanation:

Another method, using the chain rule along with the trigonometric identity $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

${\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right) = {\left(2 \sin \left(x\right) \cos \left(x\right)\right)}^{2} / 4 = {\sin}^{2} \frac{2 x}{4}$

$\implies \frac{d}{\mathrm{dx}} {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right) = \frac{d}{\mathrm{dx}} {\sin}^{2} \frac{2 x}{4}$

$= \frac{1}{4} \frac{d}{\mathrm{dx}} {\sin}^{2} \left(2 x\right)$

$= \frac{1}{4} \cdot 2 \sin \left(2 x\right) \left(\frac{d}{\mathrm{dx}} \sin \left(2 x\right)\right)$

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

$= \frac{\sin \left(2 x\right) \cos \left(2 x\right)}{2} \cdot 2$

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