Question

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

Answer 1

By the chain and product rules, `d/dx(sin^2(x)*cos^2(x)) =-2cos(x)sin(x)(sin^2(x)-cos^2(x))`

Explanation:

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

Starting with `sin^2(x)` and using the chain rule to take its derivative, we have:

`2sin(x)*cos(x)`

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

`2sin(x)*cos(x)*cos^2(x)` or `2cos^3(x)sin(x)`

Now for the right side, we use the chain rule to take the derivative of `cos^2(x)`:

`2cos(x)*(-sinx)` or `-2cos(x)sin(x)`

Similarly to with the left side, we now multiply this by our `sin^(x)` term, so the right side of the derivative is

`-2cos(x)sin(x)*sin^2(x)` or `-2sin^3(x)cos(x)`

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:

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

Answer 2

`d/dxsin^2(x)cos^2(x)=sin(2x)cos(2x)`

Explanation:

Another method, using the chain rule along with the trigonometric identity `sin(2x) = 2sin(x)cos(x)`

`sin^2(x)cos^2(x) = (2sin(x)cos(x))^2/4 = sin^2(2x)/4`

`=> d/dxsin^2(x)cos^2(x) = d/dxsin^2(2x)/4`

`=1/4d/dxsin^2(2x)`

`=1/4*2sin(2x)(d/dxsin(2x))`

`=sin(2x)/2*cos(2x)(d/dx2x)`

`=(sin(2x)cos(2x))/2*2`

`=sin(2x)cos(2x)`