Question

How do you find the derivative of `sin^2(x)cos^2(x)`?

Answer 1

I would rewrite `sin^2xcos^2x`

Explanation:

We could use the product, power and chain rules, but I'd prefer either of the two below:

Use the Pythagorean Identity

`f(x) = sin^2xcos^2x = sin^2x(1-sin^2x) = sin^2x-sin^4x`

`f'(x) = 2sinxcosx-4sin^3xcosx`

Rewrite as desired.

Of course, we could have used `f(x) = cos^2x-cos^4x`

to get `f'(x) = -2cosxsinx+4cos^3xsinx`.

Use a Double Angle Identity

`sin2x = 2sinxcosx` so

`f(x) = (sinxcosx)^2 = (1/2sin2x)^2 = 1/4 sin^2 2x`.

So, `f'(x) = 1/4[2sin2xd/dx(sin2x)]`

`= 1/2sin2x cos2x d/dx(2x)`

`= sin2x cos2x`

Rewrite if you wish.