Question

How do you find the derivative of `y=sin^2x cos^2x`?

Answer 1

`dy/dx=-2sinxcosx(sin^2x-cos^2x)`

Explanation:

Use the product rule:

If `y=f(x)g(x)`, then

`dy/dx=f'(x)g(x)+g'(x)f(x)`

So,

`f(x)=sin^2x`
`g(x)=cos^2x`

Use the chain rule to find both derivatives:

Recall that `d/dx(u^2)=2u * (du)/dx`

`f'(x)=2sinxd/dx(sinx)=2sinxcosx`

`g'(x)=2cosxd/dx(cosx)=-2sinxcosx`

Thus,

`dy/dx=2sinxcosx(cos^2x)-2sinxcosx(sin^2x)`

`=>-2sinxcosx(sin^2x-cos^2x)`

There is the identity that `2sinxcosx=sin2x`, but that identity's more confusing than helpful when simplifying answers.

Answer 2

There is something that makes the answer a lot simpler to find.

Explanation:

You can also remember that `sin(2x) = 2sin(x)cos(x)`, hence a new expression of the function.

`f(x) = sin^2(x)cos^2(x) = sin(x)cos(x)sin(x)cos(x) = (sin(2x)/2)^2 = sin^2(2x)/4` which is a lot easier to derivate (1 square instead of 2).

The derivative of `u^n` is `n*u'u^(n-1)` and the derivative of `sin(2x)` is `2cos(2x)`

So `f'(x) = (4cos(2x)sin(2x))/4 = sin(4x)/2`.

The advantage of those trigonometric identities is for physicists, they can find every piece of information in the wave that this function represents. They're also very useful when you have to find primitives of trigonometric functions.