Question

What is the derivative of? : `sin^2(x/2) \ cos^2(x/2)`

Answer 1

`(sin^2(x/2) * cos^2(x/2))' = (-cos(x/2)+2cos^3(x/2))sin(x/2)`

Explanation:

Note that `sin^2(x/2) = 1-cos^2(x/2)`

So `sin^2(x/2) * cos^2(x/2) = (1-cos^2(x/2)) * cos^2(x/2)`
`= cos^2(x/2)-cos^4(x/2)`
Finding the derivative of this is a little simpler :)

Let's first find the derivative of `cos^2(x/2)` and use the same method for the other part.

To this use the chain rule.
`(dy)/(dx) = (dy)/(du) * (du)/(dx)`
Let's say `u = cos(x/2)`
So `cos^2(x/2) = u^2`

`d/(du)(u^2) = 2u`

However to calculate `(du)/(dx)`, we would have to use the chain rule again. I'm going to skip that, but just note that it's nescesary.

`d/(dx)(cos(x/2)) = -1/2sin(x/2)`

Bringing it all together:
`(dy)/(dx) = 2u * (-1/2sin(x/2))`
`2cos(x/2) * (-1/2sin(x/2)) = -cos(x/2)sin(x/2)`

Using the same method for `cos^4(x/2)`
`u = cos(x/2)`
`d/(du)u^4= 4u^3`
`(du)/(dx) = -1/2sin(x/2)`

So `(dy)/(dx) = 4cos^3(x/2) * (-1/2sin(x/2))`
`= -2cos^3(x/2)sin(x/2)`

Now combine these in the final step.
`-cos(x/2)sin(x/2) - (-2cos^3(x/2)sin(x/2))`
`= (-cos(x/2)+2cos^3(x/2))sin(x/2)`

You could probably reduce this, but I think it's fine :)

Answer 2

`d/dx sin^2(x/2) \ cos^2(x/2) = 1/4sin2x`

Explanation:

Let:

`y = sin^2(x/2) \ cos^2(x/2)`

We could apply the product rule and chain rule but the expression can be significantly simplified using the sine double angle formula:

`sin 2A = 2sinAcosA iff sinAcosA = 1/2 sin 2A`

Thus we can write the initial expression as:

`y = (1/2sinx)^2`
`\ \ = 1/4sin^2x`

So differentiating, using the chain rule, we get:

`dy/dx = (1/4)(2sinx)(cosx)`
`" " = (1/4)(2sinxcosx)`

Again using the sine double angle formula, we have:

`dy/dx = (1/4)sin(2x)`