Question

What is the `n^(th)` derivative of `sin^2x`?

Answer 1

`f^((n)) sin^2x = { (sin^2x,n=0), ((-1)^(n/2+1) \ 2^(n-1) \ cos 2x,n gt 0 " even"), ((-1)^((n+1)/2+1) \ 2^(n-1) \ sin 2x,n gt 0 " odd") :}`

Explanation:

We have:

`f(x) = sin^2x`

Differentiating once wrt `x` (using the chain rule), we get the first derivative:

`f'(x) = 2sinxcosx`

At first glance we may suspect that to gain further derivatives we will require the product rule and their form will become increasingly more complex. However we note that:

`sin 2A -= 2sinAcosA`

Allowing us to write the first derivative as:

`f'(x) = sin2x`

So differentiating further times we get:

`f^((2))(x) = 2cos2x`
`f^((3))(x) = -2^2sin2x`
`f^((4))(x) = -2^3cos2x`
`f^((5))(x) = 2^4sin2x`
`f^((6))(x) = 2^5cos2x`
`f^((7))(x) = -2^6sin2x`
`f^((8))(x) = -2^7cos2x`
`vdots`

And a clear pattern is now forming, and the `n^(th)` derivative is:

`f^((n)) sin^2x = { (sin^2x,n=0), ((-1)^(n/2+1) \ 2^(n-1) \ cos 2x,n gt 0 " even"), ((-1)^((n+1)/2+1) \ 2^(n-1) \ sin 2x,n gt 0 " odd") :}`