What is the #n^(th)# derivative of #sin^2x#?
1 Answer
# 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
# 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
# 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") :} #
