Question

How do you find the nth derivative of `f(x) = sin^4(x) + sin^2(x)cos^2(x)`?

Answer 1

`f^n(x) = 2^(n-1)(delta_0 + cos (2x + ((n-2)pi)/2)) " " n in ZZ_0^+`

where `delta_0 = {""_(0, " otherwise")^( 1, " " n = 0)`

Explanation:

`sin^2 A + cos^2 A = 1`
`sin A = cos (A-pi/2)`
`cos 2A = 1 - 2sin^2 A`

`f(x) = sin^4 x + sin^2 x cos^2 x = sin^2 x(sin^2 x + cos^2 x) = 1/2(1-cos 2x) = 1/2(1+cos(2x-pi))`

`f'(x) = sin 2x = cos (2x-pi/2)`
`f''(x) = 2cos 2x = 2cos (2x+0pi)`
`f'''(x) = -4sin 2x = -4 cos (2x-pi/2) = 4 cos (2x+(pi)/2)`
`f''''(x) = -8cos 2x = 8 cos (2x+pi)`

`f^n(x) = 2^(n-1)(delta_0 + cos (2x + ((n-2)pi)/2)) " " n in ZZ_0^+`

where `delta_0 = {""_(0, " otherwise")^( 1, " " n = 0)`