Question

If `f(x)=sin2x` then what is the `75^(th)` derivative?

Answer 1

`f^((75))(x) = -2^75 cos2x`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = -37778931862957161709568 cos2x`

Explanation:

If we have:

`f(x) = sin2x`

Then repeatedly differentiating wrt `x` we get:

`\ \ \ \ f(x) = sin2x`
`\ \ \ f'(x) = 2cos2x`
`\ f''(x) = 2(-2sin2x) \ \ = -2^2 sin2x`
`f'''(x) = -2^2 (2cos2x) = -2^3 cos2x`
`f^((4))(x) = 2^4 sin2x`
`vdots`

After which the pattern continues with a cycle of 4, so we can represent the `n^(th)` derivative by:

`f^((n))(x) = { (2^nsin2x, n mod 4 = 0), (2^ncos2x, n mod 4 = 1), (-2^nsin2x, n mod 4 = 2), (-2^ncos2x, n mod 4 = 3) :}`

Now `75 mod 4 = 3`, and so we can deduce that:

`f^((75))(x) = -2^75 cos2x`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = -37778931862957161709568 cos2x`