Question

nth derivative of sin2x×sin4x​

Answer 1

Answer:

If n is even:

(-1)^(n-1) [ 2^(n-1) cos 2x - 3*6^(n-1) cos 6x ]

If n is odd:

f^n(x) = [-2^(n-1) sin 2x + 3*6^(n-1) sin 6x]

Or it could be

f^n(x) = [2^(n-1) sin 2x - 3*6^(n-1) sin 6x]

Step-by-step explanation:

Hope this helps