If n is an integer such that n>1 and f(x)=(sin(n*x))^n, what is f'(x)?

Let us denote sin(nx) = u(x), where u is a function of x. The equation is now therefore f(x) =(u(x))^n.

For simplicity, we will write that as f(x) = u^n

By the chain rule, we know that f'(x) = df/dx = (df/du)*(du/dx).

Firstly computing df/du, we find df/du = n*u^(n-1)

Now we need to find du/dx. Since u = sin(nx) , du/dx = ncos(nx).

Therefore, our answer is f'(x) = (df/du)(du/dx) = nu^(n-1)*ncos(nx), 

subbing in u = sin(nx) yields the final answer:

f'(x) = n(sin(nx))^(n-1)*ncos(nx)