To find the derivative of the given function y = cos^3(sin(3x)), we will use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.
Here we have a composition of three functions. The outermost function is f(u) = u^3, the middle function is g(v) = cos(v), and the innermost function is h(x) = sin(3x).
Firstly, let's differentiate the outermost function with respect to u. The outermost function f(u) = u^3. Its derivative f'(u) is 3u^2.
Secondly, differentiate the middle function with respect to v. The middle function g(v) = cos(v). Its derivative g'(v) is -sin(v).
Finally, differentiate the innermost function h(x) = sin(3x) with respect to x. By using the chain rule once again, we obtain h'(x) = 3cos(3x).
Now let's combine these results using the chain rule, we get dy/dx = f'(g(h(x))) * g'(h(x)) * h'(x).
Substitute the derivative of each function into the formula: dy/dx = 3cos^2(sin(3x)) * -sin(sin(3x)) * 3cos(3x).
So the derivative dy/dx = -9sin(sin(3x)) * cos(3x) * cos^2(sin(3x)).
Thus, the derivative of y = cos^3(sin(3x)) is -9sin(sin(3x)) * cos(3x) * cos^2(sin(3x)).