find the derivative of cos(sin(x^2-3x))

When you have

y = cos(f(x))

The derivative is:

y' = f'(x). -sin(f(x))

Apply the same thing here, twice.

y = \cos(\sin(x^2-3x))

y' = \frac{d(\sin(x^2 - 3x)}{dx} -\sin(\sin(x^2-3x))

y' = (2x-3)\cos(x^2 - 3x).-\sin(\sin(x^2-3x))

Then you simplify if necessary.

y' = -(2x-3)\cos(x^2 - 3x)\sin(\sin(x^2-3x))