There are two ways. The simpler way would be to recall that csc(x) = 1/sin(x), so cos(x)/csc(x) = cos(x)sin(x) [for all x!=(n)pi (for radians) or x!=(n)180 (for degrees) where n is any integer]. We then use the product rule for differentiation, which says that for f(x) = g(x)h(x), f'(x) = g'(x)h(x) + g(x)h'(x)
In the particular example provided, the restriction on the domain is required because, if x=n*pi (for radians) or x=n*180 (for degrees), then sin(x)=0 -> csc(x) = 1/sin(x) = 1/0. However, the argument can be made that since 1/0=oo, cos(x)/(oo) = 0 = cos(x)sin(x) = cos(x)0 at these points, thus making the domain restriction unnecessary.
Given that we have changed our function into y(x) = cos(x)sin(x), the next step is to find y'(x). Assuming we know the derivatives of the sine and cosine functions, this is relatively straightforward.
y'(x) = cos'(x)sin(x) + cos(x)sin'(x)
y'(x) = -sin(x)sin(x) + cos(x)cos(x)
y'(x) = -sin^2(x) + cos^2(x)
At this point, we could stop, but by using the trigonometric double-angle identities, we find further that...
y'(x) = -sin^2(x) + cos^2(x) = cos^2(x) - sin^2(x) = cos(2x)
Thus, for full simplification, we arrive at...
y'(x) = cos(2x).
