Here's an excellent example of how one simple trigonometric identity can spare you a significant amount of work on this derivative.
More specifically, you can use the fact that
color(blue)(sin^x + cos^2x = 1 => sin^2x = 1 - cos^2x)
to rewrite your function as
y = (1 - cos^2x)/(1-cosx) = (color(red)(cancel(color(black)((1-cosx)))) * (1 + cosx))/(color(red)(cancel(color(black)((1-cosx))))) = 1 + cosx
The derivative of y will thus be
y^' = d/dx(1 + cosx) = color(green)(-sinx)
Now, let's assume that you didn't notice you could use this identity to simplify your function.
Since your function can be written as
y = f(x)/g(x), with g(x)!=0
you can use the quotient rule to differentiate it by
color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2
Using this approach, the derivative of y would be
y^' = ([d/dx(sin^2x)] * (1 - cosx) - sin^2x * d/dx(1-cosx))/(1-cosx)^2
You can find d/dx(sin^2x) by using the chain rule to write
sin^2x = u^2, with u = sinx
This will get you
d/dx(u^2) = d/(du)u^2 * d/dx(u)
d/dx(u^2) = 2u * d/dx(sinx)
d/dx(sin^2x) = 2sinx * cosx
Take this back to your target derivative to get
f^' = (2 * sinx * cosx * (1-cosx) - sin^2x * sinx)/(1-cosx)^2
f^' = (2sinxcosx - 2sinxcos^2x - sin^3x)/(1-cosx)^2
You can write this as
f^' = (-sinx(-2cosx + 2cos^2x + sin^2x))/(1-cosx)^2
At this point, you could actually use the same identity to write
f^' = (-sinx * (-2cosx + cos^2x + overbrace(cos^2x + sin^2x)^(color(red)("=1"))))/(1-cosx)^2
f^' = (-sinx * (1 - 2cosx + cos^2x))/(1-cosx)^2
Since you know that
color(blue)( (a-b)^2 = a^2 - 2ab + b^2)
you can write
1 - 2cosx + cos^2x = (1 - cosx)^2
Finally, plug this back into the expression for f^' to get
f^' = (-sinx * color(red)(cancel(color(black)((1-cosx)^2))))/color(red)(cancel(color(black)((1-cosx)^2))) = color(green)(-sinx)
So there you have it, the same result, but significantly more work to do.
