##y^’ = -sinx##
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 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 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.