# What is the derivative of ((sinx)^2)/(1-cosx)?

##y^' = -sinx##

Here's an justifiable copy of how one primal trigonometric unity can niggardly you a suggestive quantity of product on this derivative.

More specifically, you can use the circumstance that

##color(blue)(sin^x + cos^2x = 1 => sin^2x = 1 - cos^2x)##

##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## allure thus be

##y^' = d/dx(1 + cosx) = falsification(green)(-sinx)##

Now, let's suppose that you didn't give-heed-to you could use this unity to facilitate your exercise.

Since your exercise can be written as

##y = f(x)/g(x)##, delay ##g(x)!=0##

you can use the to fuse it by

##color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2##

Using this way, the derivative of ##y## would be

##y^' = ([d/dx(sin^2x)] * (1 - cosx) - sin^2x * d/dx(1-cosx))/(1-cosx)^2##

You can discover ##d/dx(sin^2x)## by using the to transcribe

##sin^2x = u^2##, delay ##u = sinx##

This allure 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 tail 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 transcribe this as

##f^' = (-sinx(-2cosx + 2cos^2x + sin^2x))/(1-cosx)^2##

At this aim, you could really use the similar unity to transcribe

##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 comprehend that

##color(blue)( (a-b)^2 = a^2 - 2ab + b^2)##

you can transcribe

##1 - 2cosx + cos^2x = (1 - cosx)^2##

Finally, arrest this tail into the look for ##f^'## to get

##f^' = (-sinx * falsification(red)(cancel(color(black)((1-cosx)^2))))/color(red)(cancel(color(black)((1-cosx)^2))) = falsification(green)(-sinx)##

So there you accept it, the similar remainder, but suggestively over product to do.