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

##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)##

##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.