Every time you're dealing with a function that is actually the *quotient* of two other functions, let's call them #f(x)# and #g(x)#, you can differentiate said function by using the quotient rule.

For #y = f(x)/g(x)#, you have

#color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2#, where #g(x) !=0#

In your case, you have

#y = sinx/(1 - cosx)#, with

#{(f(x) = sinx), (g(x) = 1 - cosx) :}#

You alsoe need to remember that

#d/dx(sinx) = cosx#

and that

#d/dx(cosx) = -sinx#

So, the derivatives of #f(x)# and #g(x)# will be

#d/dx(f(x)) = d/dx(sinx) = cosx#

#d/dx(g(x)) = d/dx(1 - cosx)#

#d/dx(g(x)) = d/dx(1) - d/dx(cosx) = 0 - (-sinx) = sinx#

The derivative of your function #y# will thus be

#d/dx(y) = y^' = (cosx * (1 - cosx) - sinx * sinx)/(1 - cosx)^2#

#y^' = (cosx - cos^2x - sin^2x)/(1 - cosx)^2#

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

You can simplify this further by using the fact that

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

to get

#y^' = color(green)((cosx - 1)/(1 - cosx)^2)#