Question

Find F'(x) given F(x) = [Cos(2x)] / [1-Sin(x)]

I have used the quotient rule and gotten to
[ cosxcos2x-(2(1-sinx)(sin2x) ] / [(1-sinx)^2]

But the answer is :

[ 4(sin2x) - 3cos(x) + cos(3x) ]/ 2[(1-sinx)^2]

Answer 1

Since you only have cos(2x) then cos(3x) isn't possible without something like a triple angle formula.
You have:
\[f(x)=\frac{\cos(2x)}{1-\sin(x)}; f'(x)=?\]
Start with the quotient rule and chain rule.
\[f'(x)=\frac{-2\sin(2x)(1-\sin(x))-\cos(2x)(-\cos(x))}{(1-\sin(x))^2}=\frac{-2\sin(2x)+2\sin(2x)\sin(x)+\cos(2x)\cos(x))}{(1-\sin(x))^2}\]
Which is absolutely correct (unless I made an error D:)
You can use trig identities to get it into a different form.

Answer 2

The end of that second line should be "+cos(2x)cos(x)" It doesn't fit on my screen.