Question 5 (ii) for this paper: http://sixthform.m34maths.com/A-level%20exams/c3/c309ju.pdf
I done the first part, and attempted part (ii), I came up to the second to last step but got stuck. The answer is in the mark scheme (scroll down in pdf file) however I just can't wrap my head around the last bit, an explanation would be helpful. Thank you
You start with $e^{2y}=1+\sin x$, which you can represent as: $$y=\frac12(\ln(1+\sin x))$$ deriving gives: $$y'=\frac12\frac{(1+\sin x)'}{1+\sin x} = \frac{\cos x}{2(1+\sin x)}$$ but we know that $e^{2y}=1+\sin x$, so: $$y'=\frac{\cos x}{2e^{2y}}$$