Q:

What is y'' if $\sin y = y + 5x$?

I got $ 5\sin y / (\cos y - 1)^2$ as my answer, but the correct answer was given as $25\sin y / (\cos y - 1)^3$.
My thought process:
Derive the original equation to get $y'\cos y = y' +5$
$$y'(\cos y- 1 ) = 5$$
$$y' = 5/(\cos y - 1) = 5(\cos y - 1)^{-1}$$
$$y'' = 5\frac{\sin y }{ (\cos y - 1)^2}$$
Can anyone explain to me why this is wrong, and how to get the correct answer?

A:

We have $y=\sin y-5x$ so we differentiate : $y'=y'\cos y-5$ which gives $y'=\frac5{\cos y-1}$ and we differentiate again: $y''=\frac{5y'\sin y}{(\cos y-1)^2}=\frac{25\sin y}{(\cos y-1)^3}$.

