Graduate Some questions about the derivation steps in the Gravitational deflection of light section in Schutz

[MathematicalPhysicist]

TL;DR

My question is referring to some derivation from pages 293-294 of Schutz's second edition (2009) of A First Course in GR.

In the screenshots below there are the equations (11.49) and (11.53).

I don't understand how did he derive equation (11.53) from Eq.(11.49)?
From (11.49) I get: $d\phi/dy= d\phi/du du/dy = (1/b^2-u^2+2Mu^3)^{-1/2}(1+2My)$.
It seems he neglected the $2Mu^3$ since $Mu\ll 1$, so $y\approx u$, but how do we get the added term of $O(M^2u^2)$?

 

[ergospherical]

I would have thought\begin{align*}
\dfrac{d\phi}{dy} = \dfrac{d\phi}{du} \left( \dfrac{dy}{du} \right)^{-1} &= \dfrac{( 1 - 2Mu)^{-1} }{\left( b^{-2} - u^2(1-2Mu) \right)^{1/2}}
\end{align*}then because $y^2 = u^2(1-Mu)^2 \sim u^2(1-2Mu)$ and also $(1-2Mu)^{-1} \sim 1+2Mu + O(M^2u^2) \sim 1 + 2My + O(M^2u^2)$ you get $\mathrm{11.53}$...

 

[MathematicalPhysicist]

I would have thought\begin{align*}
\dfrac{d\phi}{dy} = \dfrac{d\phi}{du} \left( \dfrac{dy}{du} \right)^{-1} &= \dfrac{( 1 - 2Mu)^{-1} }{\left( b^{-2} - u^2(1-2Mu) \right)^{1/2}}
\end{align*}then because $y^2 = u^2(1-Mu)^2 \sim u^2(1-2Mu)$ and also $(1-2Mu)^{-1} \sim 1+2Mu + O(M^2u^2) \sim 1 + 2My + O(M^2u^2)$ you get $\mathrm{11.53}$...

It seems I got it wrong. Thanks for correcting me.