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

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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)##?

schutz2.pngschutz1.png
schutz1.png
schutz2.png
 
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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}##...
 
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ergospherical said:
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.
 
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