I need help understanding this derivation of the geodesic equation. I'm having a hard time seeing what 'standard techniques' from calculus yield equality $(7.10)$?
The proof is : 
This is from Schaum Series' Tensor Calculus by David C. Kay. We're using this text for a first/introductory course of differential geometry. I know this isn't the appropriate place to vent, but I don't think this is the right text to be used for our class. When we very going through this derivation in class, our instructor said of (7.10) that this is 'trivial' and although very confused about it at the time in class, like everyone else, I didn't wanted to ask the teacher to explain the trivial stuff because you obviously don't want to look stupid in front of everyone else.
 A: The "trivial result" the author is using is the Euler-Lagrange equations (plus he seems to have factored out a $1/2$ of the whole thing without telling you). As you can check elsewhere, the Euler-Lagrange equations tell us that at a critical point of the functional $L(u) = \displaystyle\int_a^b f(t,u(t),\dot u(t))dt$ we must have
$$\int_a^b \left(\frac{\partial f}{\partial u} - \frac d{dt}\frac{\partial f}{\partial \dot u}\right)dt.$$
(You should understand this separately before putting in the specific $w(t,u)$.)
Folding in the chain rule and using the fact that 
$$\frac{\partial X^i}{\partial u} = (t-a)(b-t)\phi^i(t)$$
should give you the result.
(I've never been particularly fond of that text, myself. )
A: In the following I shall present a derivation of Kay's equation 7.10.  The derivation shall proceed from the basic principles specified by David Kay, that is, "standard calculus techniques including integration by parts."  The technique given in Answer 1 above, though quite abstract, is very much valid.  In fact, Answer 1 prompted me to consult another Schaum's Outline entitled "Lagrangian Dynamics" where, in my copy, the subsection "Certain Techniques in the Calculus of Variations" of Chapter 17 presented an analysis whose problem set up is very similar to David Kay's for the development of equation 7.10.  The derivation I present below is different to that given in Schaum's "Lagrangian Dynamics." It is, I believe, simpler; it really is, as Kay states, based on just "standard calculus techniques and integration by parts".  I shall begin with the following:
$L\left(u\right)=\int\limits_a^b\sqrt{\varepsilon g_{ij} \frac{\partial X^i}{\partial t}\frac{\partial X^j}{\partial t}}\;\;dt\equiv\int\limits_a^b\sqrt{W\left(t,u\right)}\;\;dt$
Just as Kay did, I shall take $\varepsilon=1$ and assume $g_{ij} \frac{\partial X^i}{\partial t}\frac{\partial X^j}{\partial t}\geqslant 0$  (Taking $\varepsilon=1$ implies a positive definite metric).  I shall take $W\left(t,u\right)=g_{ij} \frac{\partial X^i}{\partial t}\frac{\partial X^j}{\partial t}$ and proceed as follows.
$L\left(u\right)$ has a local minimum at $u=0$.  This statement follows from the manner in which the problem has been set up; it follows a technique in the calculus of variations (refer to Dare A. Wells "Lagrangian Dynamics", Schaum's Outline Series, chapter 17, section 17.3, or any other good source on the subject) where the goal is to find a function $y$ that will minimize or maximize a definite integral of a given function of $y$.  I shall proceed by calculating $L^{\prime}\!\left(u\right)$.
$L^{\prime}\!\left(u\right)=\frac{dL}{du}=\frac{d}{du}\int\limits_a^b\sqrt{W\left(t,u\right)}\;\;dt$
$\frac{dL}{du}=\int\limits_a^b\frac{1}{2}\left(W\left(t,u\right)\right)
^{^{-1/2}}\cdot\frac{\partial W}{\partial u}\cdot dt$
Since $W\left(t,u\right)=g_{ij} \frac{\partial X^i}{\partial t}\frac{\partial X^j}{\partial t}$ then,
$\frac{\partial W}{\partial u}=\frac{\partial g_{ij}}{\partial x^{k}}\cdot
\frac{\partial x^{k}}{\partial u}\cdot\frac{\partial X^i}{\partial t}\cdot
\frac{\partial X^j}{\partial t}+
g_{ij}\frac{\partial^2 X^i}{\partial{u}\partial{t}}\cdot
\frac{\partial X^j}{\partial t}+
g_{ij}\frac{\partial X^i}{\partial t}\cdot
\frac{\partial^2 X^j}{\partial{u}\partial{t}}$
Since $x^{i}=X{^i}\left(t,u\right)\equiv x^{i}\left(t\right)+
\left(t-a\right)\left(b-t\right)u\phi^{i}\left(t\right)$ then
$\frac{\partial x^{k}}{\partial u}=\left(t-a\right)\left(b-t\right)
u\phi^{i}\left(t\right)$
NOTE: From this point forward, I shall consider $\phi^{i}$
to be understood to mean $\phi^{i}\left(t\right)$.
I now substitute the expression for $\frac{\partial x^{k}}{\partial u}$ into $\frac{\partial W}{\partial u}$ to obtain the following.
Equation 1:
$\frac{\partial W}{\partial u}=\frac{\partial g_{ij}}{\partial x^{k}}\cdot
\frac{\partial X^i}{\partial t}\cdot\frac{\partial X^j}{\partial t}
\cdot\left(t-a\right)\left(b-t\right)u\phi^{k}+
g_{ij}\frac{\partial^2 X^i}{\partial{u}\partial{t}}\cdot
\frac{\partial X^j}{\partial t}+
g_{ij}\frac{\partial X^i}{\partial t}\cdot
\frac{\partial^2 X^j}{\partial{u}\partial{t}}$
Proceeding...
$x^{i}=X{^i}\left(t,u\right)\equiv x^{i}\left(t\right)+
\left(t-a\right)\left(b-t\right)u\phi^{i}$
$\frac{dx^{i}}{dt}=\frac{\partial X^{i}}{\partial t}=\frac{dx^{i}}{dt}+
\frac{\partial}{\partial t}\left(\left(t-a\right)
\left(b-t\right)u\phi^{i}\right)$
Since $\left(t-a\right)\left(b-t\right)=\left(a+b\right)t-t^{2}-ab$, then
$\frac{\partial X^{i}}{\partial t}=\frac{dx^{i}}{dt}+
\left(a+b-2t\right)u\phi^{i}+
\left(t-a\right)\left(b-t\right)u\frac{d\phi^{i}}{dt}$
$\frac{\partial^2 X^{i}}{\partial{u}\partial{t}}=
\left(a+b-2t\right)\phi^{i}+
\left(t-a\right)\left(b-t\right)\frac{d\phi^{i}}{dt}=
\frac{\partial^2 X^{i}}{\partial{u}\partial{t}}\bigg|_{u=0}$
$\frac{\partial X^{j}}{\partial t}=\frac{dx^{j}}{dt}+
\left(a+b-2t\right)u\phi^{j}+
\left(t-a\right)\left(b-t\right)u\frac{d\phi^{j}}{dt}$
Set $u=0$ so that:
$\frac{\partial X^{j}}{\partial{t}}\bigg|_{u=0}=
\frac{dx^{j}}{dt}$
Therefore,
$\left(\frac{\partial^2 X^{i}}{\partial{u}\partial{t}}
\bigg|_{u=0}\right)\cdot
\left(\frac{\partial X^{j}}{\partial{t}}\bigg|_{u=0}\right)=
\left(a+b-2t\right)\phi^{i}\frac{dx^{j}}{dt}+
\left(t-a\right)\left(b-t\right)\frac{d\phi^{i}}{dt}\frac{dx^{j}}{dt}$
In like manner,
$\frac{\partial X^{i}}{\partial t}=\frac{dx^{i}}{dt}+
\left(a+b-2t\right)u\phi^{i}+
\left(t-a\right)\left(b-t\right)u\frac{d\phi^{i}}{dt}$
Set $u=0$ so that:
$\frac{\partial X^{i}}{\partial{t}}\bigg|_{u=0}=
\frac{dx^{i}}{dt}$
and with
$\frac{\partial^2 X^{j}}{\partial{u}\partial{t}}=
\left(a+b-2t\right)\phi^{j}+
\left(t-a\right)\left(b-t\right)\frac{d\phi^{j}}{dt}=
\frac{\partial^2 X^{j}}{\partial{u}\partial{t}}\bigg|_{u=0}$
$\left(\frac{\partial X^{i}}{\partial{t}}\bigg|_{u=0}\right)\cdot
\left(\frac{\partial^2 X^{j}}{\partial{u}\partial{t}}\bigg|_{u=0}\right)=
\left(a+b-2t\right)\phi^{j}\frac{dx^{i}}{dt}+
\left(t-a\right)\left(b-t\right)\frac{d\phi^{j}}{dt}\frac{dx^{i}}{dt}$
I now form the following sum:
$g_{ij}\left(\frac{\partial^2 X^{i}}{\partial{u}\partial{t}}
\bigg|_{u=0}\right)\cdot
\left(\frac{\partial X^{j}}{\partial{t}}\bigg|_{u=0}\right)+
g_{ij}\left(\frac{\partial X^{i}}{\partial{t}}\bigg|_{u=0}\right)\cdot
\left(\frac{\partial^2 X^{j}}{\partial{u}
\partial{t}}\bigg|_{u=0}\right)=$
$=g_{ij}\left(a+b-2t\right)\phi^{i}\frac{dx^{j}}{dt}+
g_{ij}\left(t-a\right)\left(b-t\right)\frac{d\phi^{i}}{dt}
\frac{dx^{j}}{dt}+g_{ij}\left(a+b-2t\right)\phi^{j}
\frac{dx^{i}}{dt}+g_{ij}\left(t-a\right)\left(b-t\right)
\frac{d\phi^{j}}{dt}\frac{dx^{i}}{dt}=$
$=2g_{ij}\left(a+b-2t\right)\phi^{j}\frac{dx^{i}}{dt}+
2g_{ij}\left(t-a\right)\left(b-t\right)\frac{d\phi^{j}}{dt}
\frac{dx^{i}}{dt}$
$\text{Noting that}\;\;\;
\frac{\partial X^{i}}{\partial{t}}\bigg|_{u=0}=
\frac{dx^{i}}{dt} \;\;\;\text{and}\;\;\;
\frac{\partial X^{j}}{\partial{t}}\bigg|_{u=0}=
\frac{dx^{j}}{dt}\;\text{,}$
the above three expressions are substituted into Equation 1 with $u=0$
to produce the following:
$\frac{\partial W}{\partial u}\bigg|_{u=0}=
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}
\frac{dx^{j}}{dt}\left(t-a\right)\left(b-t\right)\phi^{k}+
2g_{ij}\left(a+b-2t\right)\phi^{j}\frac{dx^{i}}{dt}+
2g_{ij}\left(t-a\right)\left(b-t\right)\frac{d\phi^{j}}{dt}
\frac{dx^{i}}{dt}$
$\text{Noting that}\;\;\;\frac{dL}{du}=\int\limits_a^b
\frac{1}{2}\left(W\left(t,u\right)\right)
^{^{-1/2}}\cdot\frac{\partial W}{\partial u}
\cdot dt\;\;\;\text{, then}$
$\frac{dL}{du_{_{0}}}\equiv\frac{dL}{du}\bigg|_{u=0}=
\int\limits_a^b\frac{1}{2}\left(W\left(t,0\right)\right)
^{^{-1/2}}\cdot\frac{\partial W}{\partial u}\bigg|_{u=0}\cdot dt
\;\;\;\text{and where}$
$W\left(t,0\right)=g_{ij}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}
\equiv W_{_{0}}$
Noting that $\left(a+b-2t\right)=\frac{d}{dt}
\left(\left(t-a\right)\left(b-t\right)\right)$,
I proceed by substituting 
$\frac{\partial W}{\partial u}\bigg|_{u=0}$
into 
$\frac{dL}{du_{_{0}}}$
to produce the following.
$\frac{dL}{du_{_{0}}}=\int\limits_a^b\left[\frac{1}{2}W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}
\left(t-a\right)\left(b-t\right)\phi^{k}+
g_{ij}W_{_{0}}^{^{-1/2}}\frac{d}{dt}\left(\left(t-a\right)\left(b-t\right)
\right)\cdot\phi^{j}\frac{dx^{i}}{dt}+
g_{ij}W_{_{0}}^{^{-1/2}}\left(t-a\right)\left(b-t\right)
\frac{d\phi^{j}}{dt}\frac{dx^{i}}{dt}
\right]dt$
$\frac{dL}{du_{_{0}}}=\int\limits_a^b\left[\frac{1}{2}W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}
\left(t-a\right)\left(b-t\right)\phi^{k}+
g_{ij}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\cdot
\frac{d}{dt}\left(\left(t-a\right)\left(b-t\right)\phi^{j}
\right)\right]dt$
$\frac{dL}{du_{_{0}}}=\int\limits_a^b \frac{1}{2}W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}
\left(t-a\right)\left(b-t\right)\phi^{k}dt+
\int\limits_a^b g_{ij}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\cdot
\frac{d}{dt}\left(\left(t-a\right)\left(b-t\right)\phi^{j}
\right)\cdot dt$
Substitute $k$ for $j$ in the second integral, so that
$\frac{dL}{du_{_{0}}}=\int\limits_a^b \frac{1}{2}W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}
\left(t-a\right)\left(b-t\right)\phi^{k}dt+
\int\limits_a^b g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\cdot
\frac{d}{dt}\left(\left(t-a\right)\left(b-t\right)\phi^{k}
\right)\cdot dt$
INTEGRATION BY PARTS (IBP) ON THE SECOND INTEGRAL
In this section the variable "$u$" does not correspond to
the variable "$u$" else where in this derivation.
$\text{Take}\;\;\;\frac{dv}{dt}=\frac{d}{dt}\left(\left(t-a\right)
\left(b-t\right)\phi^{k}\right)\text{,}\;\;\;\text{then}\;\;\;
v=\left(t-a\right)\left(b-t\right)\phi^{k}\text{.}$
$\text{Take}\;\;\;u=g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}
\text{,}\;\;\;\text{then}$
$\frac{du}{dt}=\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}
\frac{dx^{i}}{dt}\right)\;\;\;\text{so that}\;\;\;
du=\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}
\right)\cdot dt\text{,}\;\;\;\text{hence,}$
$\text{IBP}=uv\bigg|_{a}^{b}+\int\limits_a^b v\;du\text{,}\;\;\;
\text{where it is noted that}\;\;\;uv\bigg|_{a}^{b}=0\;\;\;
\text{produces the following:}$
$\text{IBP}=-\int\limits_a^b \left(t-a\right)\left(b-t\right)\phi^{k}
\cdot\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}
\frac{dx^{i}}{dt}\right)\cdot dt$
rearrangement yields:
$\text{IBP}=-\int\limits_a^b 
\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\right)
\cdot\left(t-a\right)\left(b-t\right)\phi^{k}\cdot dt$
RELATING BACK TO THE SECOND INTEGRAL
$\int\limits_a^b g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\cdot
\frac{d}{dt}\left(\left(t-a\right)\left(b-t\right)\phi^{k}
\right)\cdot dt=-\int\limits_a^b 
\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\right)
\cdot\left(t-a\right)\left(b-t\right)\phi^{k}\cdot dt$
I shall now substitute this into $\frac{dL}{du_{_{0}}}$ to produce:
$\frac{dL}{du_{_{0}}}=\int\limits_a^b \frac{1}{2}W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}
\left(t-a\right)\left(b-t\right)\phi^{k}dt-\int\limits_a^b 
\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\right)
\cdot\left(t-a\right)\left(b-t\right)\phi^{k}\cdot dt$
$\frac{dL}{du_{_{0}}}=\int\limits_a^b \left[\frac{1}{2}W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}-
\frac{d}{dt}\left(g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\right)\right]
\left(t-a\right)\left(b-t\right)\phi^{k} dt$
So, the necessary condition that $\frac{dL}{du_{_{0}}}=0$ implies
that $2\cdot\frac{dL}{du_{_{0}}}=0$.  Hence, when 
$\frac{dL}{du_{_{0}}}=0$
it follows that 
$\frac{dL}{du_{_{0}}}=2\cdot\frac{dL}{du_{_{0}}}=
\int\limits_a^b \left[W_{_{0}}^{^{-1/2}}
\frac{\partial g_{ij}}{\partial x^{k}}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt}-
\frac{d}{dt}\left(2g_{ik}W_{_{0}}^{^{-1/2}}\frac{dx^{i}}{dt}\right)\right]
\left(t-a\right)\left(b-t\right)\phi^{k} dt=0$
Q.E.D.
