Why does covariant dot contra variant basis equal 1 I'm reading up on tensors.  Page 18 of a NASA document has a line stating
$$\mathbf{e^{(1)}}\cdot \mathbf{e_{(1)}}=\left(\frac{\partial \mathbf{r}}{\partial u} \right) \cdot \left( \nabla u \right) \tag{1}$$
$$= 
\left(\frac{\partial x}{\partial u} \right) \left( \frac{\partial u}{\partial x}\right)+
\left(\frac{\partial y}{\partial u} \right) \left( \frac{\partial u}{\partial y}\right)
+
\left(\frac{\partial z}{\partial u} \right) \left( \frac{\partial u}{\partial z}\right) \tag{2}$$
$$=\frac{\partial u}{\partial u}=1 \tag{3}$$
Where the LHS of (1) states the dot product of the contra variant and covariant bases.
I've seen (2)-(3) in Griffiths' Introduction to Electrodynamics as well.
Question:
I understand:


*

*the reasoning for equation LHS and RHS in (1).

*the formulation of (2), through dot product of RHS of (1)

*$\frac{\partial u}{\partial u}=1$


I do not understand why (2) leads to (3).
To me, refactoring (2) looks like $\frac{\partial u}{\partial u}\left(\left(\frac{\partial x}{\partial x} \right)+
\left(\frac{\partial y}{\partial y} \right)
+
\left(\frac{\partial z}{\partial z} \right) \right) =\frac{\partial u}{\partial u}\left( 1+1+1\right)=3\frac{\partial u}{\partial u}=3 \neq \frac{\partial u}{\partial u}=1$ 
Please help a confused physicist.
 A: Firstly, you can't just push the parts of a (partial) derivative $\partial u/\partial x$ around as you like: $\partial u/\partial x$ is one quantity, that happens to be composed of a number of symbols. This is clearer if you write it as $u_x$ or $u_{,x}$.
The next problem is that while for total derivatives or functions of one variable
$$ \frac{dy}{dx} \frac{dx}{dy} = 1, $$
the same is not the case for partial derivatives, as the following example will illustrate: let
$$ u = x+y, \qquad v=x-y. $$
Then inverting gives
$$ x = \frac{u+v}{2} \qquad y = \frac{u-v}{2}. $$
So,
$$ \frac{\partial u}{\partial x} = 1 \qquad \frac{\partial u}{\partial y} = 1 \\
\frac{\partial x}{\partial u} = \frac{1}{2} \qquad \frac{\partial y}{\partial u} = \frac{1}{2},
$$
and therefore
$$ \frac{\partial u}{\partial x} \frac{\partial x}{\partial u} = \frac{1}{2} \qquad \frac{\partial u}{\partial y} \frac{\partial y}{\partial u} = \frac{1}{2}, $$
so the sum is $1$, but the products are not $1$ individually.
A further instructive example is the even simpler-looking example
$$ u = x + ay, \qquad v = y. $$
Then
$$ x = u-av \qquad y = v. $$
So $ \frac{\partial v}{\partial x} = 0 $, but
$$ \frac{\partial x}{\partial v} = -a. $$
This tells us two interesting things: firstly, what we already know about the product not being $1$. And secondly, that although $y=v$, $\partial x/\partial y = 0 $ ($x$ and $y$ are meant to be independent to start with, after all!), so $ \frac{\partial x}{\partial v} \neq \frac{\partial x}{\partial y} $. What does this mean? It means that the partial derivative very much depends on what is being held constant, in addition to what is allowed to vary.
In the case of $\frac{\partial x}{\partial y}$, the coordinates other than $y$ (namely $x$) is being held constant, so of course this gives zero. But in $\frac{\partial x}{\partial v}$, $u$ is being held constant, and since this is not the same as $x$, the answer is different. This is one of the most difficult, and most important, things to understand about partial derivatives. There's a nice illustration of this on p.190 of Penrose's Road to Reality.
The correct approach is to apply the chain rule for partial derivatives, namely that if $f$ is a function of $x,y,z$, which in turn are functions of $u$ and other variables, then
$$ \frac{\partial f}{\partial u} = \frac{\partial x}{\partial u}\frac{\partial f}{\partial x} + \frac{\partial y}{\partial u}\frac{\partial f}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial f}{\partial z}. $$
If then $f=u$, the result follows.
