Partial Derivative of an Unknown Function with Multiple Parameters I'm reading John Taylor's Mechanics, and I'm in Chapter 6, Calculus of Variations. At one point when deriving the Euler-Lagrange equation he does a partial derivative of an unknown function f:
$$ \frac{\partial f(y+\alpha\eta,\space y'+\alpha\eta',\space x)}{\partial\alpha} $$
and ends up getting:
$$\eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}$$
through the chain rule, but I don't understand how exactly this happens. Can someone make it more clear how to go about getting this?
 A: The first thing to observe is that it is slightly inexact to obtain the term
$$\eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}$$
as $$ \frac{\partial f(y+\alpha\eta,\space y'+\alpha\eta',\space x)}{\partial\alpha} $$
The right relation between these two terms is the following one
$$
\frac{\partial f(y+\alpha\eta,\space y'+\alpha\eta',\space x)}{\partial\alpha}{\color{red}{\Bigg|_{\alpha=0}}} = \eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}
$$
Said that, now everything follows in the standard way. To see this, let's "mask" the arguments of $f$ as follows
$$
\begin{eqnarray}
x & \mapsto & x, \\
y & \mapsto & y,\\
y^\prime & \mapsto & z.
\end{eqnarray}
$$
We thus can write
$$
f(y,y^\prime, x)\equiv f(y,z,x)\equiv f(x,y,z),
$$
having reordered the arguments of $f$ just for the sake of simplicity. Then
$$
\begin{split}
\frac{\partial f(y+\alpha\eta,\space y'+\alpha\eta',\space x)}{\partial\alpha}{\color{red}{\Bigg|_{\alpha=0}}} & \equiv \frac{\partial f(x, y+\alpha\eta, z+\alpha\eta')}{\partial\alpha}\\
& = \lim_{\alpha \to 0} \frac{ f(x, y+\alpha\eta, z+\alpha\eta')-f(x,y,z)}{\alpha}\\
& =\lim_{\alpha \to 0} \left[\frac{ f(x, y+\alpha\eta, z+\alpha\eta')- f(x, y, z+\alpha\eta')+ f(x, y, z+\alpha\eta')- f(x,y,z)}{\alpha}\,\right]\\
& =\lim_{\alpha \to 0} \frac{ f(x, y+\alpha\eta, z+\alpha\eta')- f(x, y, z+\alpha\eta')}{\alpha} + \lim_{\alpha \to 0}\frac{f(x, y, z+\alpha\eta')-f(x,y,z)}{\alpha}\\
& = \eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial z} \\
& \equiv \eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}
\end{split}
$$
