How does one evaluate this function of several variables? In deriving the Euler-Lagrange equation, one step involves evaluating this:
$$\frac{\partial f(y(x)+\alpha\eta(x), y'(x)+\alpha\eta'(x), x)}{\partial \alpha}$$
(this is from pg. 220 of 'Classical Mechanics' John Taylor). The result is $$\eta \frac{\partial f}{\partial y}+\eta' \frac{\partial f}{\partial y'}$$ but I'm unsure of the intermediate steps. Any pointers would be appreciated, thank you.
 A: This is simply the chain rule. We have $$\frac{\partial f\Big(u(\alpha), v(\alpha)\Big)}{\partial \alpha} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial \alpha} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial \alpha}$$
Put in your $u,v$ and you should be able to work it out.
A: So, functional notation is a bit confusing, but there's method to the madness. Let's figure it out:
There's a function on three parameters, $f(z_1,z_2,z_3)$. By default, the parameters are simply
$$
z_1=y\qquad 
z_2=y'\qquad 
z_3=x
$$
So, we normally write the function as $f(y,y',x)$.
We then change the parameters of this function. We now have
$$
z_1=y+\alpha\eta(x)\qquad 
z_2=y'+\alpha\eta'(x)\qquad 
z_3=x
$$
The chain rule gives us the following result:
$$
\begin{align}
\frac{\partial f}{\partial \alpha}&= 
\frac{\partial f}{\partial z_1} \frac{\partial z_1}{\partial \alpha} +
\frac{\partial f}{\partial z_2} \frac{\partial z_2}{\partial \alpha} +
\frac{\partial f}{\partial z_3} \frac{\partial z_3}{\partial \alpha}
\\&=
\frac{\partial f}{\partial z_1} \eta(x) +
\frac{\partial f}{\partial z_2} \eta'(x) +
\frac{\partial f}{\partial z_3} \cdot 0
\\&=
\frac{\partial f}{\partial z_1} \eta +
\frac{\partial f}{\partial z_2} \eta'
\end{align}
$$
Now, notice that
$$
\begin{align}
\frac{\partial f}{\partial y} &= 
\frac{\partial f}{\partial z_1} \cdot 1 
+\frac{\partial f}{\partial z_2}\cdot 0
+ \frac{\partial f}{\partial z_3}\cdot 0 \\&= 
\frac{\partial f}{\partial z_1}
\end{align}
$$
and, along similar lines, $\displaystyle \frac{\partial f}{\partial z_2} = \frac{\partial f}{\partial y'}$.  We can substitute into the previous equation to get:
$$
\frac{\partial f}{\partial \alpha} = 
\frac{\partial f}{\partial y} \eta +
\frac{\partial f}{\partial y'} \eta'
$$
I hope that clears up something (that I certainly found) confusing.
