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.
On
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.
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.