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?
Best Answer
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 $$ \begin{split} \frac{\partial f(y+\alpha\eta,\space y'+\alpha\eta',\space x)}{\partial\alpha}{\color{red}{\Bigg|_{\alpha=0}}} & \equiv \frac{\partial f(x, y+\alpha\eta, z+\alpha\eta')}{\partial\alpha}\\ & = \lim_{\alpha \to 0} \frac{ f(x, y+\alpha\eta, z+\alpha\eta')-f(x,y,z)}{\alpha}\\ & =\lim_{\alpha \to 0} \left[\frac{ f(x, y+\alpha\eta, z+\alpha\eta')- f(x, y, z+\alpha\eta')+ f(x, y, z+\alpha\eta')- f(x,y,z)}{\alpha}\,\right]\\ & =\lim_{\alpha \to 0} \frac{ f(x, y+\alpha\eta, z+\alpha\eta')- f(x, y, z+\alpha\eta')}{\alpha} + \lim_{\alpha \to 0}\frac{f(x, y, z+\alpha\eta')-f(x,y,z)}{\alpha}\\ & = \eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial z} \\ & \equiv \eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'} \end{split} $$