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I am following a Physics textbook where the author has written the following using the chain rule:

$\begin{align} \frac{\partial f(y+\alpha \eta, y'+\alpha\eta', x)}{{\partial\alpha}} &=\eta\frac{\partial f}{\partial y}+\eta '\frac{\partial f}{\partial y'} \end{align}$

Here's how I have gone about it:

$ \begin{align} \frac{\partial f(y+\alpha \eta, y'+\alpha\eta', x)}{{\partial\alpha}} &= \frac{\partial f(y+\alpha \eta, y'+\alpha\eta', x)}{{\partial (y+\alpha \eta)}}\frac{\partial (y+\alpha \eta)}{\partial \alpha} \\& + \frac{\partial f(y+\alpha \eta, y'+\alpha\eta', x)}{{\partial (y'+\alpha \eta')}}\frac{\partial (y'+\alpha \eta')}{\partial \alpha} \\&+ \frac{\partial f(y+\alpha \eta, y'+\alpha\eta', x)}{{\partial x}}\frac{\partial x}{\partial \alpha} \\&= \eta\frac{\partial f}{\partial (y+\alpha \eta)}+\eta '\frac{\partial f}{\partial (y'+\alpha \eta')} + 0 \end{align} $

But how did $\partial (y+\alpha \eta)$ in the denominator become $\partial y$? Likewise for the other denominator $\partial (y'+\alpha \eta')$?

Thanks

Edit:

$\begin{align} \eta' &= \frac{d\eta}{ dx}\\y & = y(x)\end{align} $

Also, $\alpha$ is an independent parameter

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  • $\begingroup$ $\alpha \to 0$? $\endgroup$
    – Gregory
    Commented May 21, 2017 at 19:58
  • $\begingroup$ What do you mean by $\partial(y_\alpha\eta)$ in the first place? $\endgroup$
    – amd
    Commented May 21, 2017 at 19:59
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    $\begingroup$ is this for finding the Euler-Lagrange equations? $\endgroup$
    – Gregory
    Commented May 21, 2017 at 20:16
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    $\begingroup$ Figured. It is a standard way to arrive at it. At the very least $\alpha \ll 1$ is a small parameter and one could neglect its variation immediately (usually with very general conditions such as smoothness, etc.). This is why $\alpha = 0$ soon after this. $\endgroup$
    – Gregory
    Commented May 21, 2017 at 20:18
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    $\begingroup$ Personally, I don't like the notation being used. I'm hesitant to just flat out say yes to this. I will put another approach in the answers $\endgroup$
    – Gregory
    Commented May 21, 2017 at 20:24

2 Answers 2

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I would write $$ F(\alpha)=f(y+\alpha\eta,y'+\alpha\eta',x) $$ Then \begin{align} \frac{dF}{d\alpha}(0)=\Bigl(&\frac{\partial f}{\partial y}(y+\alpha\eta,y'+\alpha\eta',x)\frac{d(y+\alpha\eta)}{d\alpha}+\\ &\frac{\partial f}{\partial y'}(y+\alpha\eta,y'+\alpha\eta',x)\frac{d(y'+\alpha\eta')}{d\alpha}+\\ &\frac{\partial f}{\partial x}(y+\alpha\eta,y'+\alpha\eta',x)\frac{dx}{d\alpha}\Bigr)\Bigr|_{\alpha=0} \end{align}

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  • $\begingroup$ So, are we allowed to put $\alpha =0$ within the differentials as well, like you have done in the denominators? $\endgroup$
    – Vibhu
    Commented May 21, 2017 at 21:09
  • $\begingroup$ @Vibhu: I suppose it is a different notation, I don't like to write, for a function $f(x,y)$, say, $\frac{\partial f}{\partial x_0}$, but I think it is better and clearer to write $\frac{\partial f}{\partial x}(x_0,y)$, where the $x$ in the differential only mean "take the derivative with respect to first variable", and does not mean "where" to evaluate the derivative $\endgroup$ Commented May 21, 2017 at 21:13
  • $\begingroup$ Oh my God, I have never seen this practice earlier! $\endgroup$
    – Vibhu
    Commented May 21, 2017 at 21:20
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Using the chain rule we note that $\frac{\partial}{\partial \alpha} = \frac{\partial y}{\partial \alpha} \frac{\partial}{\partial y} + \frac{\partial y'}{\partial \alpha} \frac{\partial}{\partial y'} = \eta \partial_y + \eta' \partial_{y'}$ \begin{align} \frac{\partial f}{{\partial\alpha}} &=\eta\frac{\partial f}{\partial y}+\eta '\frac{\partial f}{\partial y'} \\ \end{align}

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  • $\begingroup$ This is wrong. $y=y(x)$. Hence, $\frac{\partial y}{\partial \alpha} = 0$ $\endgroup$
    – Vibhu
    Commented May 21, 2017 at 20:36
  • $\begingroup$ It is not wrong, you need to look at what you're actually doing. Taken from the context of E-L equations. We have a functional $\mathcal F[y]$ for which there is a $y = y_0$ which minimizes the functional. To find it we consider perturbations $y = y_0 + \alpha \eta$ and $y = y_0' + \alpha \eta'$ where $\alpha \ll 1$ and $(\eta, \eta') \in L_2$. This may be more formal than you were hoping for though. $\endgroup$
    – Gregory
    Commented May 21, 2017 at 20:40
  • $\begingroup$ Okay then, I will be patient. Maybe it will get clear in future when I have a deeper knowledge of the theory. Thanks! $\endgroup$
    – Vibhu
    Commented May 21, 2017 at 20:43
  • $\begingroup$ I should clarify that $\alpha(\eta, \eta')$ should be $L_2$ close to $y_0$, meaning that for some small $\epsilon$, we have $\int (y_0 - \alpha \eta)^2 \, dx < \epsilon$. But yes, reading a little further should help clarify some of the details (what is meant by close to, etc.) Good luck. $\endgroup$
    – Gregory
    Commented May 21, 2017 at 20:55

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