# An elementary problem with partial derivatives and chain rule

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

• $\alpha \to 0$? Commented May 21, 2017 at 19:58
• What do you mean by $\partial(y_\alpha\eta)$ in the first place?
– amd
Commented May 21, 2017 at 19:59
• is this for finding the Euler-Lagrange equations? Commented May 21, 2017 at 20:16
• 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. Commented May 21, 2017 at 20:18
• 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 Commented May 21, 2017 at 20:24

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}
• So, are we allowed to put $\alpha =0$ within the differentials as well, like you have done in the denominators? Commented May 21, 2017 at 21:09
• @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 Commented May 21, 2017 at 21:13
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}
• This is wrong. $y=y(x)$. Hence, $\frac{\partial y}{\partial \alpha} = 0$ Commented May 21, 2017 at 20:36
• 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. Commented May 21, 2017 at 20:40
• 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. Commented May 21, 2017 at 20:55