An elementary problem with partial derivatives and chain rule

Question

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

Answer 1

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}

Answer 2

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}