Partial derivative of a function with respect to its derivative

Question

While going through some introductory notes on Lagrangian and Hamiltonian Mechanics, I was a bit surprised by the notion of a functional
$$ J(y):=\int_{a}^{b} F\left(x, y, y_{x}\right) \mathrm{d} x $$

and how it depends on $x, y$ and $y_x$. Given that $y = y(x)$, it follows that $ F = F(x)$, and I didn't understand the need to show an explicit dependence on $ y $ and $y_x$. The confusion that I specifically have is with respect to a solution of the brachsitochrome problem, $$ \begin{aligned} \frac{\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}}{\left(y_{1}-y\right)^{\frac{1}{2}}}-y_{x} \frac{\partial}{\partial y_{x}}\left(\frac{\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}}{\left(y_{1}-y\right)^{\frac{1}{2}}}\right) &=c \\ \frac{\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}}{\left(y_{1}-y\right)^{\frac{1}{2}}}-\frac{y_{x}}{\left(y_{1}-y\right)^{\frac{1}{2}}} \cdot \frac{\partial}{\partial y_{x}}\left(\left(1+\left(y_{x}\right)^{2}\right)^{\frac{1}{2}}\right) &=c \end{aligned} $$

What does it mean to take the partial of a function with respect to its derivative, that is, what does $ \frac{\partial y}{\partial y_{x}} $ mean and why is it equal to zero? In other words, why were able to bring $ (y_1 - y)^{\frac{1}{2}} $ outside the $\frac{\partial}{\partial y_x}$ ?

Answer 1

An ODE as

$$ y''y'+3y'^2+x y=0 $$

can be represented as

$$ F(x,y,y',y'')=0 $$

or

$$ F(x,u_1,u_2,u_3)=0,\ \ \cases{u_1=y\\ u_2=\frac{d u_1}{dx}\\ u_3=\frac{d u_2}{dx}} $$

now if we want to derive it regarding $x$ we can proceed as follows

$$ \frac{d F}{dx}=\frac{\partial F}{\partial x}\frac{dx}{dx}+\frac{\partial F}{\partial u_1}\frac{d u_1}{dx}+\frac{\partial F}{\partial u_2}\frac{d u_2}{dx}+\frac{\partial F}{\partial u_3}\frac{du_3}{dx} $$

or

$$ \frac{d F}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx}+\frac{\partial F}{\partial y'}\frac{dy'}{dx}+\frac{\partial F}{\partial y''}\frac{dy''}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}y'+\frac{\partial F}{\partial y'}y''+\frac{\partial F}{\partial y''}y''' $$

now applying to our example

$$ \cases{ \frac{\partial F}{\partial x}=y\\ \frac{\partial F}{\partial y}=x\\ \frac{\partial F}{\partial y'}=y''+6y'\\ \frac{\partial F}{\partial y''}=y' } $$

and the result

$$ \frac{dF}{dx}=y + x y'+(y''+6y')y''+y'y'''=0 $$

Answer 2

This notation can be defined in the following more rigorous way $$ \partial_{y_x}F(x,y,y_x) := (\partial_3F)(x,y,y_x) $$ where $\partial_3$ is the derivation with respect to the third variable of $F$. If you prefer, you can also write $F: (x,Y,Z)↦F(x,Y,Z )$ and then $$ \partial_{y_x}F(x,y,y_x) = (\partial_ZF)(x,y,y_x). $$ In pratice, you first differentiate assuming your functions inside are independent variables, and then you evaluate.