I Extract a derivative from an equation

[Malamala]

Hello! This if from a physics paper but I will write it as abstract as I can. We have a function $f(g(a),a)$ and we know that f is minimized with respect to g for any given a i.e. $$\frac{df}{dg}|_a=0$$ As this is true for any a, we have $$\frac{d}{da}\frac{df}{dg}|_a=0$$ from which we get: $$\frac{\partial}{\partial a}(\frac{df}{dg})+\frac{d^2f}{dg^2}\frac{dg}{da}|_a=0$$ And from here $\frac{dg}{da}$ can be easily extracted (which is what I need), in terms of the other derivatives (which are assumed to be known). (Here is the actual physics paper from which I simplified the equations, for reference, equations 20-23). Now I want to extent the same analysis to one more variable i.e. $f(g(a,b),a,b)$ and find an expression for $\frac{\partial g}{\partial a}$ and $\frac{\partial g}{\partial b}$ (with the same minimization assumptions) but I am kinda stuck. I tried to follow the same approach i.e. $$\frac{df}{dg}|_{a,b}=0$$ from which I would get $$\frac{\partial}{\partial a}\frac{df}{dg}|_{a,b}=0$$ and $$\frac{\partial}{\partial b}\frac{df}{dg}|_{a,b}=0$$ But I am not sure what to do from here. I guess I am writing the new equations the wrong way, but I am not sure what to do. Can someone help me? Thank you!

 

[andrewkirk]

The problem is difficult to express with the Leibniz-style notation for partial derivatives: $\frac d{dx}, \frac\partial{\partial x}$. So let's switch notation to one that specifically indicates the number of which argument we are differentiating against. The operator $D{}_k$ denotes partial differentiation of a function wrt its $k$-th argument. Where there is only one argument we can drop the subscript and just write $D{}$. That notation makes writing the problem easier.

First let's rewrite the first problem:

We have functions $f:\mathbb R^2\to\mathbb R$ and $g:\mathbb R\to\mathbb R$.

The first equation in the OP can be written as:

$$\forall a:\ \ (D{}_1 f) (g(a),a)= 0$$

and the second equation is:

$$\forall a:\ \ (D{} h)(a) = 0$$

where $h:\mathbb R\to \mathbb R$ is defined as $h(x)=(D{}_1 f)(g(x),x)$, that is, the function $(D{}_1 f):\mathbb R^2\to\mathbb R$ applied to the arguments $g(x)$ and $x$.

We can rewrite that last equation by applying the total derivative rule, as:

$$0 = (D{} h)(a)
= (D{}_1 (D{}_1 f))(g(a),a)\cdot (D{} g)(a)
+(D{}_2(D{}_1 f))(g(a),a)
$$
whence we deduce the wanted result:
$$(D{} g)(a)
= - \frac{(D{}_2D{}_1 f)(g(a),a)}
{(D{}_1 D{}_1 f)(g(a),a)}
$$

Now let's apply the same procedure to the extended case, using functions $f:\mathbb R^3\to\mathbb R$ and $g:\mathbb R^2\to\mathbb R$.

First equation is:
$$\forall a,b:\ \ (D{}_1 f) (g(a,b),a,b)= 0$$
The second equation is a pair:
$$\forall a:\ \ (D{} h_a)(a) = 0$$
$$\forall b:\ \ (D{} h_b)(b) = 0$$
where functions $h_a,h_b:\mathbb R\to\mathbb R$ are defined by:
$$h_a(x) = (D{}_1 f)(g(x,b),x,b)$$
that is, the function $(D{}_1 f):\mathbb R^3\to\mathbb R$ applied to the arguments $g(x,b)$, $x$ and $b$;

and

$$h_b(x) = (D{}_1 f)(g(a,x),a,x)$$
that is, the function $(D{}_1 f):\mathbb R^3\to\mathbb R$ applied to the arguments $g(a,x)$, $a$ and $x$.

We expand the pair of equations using the total derivative rule, to get:
$$0 = (D{} h_a)(x)
= (D{}_1 (D{}_1 f))(g(x,b),x,b)\cdot (D{}_1 g)(x,b)
+(D{}_2(D{}_1 f))(g(x,b),x,b)
$$
and
$$0 = (D{} h_b)(x)
= (D{}_1 (D{}_1 f))(g(a,x),a,x)\cdot (D{}_2 g)(a,x)
+(D{}_3(D{}_1 f))(g(a,x),a,x)
$$
whence we deduce the wanted results:
$$(D{}_1 g)(a,b)
= - \frac{(D{}_2D{}_1 f)(g(a,b),a,b)}
{(D{}_1 D{}_1 f)(g(a,b),a,b)}
$$
and
$$(D{}_2 g)(a,b)
= - \frac{(D{}_3D{}_1 f)(g(a,b),a,b)}
{ (D{}_1 D{}_1 f)(g(a,b),a,b)}
$$

noting that we have replaced $x$ by $a$ in the first equation and $b$ in the second.

The insights article here explains the confusion that is frequently created by the Leibniz notation for partial derivatives, and proposes the system used in this post as a means of clearing up the confusion.