# Interpreting a notation in calculus of variations (differentiating with respect to a derivative)

Consider a functional $$J[y]$$ defined by: $$J[y] = \int_a^b F(x, y, y') dx \tag{1}$$

Here, $$F$$ is a function that depends on the independent variable $$x$$, the function $$y(x)$$, and its derivative $$y' = \frac{dy}{dx} \tag{2}$$.

In the calculus of variations, the operation of differentiating $$F$$ with respect to $$y'$$ is involved:
$$\frac{\partial F}{\partial y'} \tag{3}$$

This represents the rate of change of the function $$F$$ with respect to the derivative of $$y$$, $$y'$$. Operationally, since $$y'$$ is $$\frac{dy}{dx}$$, this differentiation is investigating how sensitive $$F$$ is to changes in the rate at which $$y$$ changes with respect to $$x$$.

I find the notation $$\frac{\partial F}{\partial y'}$$ a bit confusing in the sense that we are differentiating a function with respect to the derivative. If we just think $$y'$$ is just another variable symbol and proceed normally as most books do it does not cause much problems, but my question is :
What is the mathematical meaning of $$\frac{\partial F}{\partial y'}$$ in terms of limits?

In ordinary differential calculus, we don't encounter differentiation with respect to a derivative itself. Thanks.

• In calculus of variations $y'$ itself is treated as a variable, instead of being the derivative of $y$. Apr 13 at 16:10
• Related but not quite a duplicate: math.stackexchange.com/questions/1205829/… Apr 13 at 21:51

Consider a functional $$J[y]$$ defined by: $$J[y] = \int_a^b F(x, y, y') dx \tag{1}$$

I think it will be helpful to remind you right away that your notation implies that you have already agreed to consider the function $$F$$ to be a function of three independent variables. $$F=F(a,b,c)\;.$$

It may also be helpful to consider an explicit example $$F_{ex}$$. You could, for example, have a function like: $$F_{ex}(a,b,c) = a^2 + b^5 + c^9\;,$$ which means you have: $$F_{ex}(x, y(x), y'(x)) = x^2 + (y(x))^5 + (y'(x))^9\;.$$

The notation $$F(x,y,y')$$ simply means that you are evaluating $$F(a,b,c)$$ at $$a=x$$, $$b=y$$, and $$c=y'$$.

Although, as you correctly note, $$y' = \frac{dy}{dx}$$ and therefore $$y'$$ is not generally independent of $$y$$, we do not really care about that. We only care that $$\frac{\partial F}{\partial c}$$ is supposed to remind us we are differentiating $$F$$ with respect to its third argument.

This is also what $$\frac{\partial F}{\partial y'}$$ is supposed to mean.

...I find the notation $$\frac{\partial F}{\partial y'}$$ a bit confusing in the sense that we are differentiating a function with respect to the derivative.

Yes, it is a bit confusing. But remember, what $$\frac{\partial F}{\partial y'}\tag{A}$$ really means is just differentiation with respect to only the third argument. It might be easier to understand if written like this: $$\left.{\left(\frac{\partial F}{\partial c}\right)}\right|_{a=x, b=y(x), c=y'(x)}\;.\tag{B}$$ But, of course, it takes a lot more pencil marks to write B than to write A.

If we just think $$y'$$ is just another variable symbol and proceed normally as most books do it does not cause much problems, but my Doubt/Question is :
What is the mathematical meaning of $$\frac{\partial F}{\partial y'}$$ in terms of limits?

It is the usual definition of the partial deriviative: $$\frac{\partial F}{\partial y'}\equiv \left.{\left(\lim_{h\to 0}\frac{F(a,b,c + h) - F(a,b,c)}{h}\right)}\right|_{a=x,b=y(x), c=y'(x)}$$

For example, using our example $$F_{ex}$$ specified above: $$\frac{\partial F}{\partial y'} = 9(y'(x))^8\;.$$

The "functional" is a way to convert functions to real numbers : It maps a function (of 1 variable or 2 variables or more variables) to a real number.

Here the functional is $$J[y]$$ to convert the given function $$y$$ to a real number.
Current functional is to take the Integration of some $$F$$ which is a "common" or "ordinary" function of 3 independent variables.

When we consider $$F$$ in isolation , it is a "common" or "ordinary" function of 3 variables , like $$F(V_1,V_2,V_3)=V_1+V_2^2+ 3V_3$$ : nothing confusing there.

We then "Substitute" those variables with terms involving the "Input" to the functional : given the function $$y$$ , we extract the independent variable $$x$$ & then evaluate the derivative $$y'$$.
With that "Substitution" , we will have only $$x$$ terms in the Integral & we can get the real number for the functional.

Coming to the confusing $$\partial F / \partial y'$$ , it is a short-hand for $$\partial F / \partial V_3$$ , which is straight-forward & not confusing.

Like-wise , $$\partial F / \partial y$$ is a short-hand for $$\partial F / \partial V_2$$

Of course , $$\partial F / \partial x$$ is a short-hand for $$\partial F / \partial V_1$$

In each case [ $$\partial F / \partial V_1$$ , $$\partial F / \partial V_2$$ , $$\partial F / \partial V_3$$ ] the other 2 variables are held "Constant" to get the Partial Derivatives.

With that , there is no over-all confusion.

OP : What is the mathematical meaning of $$\frac{\partial F}{\partial y'}$$ in terms of limits?

There is no meaning in terms of limits involving $$x$$ & $$y$$ & $$y'$$ simultaneously.
It is simply a short-hand notation for $$\partial F / \partial V_3$$ where $$V_3$$ is the third "Input" variable to the function $$F$$ , keeping the other variables unchanged.

Later , we will be setting it to $$V_3=y'$$ & we will have a Differential Equation involving $$y',y,x$$. Solving that , we will get $$y$$ in terms of $$x$$.

• I am a huge fan of introducing temporary auxiliary variables in situations like these. +1. Apr 14 at 18:43
• Thanks for the "Supportive Comment" , @JonathanZ , though I wonder whether you forgot the +1 at the end !
– Prem
Apr 14 at 18:59
• Hmmm, I'm seeing an up arrow registered at my end. Maybe it's taking a while to update on some server? (When I was as close to 10k rep as you are, I counted every up vote. 🙂) Apr 14 at 19:22
• I think that vote disappeared into the void , since there is no increment on my side. No worries though , @JonathanZ , It is the thought that counts ! More-over the other answer is very similar to mine & came up 80 minutes later , yet it has got lots of upvotes : unpredictable !
– Prem
Apr 15 at 6:45

$$F$$ is a function from $$\mathbb R^3$$ to $$\mathbb R$$. If you label points in $$\mathbb R^3$$ as $$(x,y,z)$$, then $$\frac{\partial F}{\partial x}$$, $$\frac{\partial F}{\partial y}$$ and $$\frac{\partial F}{\partial z}$$ make sense as functions from $$\mathbb R^3$$ to $$\mathbb R$$.

Now, among all points in $$\mathbb R^3$$, you only look at the points that are of the form $$(x,y(x),y^\prime(x))$$ for some $$x$$. The notation $$\frac{\partial F}{\partial y^\prime}$$ simply means $$\frac{\partial F}{\partial z}$$ evaluated at $$(x,y(x),y^\prime(x))$$ (remember $$\frac{\partial F}{\partial z}$$ is a function from $$\mathbb R^3$$ to $$\mathbb R$$).

Hope this helps. :)