# Proper notation in chain rule

• I
• Happiness
In summary: I think the notation ##V(t)## is not good either.)In summary, the conversation discusses the correct version of the chain rule and the naming of a function after substituting a variable in a given function. It is concluded that the function should still be referred to as V, but now as a function of t instead of r. There is also a discussion about the ambiguity of the notation V(2), which could refer to either V(r=2) or V(t=2), and the need for better notation to avoid confusion.

#### Happiness

Is the chain rule below wrong?

What I propose is as follows:

Given that ##x_i=x_i(u_1, u_2, ..., u_m)##. If we define the function ##g## such that ##g(u_1, u_2, ..., u_m)=f(x_1, x_2, ..., x_n)##, then

##\frac{\partial g}{\partial u_j}=\sum_{i=1}^n\frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial u_j}##.

This version of chain rule is what is being used, it seems, in the example below when the answer given replaces ##f## with ##g## in the last line.

A related question is as follows:

Consider the function ##V(r)=\frac{1}{3}\pi r^2h##, where ##h## is a constant. Suppose ##r## is a function of ##t## such that ##r(t)=at^2##, where ##a## is a constant.

What do we call the function after substituting ##r## with ##at^2##, which gives ##\frac{1}{3}\pi a^2t^4h##?

I guess we have to give it a different name: ##W(t)=\frac{1}{3}\pi a^2t^4h##, because ##V(t)## would give ##V(t)=\frac{1}{3}\pi t^2h##. Then ##\frac{\partial V(t)}{\partial t}=\frac{\partial V(r)}{\partial r}##. Am I right?

If we still call it ##V## as follows: ##V(t) = \frac 1 3 \pi a^2 t^4 h##, we will run into a problem.

Since ##V(r)=\frac{1}{3}\pi r^2h##, when ##r=2##, we would write ##V(2)=\frac{1}{3}\pi\,2^2\,h##. But if we write ##V(t)=\frac{1}{3}\pi a^2t^4h##, when ##t=2##, we have ##V(2)=\frac{1}{3}\pi a^2\,2^4\,h##. Then ##V(2)\neq V(2)##.

Last edited:
About the first question I think the answer is that it is correct (the formula 5.17), but about
Happiness said:
. Am I right?
The correct answer is the same, this one
##\frac{\partial V(t)}{\partial t}=\frac{\partial V(r)}{\partial r}\frac{\partial r}{\partial t}##
I did not check the last equation but I think it is not too difficult

Happiness said:
Is the chain rule below wrong?
View attachment 102183

What I propose is as follows:

Given that ##x_i=x_i(u_1, u_2, ..., u_m)##. If we define the function ##g## such that ##g(u_1, u_2, ..., u_m)=f(x_1, x_2, ..., x_n)##, then

##\frac{\partial g}{\partial u_j}=\sum_{i=1}^n\frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial u_j}##.

This version of chain rule is what is being used, it seems, in the example below when the answer given replaces ##f## with ##g## in the last line.

A related question is as follows:

Consider the function ##V(r)=\frac{1}{3}\pi r^2h##, where ##h## is a constant. Suppose ##r## is a function of ##t## such that ##r(t)=at^2##, where ##a## is a constant.

What do we call the function after substituting ##r## with ##at^2##, which gives ##\frac{1}{3}\pi a^2t^4h##?
You still call it V, but now instead of V being a function of r, it's a function of t. You could refer to it as ##V(t) = \frac 1 3 \pi a^2 t^4 h##, or you could refer to it as ##V(r(t))##
Partials really don't have a place here. The first definition of V has it as a function of r alone (h is a constant, you said), so writing ##\frac{\partial V(r)}{\partial r}## is an overcomplication. ##\frac{d V(r)}{dr}## is appropriate.

Now, since r is a function of t alone, then you can refer to ##\frac{d V}{dt}##, and can calculate it using the chain rule for functions of a single variable.
Happiness said:
I guess we have to give it a different name: ##W(t)=\frac{1}{3}\pi a^2t^4h##, because ##V(t)## would give ##V(t)=\frac{1}{3}\pi t^2h##. Then ##\frac{\partial V(t)}{\partial t}=\frac{\partial V(r)}{\partial r}##. Am I right?

Mark44 said:
You still call it V, but now instead of V being a function of r, it's a function of t. You could refer to it as ##V(t) = \frac 1 3 \pi a^2 t^4 h##.

Since ##V(r)=\frac{1}{3}\pi r^2h##, when ##r=2##, we would write ##V(2)=\frac{1}{3}\pi\,2^2\,h##. But if we write ##V(t)=\frac{1}{3}\pi a^2t^4h##, when ##t=2##, we have ##V(2)=\frac{1}{3}\pi a^2\,2^4\,h##. Then ##V(2)\neq V(2)##.

Happiness said:
Given that ##x_{i}=x_{i}(u_1,u_2,...,u_m)x_i=x_i(u_1, u_2, ..., u_m)##. If we define the function ##g## such that ##g(u_1,u_2,...,u_m)=f(x_1,x_2,...,x_n)##, then

I think it is understood also that you can express ##u_i=u_i(x_{1},\ldots,x_{n})##...

Happiness said:
Since ##V(r)=\frac{1}{3}\pi r^2h##, when ##r=2##, we would write ##V(2)=\frac{1}{3}\pi\,2^2\,h##. But if we write ##V(t)=\frac{1}{3}\pi a^2t^4h##, when ##t=2##, we have ##V(2)=\frac{1}{3}\pi a^2\,2^4\,h##. Then ##V(2)\neq V(2)##.
No, that's not right. You're comparing apples and oranges. If r = 2, then V(2) = ##\frac{1}{3}\pi r^2h = \frac{1}{3}\pi 2^2h = \frac{4}{3}\pi h##. It's understood here that V(2) means that you evaluate things for r = 2.
V(r(2)) = V(2), but when r = 2 you're going to get a different value of V than for t = 2.

If r = 2, then ##t = \pm \sqrt{\frac 2 a}##, so V(r = 2) using the first formula is exactly equal to V(##t = \pm \sqrt{\frac 2 a}##) using the second formula.

This is basic function composition.

Mark44 said:
No, that's not right. You're comparing apples and oranges. If r = 2, then V(2) = ##\frac{1}{3}\pi r^2h = \frac{1}{3}\pi 2^2h = \frac{4}{3}\pi h##. It's understood here that V(2) means that you evaluate things for r = 2.
V(r(2)) = V(2), but when r = 2 you're going to get a different value of V than for t = 2.

If r = 2, then ##t = \pm \sqrt{\frac 2 a}##, so V(r = 2) using the first formula is exactly equal to V(##t = \pm \sqrt{\frac 2 a}##) using the second formula.

This is basic function composition.

So ##V(2)## could either mean ##V(r=2)## or ##V(t=2)##? And there is no universally accepted notation? It seems like this ambiguity can be avoided if we use good notations.

( ##V(r=2)## and ##V(t=2)## are themselves not good notations.)

Last edited:
Happiness said:
So ##V(2)## could either mean ##V(r=2)## or ##V(t=2)##?
It should be clear from the context in which this is written.
Happiness said:
And there is no universally accepted notation? It seems like this ambiguity can be avoided if we use good notations.
Yes, of course.
If you write ##V(r) = \frac{1}{3}\pi r^2h##, and then also write ##V(t) = \frac{1}{3}\pi (at^2)^2h## (as you have done here), then writing V(2) is ambiguous. Does 2 represent a value of r or is it a value of t?

The first formulation of V above could be written as ##V(r(t)) = \frac{1}{3}\pi (r(t))^2h##, where ##r(t) = at^2##. That would clear up any ambiguity.

Mark44 said:
If you write ##V(r) = \frac{1}{3}\pi r^2h##, and then also write ##V(t) = \frac{1}{3}\pi (at^2)^2h## (as you have done here), then writing V(2) is ambiguous. Does 2 represent a value of r or is it a value of t?

If ##V## is a function that maps ##r## to ##\frac{1}{3}\pi r^2h##, then shouldn't it map ##t## to ##\frac{1}{3}\pi t^2h##? Then isn't ##V(t)=\frac{1}{3}\pi t^2h##?

This is the main reason why I believe we shouldn't write ##V(r(t))## as ##V(t)##. We should name it differently, for example, as ##W(t)##. Then ##V(2)## means ##r=2## and ##W(2)## means ##t=2##.

Happiness said:
If ##V## is a function that maps ##r## to ##\frac{1}{3}\pi r^2h##, then shouldn't it map ##t## to ##\frac{1}{3}\pi t^2h##?
Yes, and V maps x to ##\frac{1}{3}\pi x^2h##, and it maps z to ##\frac{1}{3}\pi z^2h##, but so what?

What you seem to be forgetting is that there is a function composition going on, with V being a function of r, and r being a function of t. In an abuse of notation, we have V = V(r(t))
For a given value of t, find r(t), and then find V(r(t)). So if t = 2, r(2) = 4a, and V(r(2)) = ##\frac 4 3 \pi h##
Happiness said:
Then isn't ##V(t)=\frac{1}{3}\pi t^2h##?

This is the main reason why I believe we shouldn't write ##V(r(t))## as ##V(t)##. We should name it differently, for example, as ##W(t)##. Then ##V(2)## means ##r=2## and ##W(2)## means ##t=2##.
There's no need to give it a different name if you understand function composition.

Happiness said:
I guess we have to give it a different name: ##W(t)=\frac{1}{3}\pi a^2t^4h##, because ##V(t)## would give ##V(t)=\frac{1}{3}\pi t^2h##. Then ##\frac{\partial V(t)}{\partial t}=\frac{\partial V(r)}{\partial r}##.
I don't know why you're using partial derivatives. As this thread is about the chain rule, and your example is the composition of two functions of one variable, ordinary derivatives suffice. Given V in terms of r and r in terms of t, ##V'(t_0) = V'(r(t_0))*r'(t_0)##, or ##V'(t_0) = \left.\frac{dV}{dr}\right |_{r(t_0)} \left.\frac{dr}{dt} \right|_{t_0}##

Consider ##z## to be a function of ##y##, which is itself a function of ##x##. The chain rule may be written as ##\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}##.

The ##z## on the LHS means ##z(y(x))##, which is equivalent to ##(z\circ y)(x)##, whereas the ##z## on the RHS means ##z(y)##. So strictly speaking, they are not the same, right? ##(z\circ y\neq z)##

Happiness said:
Consider ##z## to be a function of ##y##, which is itself a function of ##x##. The chain rule may be written as ##\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}##.
Sure. Here the context is that on the left side, we're talking about a map from x value to z values, and on the right side, the left derivative refers to a map from y values to z values. Certainly the two map formulas are different.
Happiness said:
The ##z## on the LHS means ##z(y(x))##, which is equivalent to ##(z\circ y)(x)##, whereas the ##z## on the RHS means ##z(y)##. So strictly speaking, they are not the same, right? ##(z\circ y\neq z)##

Some textbooks use a different function name. If x = g(t), and y = f(x), then they will define z = h(t) = f(g(t)). My point was that, as long as the context was understood, it wasn't necessary to introduce another function. If you have defined V in terms of r, then it's understood that V(2) means to evaluate V at r = 2. But if there is uncertainty whether you mean V as a function r or V as a function of t (with different formulas), then it's not clear what V(2) is supposed to mean.

## 1. What is the proper notation for the chain rule?

The proper notation for the chain rule is (f ∘ g)'(x) = f'(g(x)) * g'(x), where f and g are functions and x is the variable.

## 2. How is the chain rule used in calculus?

The chain rule is used in calculus to find the derivative of a composite function, where one function is nested inside another. It allows us to break down complex functions into simpler ones and calculate their derivatives separately.

## 3. Can you provide an example of the chain rule?

Sure, let's say we have the function f(x) = (x^2 + 2x)^3. To find its derivative, we can use the chain rule by setting g(x) = x^2 + 2x and f(x) = g(x)^3. Applying the chain rule, we get f'(x) = 3(x^2 + 2x)^2 * (2x + 2) = 3(x^2 + 2x)^2 * (2x + 2).

## 4. What happens if there are multiple functions nested inside each other?

If there are multiple functions nested inside each other, we can apply the chain rule multiple times. For example, if we have the function f(x) = (x^2 + 2x)^3 * sin(x^2 + 2x), we can set g(x) = x^2 + 2x and h(x) = sin(x) and use the chain rule twice to find the derivative of f(x).

## 5. Is there a visual representation of the chain rule?

Yes, the chain rule can be visualized using a tree diagram. The innermost function is at the bottom of the tree, with each subsequent layer representing the nested functions. The branches of the tree represent the derivatives of each function, and the final derivative is calculated by multiplying the branches along the path from the top of the tree to the bottom.