# I Proper notation in chain rule

1. Jun 18, 2016

### 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: Jun 18, 2016
2. Jun 18, 2016

### PabloAMC

About the first question I think the answer is that it is correct (the formula 5.17), but about
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

3. Jun 18, 2016

### Staff: Mentor

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.

4. Jun 18, 2016

### Happiness

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 $V(2)\neq V(2)$.

5. Jun 18, 2016

### Ssnow

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

6. Jun 18, 2016

### Staff: Mentor

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.

7. Jun 18, 2016

### Happiness

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: Jun 18, 2016
8. Jun 18, 2016

### Staff: Mentor

It should be clear from the context in which this is written.
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.

9. Jun 18, 2016

### Happiness

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 $t=2$.

10. Jun 18, 2016

### Staff: Mentor

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)) = $\frac 4 3 \pi h$
There's no need to give it a different name if you understand function composition.

11. Jun 18, 2016

### Staff: Mentor

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 $V'(t_0) = \left.\frac{dV}{dr}\right |_{r(t_0)} \left.\frac{dr}{dt} \right|_{t_0}$

12. Jun 18, 2016

### Happiness

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? $(z\circ y\neq z)$

13. Jun 18, 2016

### Staff: Mentor

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.
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.

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