I Is it necessary to use a different function name in the chain rule?

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

 

[PabloAMC]

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

. 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

 

[Mark44]

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.

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?

 

[Happiness]

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

 

[Ssnow]

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 $u_i=u_i(x_{1},\ldots,x_{n})$...

 

[Mark44]

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.

 

[Happiness]

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.)

 

[Mark44]

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.

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.

 

[Happiness]

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 $t=2$.

 

[Mark44]

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)) = $\frac 4 3 \pi h$

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.

 

[Mark44]

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

 

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

 

[Mark44]

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.

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.