# Fundamental confusions of calculus

Discussion in 'Physics & Math' started by arfa brane, Feb 11, 2012.

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1. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Here's my take (Tach, I know you think this is wrong, and you've given your reasons why. I'm posting this so that if I'm wrong, I can be corrected by others in the thread.)

If x is a function of $\theta$...
$f(\theta)=3 \theta+ sin^2(\theta)+ln(x(\theta)) \\ \frac{\partial }{\partial \theta} f(\theta) = \frac{d}{d \theta} f(\theta) = 3+2 sin(\theta) cos (\theta) + \frac{1}{x}\frac{dx}{d\theta}$​

If x and $\theta$ are independent variables...
$f(\theta, x) =3 \theta+ sin^2(\theta)+ln(x) \\ \frac{d}{d \theta}f(\theta, x) \mbox{ is meaningless.} \\ \frac{\partial }{\partial \theta} f(\theta, x)= 3 + 2 \sin(\theta) \cos (\theta)$​

Last edited: Feb 12, 2012

3. ### przyksquishyValued Senior Member

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3,203
I'll use one of Tach's examples to clarify things. This one's the most interesting because $f$ has both an explicit dependence on $\theta$ and an implicit dependence through $u$:
For convenience I'll shorten that to $f(\theta,\,u(\theta),\, v)$. Then strictly speaking, the function $f$ is a function of $\theta$, $u$ and $v$ and technically the partial derivative $\frac{\partial f}{\partial \theta}$ is just what Tach said it is: the derivative of $f$ only through its explicit dependence on $\theta$, with $u$ and $v$ held constant. Strictly speaking if you substitute $u = u(\theta)$ you've defined a different function:
$f'(\theta,\, v) \,=\, f(\theta,\, u(\theta),\, v) \,,$​
which you may also want to take partial derivatives of. Then the partial derivatives with respect to $\theta$ are related by
$\frac{\partial f'}{\partial \theta} \,=\, \frac{\partial f}{\partial \theta} \,+\, \frac{\partial f}{\partial u} \, \frac{\mathrm{d}u}{\mathrm{d}\theta} \,.$​
Obviously, neither $\frac{\partial f'}{\partial \theta}$ nor $\frac{\partial f}{\partial \theta}$ are total derivatives.

With that said, while $f$ and $f'$ are strictly speaking different functions in the mathematical sense, in a physical application you may still want to think of $f$ and $f'$ as the same quantity, just expressed in two different ways, or to think about $f(\theta,\, u(\theta),\, v)$ as a redefinition of $f$ rather than an entirely new function that needs a different name.

I remember from my thermodynamics classes that we were often taking partial derivatives of the same quantity but in terms of different variables. Apparently this happens routinely enough that researchers in the field developed their own notation for it. For example, when we write the heat capacity at constant volume as
$C_{\mathrm{V}} \,=\, \Bigl( \frac{\partial Q}{\partial T} \Bigr)_{V} \,,$​
the subscript $V$ on the partial derivative indicates that we're thinking of $Q$ as a function of $V$ and $T$, and taking the partial derivative with $V$ held constant. Similarly for the heat capacity at constant pressure:
$C_{\mathrm{p}} \,=\, \Bigl( \frac{\partial Q}{\partial T} \Bigr)_{p} \,,$​
the subscript $p$ indicates we're now thinking of $Q$ as a function of $p$ and $T$, and taking the partial derivative with $p$ held constant. Normally in thermodynamics $T$, $p$ and $V$ are related to one another via a state equation, eg. $V = V(p,\, T)$, so $Q(V,\, T)$ and $Q(p,\, T)$ are related by variable substitution, similar to the example above.

So in short, take whatever partial or total derivatives you want. The important thing is to know which derivative you're taking under what conditions, and why it is you need precisely that derivative.

5. Yes, do you have a problem with this?

7. First off, x is a variable, not a function of $\theta$, so you got that wrong.
Second off, the total derivative is not equal to the partial derivative.
Third off, even if x were a function of $\theta$, you STILL got the total derivative wrong because you are missing the term in $2 sin(\theta) cos (\theta)$.
Three basic mistakes in a post on basic calculus.

8. Correct. (except that x is an independent variable, not a function of $\theta$), which brings us to the second part:

Your claim that $\frac{d}{d \theta}f(\theta, x)$ "is meaningless" is not only wrong but borders on pettiness.
Your claim that $\frac{\partial }{\partial \theta} f(\theta, x)= 3 + 2 \sin(\theta) \cos (\theta)$ is outright wrong because it shows that you do not understand the meaning of partial differentiation.

9. ### przyksquishyValued Senior Member

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3,203
Er, nope. As written, $\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$.

10. Err, wrong. Partial derivative means derivative wrt the explicit variable only. What you took is the total derivative wrt $\theta$. Think about:

$f(\theta,u,x)=3 \theta +u^2+ln(x)$

where $u=sin(\theta)$.

This means $\frac{\partial f}{\partial \theta}=3$. Period.

Last edited: Feb 12, 2012
11. It is not only what I said it is, it is what standard calculus books say it is.

Last edited: Feb 12, 2012
12. ### przyksquishyValued Senior Member

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3,203
I said as written. There is no intermediate variable $u$ in the expression $3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x)$. As written that's an expression that depends only on two independent variables, and its partial derivative with respect to $\theta$ is $3 \,+\, \sin(2\theta)$.

The way you define $f$ here is, incidentally, ambiguous:
On its own, that is just defining the function $f(\theta,\,x)$ to be equal to the expression above. There is no mention of $u$. The only hint that there should be such an intermediate variable at all is from earlier in your post, in the explanation that's supposed to motivate the example. But even there you never say how $f$ is supposed to depend on $u$. You may have meant
$f(\theta,\, u,\, x) \,=\, 3\theta \,+\, u \,+\, \ln(x) \,,$​
with $u \,=\, \sin(\theta)^{2}$, in which case $\frac{\partial f}{\partial \theta} = 3$. But you never said that. From what you wrote, $f$ could just as well have been
$f(\theta,\, u,\, x) \,=\, u \,+\, \sin(\theta)^{2} \,+\, \ln(x) \,,$​
with $u \,=\, 3\theta$, or
$f(\theta,\, u,\, x) \,=\, \theta \,+\, u \,+\, \sin(\theta)^{2} \,+\, \ln(x) \,,$​
with $u \,=\, 2 \theta$. In fact if you take your example literally and add that $f$ is supposed to depend on $u$, then your example just reads like
$f(\theta,\, u,\, x) \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \ln (x) \,,$​
i.e. with $\frac{\partial f}{\partial u} \,=\, 0$.

Either way, the way you wrote it, saying just
$f \,=\, 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \,,$​
anyone who sees that is going to tell you $\frac{\partial f}{\partial \theta} \,=\, 3 \,+\, \sin(2\theta)$.

And as others have pointed out, you certainly are wrong to say that
$\frac{\mathrm{d}f}{\mathrm{d}\theta} \,=\, 3 \,+\, \sin(2\theta)$​
because $x$ is still an independent parameter that needs to be kept constant. The derivative when $\theta$ is allowed to vary through $u$ is still a partial derivative, albeit a different one than when $u$ is kept constant. As I said in post #22, if you want to be technically correct you should really state explicitly that you're defining a new function $f'(\theta,\, x) \,=\, f(\theta,\, u(\theta),\, x)$ and then say you're taking the partial derivative $\frac{\partial f'}{\partial \theta}$, if you really need to make it unambiguous which partial derivative you're taking.

13. What in $f=f(\theta,u,v)$ is that you fail to see? Or is not clear enough to you that $sin(\theta)$ is not a variable but a function? It was not clear enough that $u=u(\theta)=sin(\theta)$ and $v=v(x)=ln(x)$?
Either way , your claim that $\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$ is false since the partial derivative is always taken wrt the direct variable (in this case, $\theta$) and not through a function (in this case $\sin(\theta)$) . The moment you start using chain differentiation you cease calculating partial differentials and you are starting to calculate total differentials.

Last edited: Feb 12, 2012
14. ### Guest254Valued Senior Member

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1,056
Tach is completely wrong. In fact, I'd go further than that. What Tach has said here is not only wrong, it's down right stupid.
Are you serious? You don't realise $f(\theta)=\theta$ is also a function? Hang your head in shame -- you don't understand calculus.

15. Of course, the thread would not be complete without your trolling. I take it that you skipped this class, the definition is not difficult at all. See how the differeniation is done wrt the independent variable t and through the chain rule? Do you think you can substitute $\theta$ for $t$ all by yourself ?

16. ### prometheusviva voce!Registered Senior Member

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2,045
There's quite enough here to show that Tach is completely wrong. The only place this thread can now go is into trolling, hence it is closed.

17. ### AlphaNumericFully ionizedRegistered Senior Member

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6,702
Prom, I think you've been a bit quick in closing the thread, plus someone via PM asked if it could be reopened if we kept an eye on it so I hope you don't mind but I'm going to reopen it.

As a general announcement I will say, however, that if anyone becomes insulting or the like then immediate infraction warnings will be handed out. I haven't yet suspended anyone since I got my mod powers so let's try not to force my hand okay everyone?

That simply isn't true and there's a great many classic examples to counter your claim.

For example, in Lagrangian/Hamiltonian mechanics partial derivatives are used all the time. For example, the kinetic energy of an object might be $T = \frac{1}{2}m\dot{q}^{2}$. By your logic since $\dot{q}^{2}$ is a function of $\dot{q}$ then $\frac{\partial T}{\partial \dot{q}} = 0$ identically. Except it doesn't, it equals $m\dot{q}$ and is the definition of the canonical momentum associated to q.

This can be elaborated further by considering a specific example, the pendulum. The Lagrangian is $L = \frac{1}{2}m l^{2} \dot{\theta}^{2} + mgl \cos \theta$. The Euler-Lagrange equations are $\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0$ with $q = \theta$ here. If we used your logic then $\frac{\partial L}{\partial \dot{q}} = 0$ because the only time $\dot{q}$ appears is as a function of $\dot{\theta}$, not a lone linear term. Furthermore your logic says $\frac{\partial L}{\partial \theta} = 0$. So by your logic the EL equation says 0 = 0. While it's true that 0 = 0 anyone who has done such mechanics knows the EL equation of a pendulum is not such a trivial pointless result. Instead we have $\frac{\partial L}{\partial \dot{\theta}} = ml^{2}\dot{\theta}$ and $\frac{\partial L}{\partial \theta} = -mgl\sin \theta$. Thus the EL equation says $ml^{2}\ddot{\theta} = -mgl\sin\theta$. Some clearing up and using the small angle approximation indeed gives us the well known SHM equation $\ddot{\theta} = -\frac{g}{l}\theta$.

Or how about the wave equation, $\partial_{t}^{2}A = \lambda \Delta A$ where $\Delta = \sum_{i} \partial_{i}^{2}$. Clearly $A = t^{3} + \sum_{i}x_{i}^{3}$ isn't a wave phenomenon yet by your logic it solves the equation since by your logic $\partial_{i} x_{i}^{2} = 0$.

Yes, it's essential to distinguish between total and partial derivatives and its easy for a novice to fall foul of some subtle properties but what you're claiming simply isn't true. A partial derivative of a multivariable function can be viewed as taking a normal derivative if you fixed all other variables. This is formally defined in much the same way as the usual derivative, ie $\frac{\partial}{\partial x}f(x,y) \equiv \lim_{h \to 0} \frac{f(x+h,y)-f(x,y)}{h}$

If you put in $f(x,y) = x + \sin x + \ln y = g(x) + h(y)$ then this clearly reduces to $\partial_{x}f(x,y) = g'(x)$, ie $\partial_{x}(x+\sin x) = 1 + \cos x$. This is the definition of the partial derivative. You're telling people "Oh you must have skipped this class" but multiple people have walked you through this. Hell, you even posted it yourself but it seems you didn't actually check the formula because you mistakenly give it as the definition for the total derivative. If you checked your own Wiki link you'd see it's nowhere to be seen. It is, however, on the partial derivative page.

Tach, there's a fine line between being confident and being abrasive, even when you're factually correct. In this thread you have been factually incorrect on several things, you've ignored or failed to understand multiple explanations and generally been abrasive throughout. If you want to wail against things you view as ignorant then spend more time in the pseudo forums. If you want to 'discuss' specifics of mainstream maths and physics in this forum you need to tone it down. A lot. I'm saying this because this is by no means the first time this has happened. You charge in and assert factually incorrect things. Now everyone does that from time to time but when you repeatedly ignore multiple explanations from multiple people you begin to wander out of the realm of friendly discussion to disruptive trolling. In just the last 20 posts I count 5 people who are either working on or who have a PhD correcting you on something. It's a pretty fine line to reopen the thread given previous track records so consider this a warning, next time this sort of thing happens more than just thread locking will be done.

This thread will be left open so you can respond. You don't need to respond to the last paragraph but you should take it on board. If you do respond to the mathematics then you need to do more than just post a link to a Wiki page or Google book and say "Read it", you need to explain why the definition of the partial derivative doesn't apply. Please address the points directly, don't just deflect. It'd be an added bonus if you could respond to my last post, where I explained how $\partial_{i}$ can be a vector in the appropriate space, you sorta skipped over it once I laid it all out. I'll assume it was because the thread went in another direction rather than you just not wanting to say "Okay, I was mistaken". I'm sure you wouldn't want to give people that impression.

Last edited: Feb 12, 2012
18. I think that you missed (like the others) what post 18 is telling you:

$f=f(\theta, u(\theta), v(x), w....)$

then:

The total derivative wrt $\theta$is:

$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$

In the example:

$u(\theta)=sin^2(\theta)$
$v(x)=ln(x)$

and

$f=3 \theta +u +v$

So, now I suggest that you pay attention to this (Differentiation of Composite Functions). Equations (1)(2) are exactly what I have been posting about. It comes from one of the best books on calculus, "Problems in Mathematical Analysis" by B. Demidovich. So, if you take on board what eq(2) is telling you, then:

-the partial derivative is exactly what I have posted

-the total derivative is exactly what I have posted

-you have all (with przyk's exception) missed what the debate with Pete is all about, the absence of $\frac{\partial f}{\partial v}$ in the expression for $\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$. THIS is what is giving Pete troubles, maybe, instead of rushing to "correct" me, one of you can explain to Pete why there is no $\frac{\partial f}{\partial v}$ in the calculation of the total derivative. And, no, it isn't because $\frac{dv}{d \theta}$ "doesn't make sense", as Pete has been claiming.

Your argument from authority , that several PhD have "corrected" me is disproved by the simple application of eq(2) in the reference. It is interesting that of all the responders, przyk is the only one that got it initially correct, only to reverse himself later.

BTW: What gives you the impression that I am "wailing"?

Last edited: Feb 13, 2012
19. ### AlphaNumericFully ionizedRegistered Senior Member

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The thing I was explicitly and clearly responding to was your claim that for something like $f(x,y) = 3x + \sin^{2}x + g(y)$ you have $\partial_{x}f(x,y) = 3$, ie what you claim here. That's not true. I gave the definition of the partial derivative, which clearly disagrees with your assertion here. You ignored that. Please respond to it. You also ignored how if your logic were right then pretty much all of Lagrangian and Hamiltonian mechanics would collapse down to 0=0.

I can see why you're getting confused, it's to do with which variables you consider independent or dependent. If you have f(x,y) and then realise y is actually a function of x then you have to alter how the system behaves under $x \to x+\delta x$ via the expression you give. This is an example of subtleties which arise in PDEs compared to ODEs. However, you have not negated my explanation of why you're mistaken. The problem is coming from the fact you're just pitching expressions without, it seems, understanding their origin or context.

Rather than having a 'quote off' why don't we just stick with the definition of the partial derivative. Please use the definition of the partial derivative to prove your claim from here step by step.

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,167
I really need to check my understanding, so please be patient with the long post.
Should that be:
$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin^2\theta \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, 2\sin\theta\cos\theta$​
Or am I missing something? (edit - never mind, I see it now)

In the case of an expression with direct and indirect dependencies on the variable of interest and no other direct dependencies, my understanding is that it is implied that the indirect dependencies are ignored, eg:
\begin{align} y &= g(x) ; \frac{\partial}{\partial x} f(x, y) &= \frac{\partial}{\partial x} f(x, y)|_y \end{align}​
Is that right?

And in the case of an expression with only indirect dependencies on the variable of interest, my understanding is that it is implied that the chain rule is applied, eg:
\begin{align} y &= g(x) \\ \frac{\partial}{\partial x} f(y, z) &= \frac{\partial}{\partial x} f(y, z)|_z \\ &= \frac{\partial}{\partial y} f(y, z) \, \frac{dy}{dx} \end{align}​
Is that right?

But in the case of an expression with both direct and indirect dependencies on the variable of interest, and other direct dependencies, then there is ambiguity, eg
$y = g(x) \\ \frac{\partial}{\partial x} f(x, y, z)$​
could potentially mean either
$\frac{\partial}{\partial x} f(x, y, z)|_z$ or $\frac{\partial}{\partial x} f(x, y, z)|_{y,z}$​

So it seems to me that for an expression like that you should be explicit, and that the two meanings are related like this:
$\frac{\partial}{\partial x} f(x, y, z)|_z = \frac{\partial}{\partial x} f(x, y, z)|_{y,z} \, + \, \frac{\partial}{\partial y} f(x, y, z) \, \frac{dy}{dx}$​
Is that right?
Or am I just hopelessly confused?

Watching this vid helped a lot.

Last edited: Feb 13, 2012
21. This is why you need to read post 18 before jumping in and claiming that I made an error.

I can assure you that I am not confused at all. The reference I gave lines up with my posts and disproves yours (and the others) claims.

Please stop talking down to me. Especially in the context of my having gone out of the way and having explained the above to Pete from base principle.

22. ### AlphaNumericFully ionizedRegistered Senior Member

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6,702
You're updating your post after I posted...

As I said, I don't deny such equations. Their context and application is the issue. I specifically said not for you to just post a link and say "Read that" for this reason. To use an example, Reiku often will link to a video or pdf and say "Look, it says so right there!" only for it to transpire that he's misapplying or misunderstanding the reference.

The whole "Let's write $\sin^{2}x = y$!" thing is to do with implicit parametrisation. If you started with f(x,y) then you have, without any other restrictions, a 2 dimensional system and you can vary x and y freely and independently. However, if you then declare "I can write y in terms of x!" then you're restricting from $\mathbb{R}^{2}$ to some subset with a parametrisation. It's a bit like there being a difference between explicit and implicit curvature, it depends what you view as the free parameters.

All of this comes up regularly in dynamical systems, because they involve multivariable functions, implicit and explicit parametrisations and constraints.

Firstly it isn't an argument from authority when the people have relevant experience and provide explanations. Secondly your link doesn't disprove what you've been corrected on. Please, I really don't want you to reduce your level of discussion to that of Reiku, just throwing out "Read this!" comments and not actually engaging in rational informed discussion.

Several of the people who have corrected you have working experience with dynamical systems, be it classical dynamics, quantum mechanics or relativity (or a mixture of all of them!). Such experience is relevant. If I were saying "Vanilla is the best ice cream flavour because 5 physics PhDs say so!" then it would be a misuse of authority but you're literally telling us that equations we not only have read but have worked with day to day are wrong. Your logic says Hamiltonian and Lagrangian mechanics doesn't work. Your logic contradicts the definition. If you were to put a moment's thought to this you'd realise it. Instead you appear to be working from the principle "Here's something I think supports my case, I'll use that". Instead you should be thinking "Why are their arguments false". I mean, come on, the first thing you should check is the definition of partial derivative. Clearly the definition says $\partial_{x}(3x+ \sin^{2}x + \ln y) = 3+\sin 2x$. That should be enough to make you think "Perhaps I'm not understanding the context and application of that equation I'm citing".

23. Yep, you keep missing $f=f(\theta, u(\theta),v(x))$

where $u=sin^2(\theta)$ and $f=3 \theta+u+v$.

The worst thing is not the quibble about the partial derivative, it is the fact that you do not understand the total derivative:

$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$

...which is nothing but eq (2) in the reference I attached earlier.

Last edited: Feb 13, 2012