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The total derivative as I understand it would be a linear map along the lines on the wikipedia page. https://en.wikipedia.org/wiki/Total_derivative

However in some instances it looks alot like a chain rule or people use the chain rule to compute it.

I dont see why we would assume that the variables are functions of some common parameter $t$ when we talk about total derivative.

Why are these thing being mixed everywhere? Are they related somehow?

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    The first sentence in the Wikipedia article mentions that all the variables of $f$ that are not $t$ are still assumed to be functions of $t$. The second paragraph, right before the chain-rule-using formula says explicitly $f(t, x(t), y(t))$. – Arthur Jan 10 '18 at 08:54
  • @Arthur right, and I wounder why they even consider this case. It donst seem to be the general situation when we think of a function of several variables. –  Jan 10 '18 at 08:58
  • @Arthur it looks more like a chain rule discussion. The chain rule is the chain rule and it treats derivaitves of composite functions while a total derivaitive should discribe complete behaviuor of some function. I dont see why they keep mixing the two ideas and I wounder if I am missing something which makes it reasonable to mix them. –  Jan 10 '18 at 09:00
  • It's a pretty common thing in different applications. For instance, the entire field of Lagrangian mechanics in physics cares about integrating and differentiating certain functions of time, position and velocity ("kinetic energy minus potential energy" is a common one). But position and velocity are also functions of time, and thus you have this exact setup. – Arthur Jan 10 '18 at 09:01
  • @l33tg33k: The notion you seem to be looking for is the differential $\mathrm{d}f(t, x, y)$, which is the most natural notion of derivative for a multivariable function, and will be a linear combination of $\mathrm{d}t$, $\mathrm{d}x$, and $\mathrm{d}y$. (the coefficients of the linear combination can be computed by partial derivatives). e.g. $\mathrm{d}(xy) = x \mathrm{d}y + y \mathrm{d}x$, no matter what $x$ and $y$ are and how they might be related to each other and other variables (so long as they are differentiable, of course). –  Jan 10 '18 at 09:44
  • @Hurkyl I dont do differentials after this post https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio –  Jan 10 '18 at 10:08
  • @l33tg33k: Too bad. Avoiding mathematical objects that directly express what you are trying to work with in favor of more awkward/cumbersome approaches is just as bad of a mistake as the one of using objects wrongly in ignorance. –  Jan 10 '18 at 10:20
  • @Hurkyl not imo. I get stuck if I dont know what I am thinking about. –  Jan 10 '18 at 10:31
  • @l33tg33k: Sure, but cast in that light my point was that differentials are something worth learning about how to use correctly, rather than a topic to be avoided. –  Jan 10 '18 at 11:21

1 Answers1

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It is just a special case for the chain rule with

$$f=f(t,x(t),y(t),...)$$

More precisely, total derivative is a special case of composition $f\circ g:\mathbb{R}\to\mathbb{R} $ with $f:\mathbb{R^n}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R^n}$.

Indeed, in a more general case with

$$f=f(x(u,v),y(u,v),z(u,v)....)$$

we can apply chain rule to evaluate partial derivatives of $f$ with respect to $u$ and $v$

$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u}+ \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial u}+...$$

$$\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial v}+ \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial v}+...$$

In this special case, since $x,y,...$ are function of one variable $t$, we have

$$\frac{d f}{d t} =\frac{\partial f}{\partial t} = \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial t} + \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial t}+...=\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \cdot \frac{dx}{dt}+ \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}+...$$

we denote this special case as total derivative of $f$ with respect to $t$.

user
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  • not evey function of several variables looks like that? –  Jan 10 '18 at 09:03
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    I would rather say that one can prove the chain rule using total derivatives. –  Jan 10 '18 at 09:04
  • It is a special case in which f depends by several variables t,x,y,... abd these variables can all be expressed as fuctions of t – user Jan 10 '18 at 09:07
  • why would that be on a page on the total derivative, that is what I dont understand. And my question is why this is the case –  Jan 10 '18 at 09:11
  • Note that chain rule is a more general result for function composition. – user Jan 10 '18 at 09:12
  • well yea I suppose we need a chain rule for vector valued functions of vector arguments. –  Jan 10 '18 at 09:13
  • You need to better clarify the general case to understand this special one. I will give some other details as soon I can. – user Jan 10 '18 at 09:15
  • $f$ is scalar valued here? And the total derivative is regarding after we taken the scalar product with some other vector? –  Jan 10 '18 at 09:46
  • Yes for this case $f:\mathbb{R^n}\to\mathbb{R}$ with respecto to $t,x,y,z,...$ and it becames $f:\mathbb{R}\to\mathbb{R}$ when composed with $g:\mathbb{R}\to\mathbb{R^n} \quad t\to(t,x(t),y(t),...)$ – user Jan 10 '18 at 09:50
  • so this paritcular total derivative is the compostion of two total derivatives which is a new one..thanks! I dont see how that makes it simpler tho, it is really messy. It makes one wounder what the total derivaive really is since it looks like something which is not a matrix i.e a sum when presented in this way. –  Jan 10 '18 at 09:53
  • More precisely: total derivative is a special case of composition $f°g:\mathbb{R}\to\mathbb{R} $ with $f:\mathbb{R^n}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R^n}$. – user Jan 10 '18 at 09:56
  • But of course you can also see it as the application of chain rule for function of one variable: $f(t,x(t),y(t),z(t),...)$ if you are not skilled with multivariable calculus. – user Jan 10 '18 at 09:58
  • regarding the "more precisely" comment. Can one always consider vector valued function of a vector arguement in this way? –  Jan 10 '18 at 10:01
  • @l33tg33k Sorry, I didn't understand the last question – user Jan 10 '18 at 10:03
  • well as you say that total derivaives are a special case of the compostion $f g$. Do you think then that any variables of $g$ can be parametrized by some $t$ and then take $f$ to be a scalar function. in this way "inbedd" the vector function? –  Jan 10 '18 at 10:06
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    Yef if $f$ is $f:\mathbb{R^n}\to\mathbb{R}$ you can alway parametrize as a function of a single variable t. – user Jan 10 '18 at 10:11