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I am currently reading up on partial derivatives and differentials in general. And there are a few points that seem unlcear to me (notation-wise).

  1. For example, if $f:\mathbb R\to\mathbb R,x\mapsto f(x)$ is a function, then is the following notation correct? $$\frac{d}{dx}f(x)=\frac{df}{dx}(x)=\frac{\partial}{\partial x}f(x)=\frac{\partial f}{\partial x}(x)=f'(x)$$

  2. Now, in one of my lectures we wrote $$g(x):=\frac{\partial \log (f(x))}{\partial x}=\log'f(x)\cdot f'(x)=f'(x)/f(x)$$ The part about differentiating the function is clear, using the normal chain rule $(h\circ f)=(h'\circ f)\cdot f'$. What confuses me a bit is the notation since $x$ seems to have more than one meaning. Would it be "more" correct to write $$g(x):=\frac{\partial \log (f(y))}{\partial y}\Bigg|_{y=x}$$ as in we differentiate w.r.t. $y$ and then plug in $x$ for $y$? Also, in this particular case, could we replace $\partial$ by $d$, or would this lead to different implications?

  3. And lastly, what does $df$ or $dy$ even mean? I read that the differential of $y=f(x)$ is defined to be $dy=f'(x)dx$. But how can we interpret this formula? This becomes specifically confusing when I look at the differential chain rule $$\frac{dh}{df}=\frac{dh}{dg}\cdot\frac{dg}{df}$$ What I find strange is that one seems to be using $g$ as a function as well as a variable. How does this work exactly?

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  • $\begingroup$ Ooops! It was confusing me for mostly 2 or 3 years! :D I am so glad that someone had the same question as mine! :) $\endgroup$ – H. R. Nov 10 '15 at 12:52
  • $\begingroup$ See this post for the third question. :) $\endgroup$ – H. R. Nov 10 '15 at 12:59
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First:

Yes, when you are dealing with a function $f$ of one real variable $x$, the partial derivative $\frac{\partial f}{\partial x}$ coincides with the total derivative of $f$ with respect to $x$. Be ware that those are generally two different things. They only coincide for functions that have purely explicit relations to the variable $x$ in question. That is, the other variables do not depend on $x$.

Intuitively the total derivative of $f$ measures how $f$ changes along the direction $x$ as all its variables vary. The partial derivative measures how $f$ changes along $x$ when the other parameters do not vary. You will soon learn about the derivative along a vector field or Li derivative. This is the real nice one as it does not depend on your choice of coordinates $x_1,\dots,x_n$

Second:

After a brief council with Git Gud I hereby revise the second section.

The symbol $\frac{\partial f(x)}{\partial x}$ is not meaningless. The following abstract nonsense is well defined. Consider $\frac{\partial f(x)}{\partial x}$ to be the derivative of $f$ with respect to a constant $x$ evaluated at that constant. Note that the derivative of any function with respect to a constant always turns out to be the empty function. In other word each value is the empty set. Now the symbol $\frac{\partial f(x)}{\partial x}$ stands for the empty set, hence the formula in question is well formed.

Your lector must have had something similar in mind or the formula he wrote is not well formed. Even within this context it is false. You have the right to be confused, but only ask him politely about it.

Third:

It is strange at first but when you think about it, you realize the need for your parameters to depend further on some other quantities. Consider a change of variables. Your new and your old variables are related to each other... Now if you define the differential $df$ by $$df(x_1,x_2,\dots,x_n)=\sum_{i=1}^n\frac{\partial f}{\partial x_i}dx_i$$ Then this is what the differential means. This meaning becomes more sensible when you study differential forms in differential geometry. In laymen’s terms, the differentials also called differential $1$-forms are just the smooth covector fields.

EDIT: (Answer to the comments)

The symbol $\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial x}$ denotes the product of the partial derivative of $f$ with respect to $x_1$ with the partial derivative of $x_1$ with respect to $x$. Here $f$ is a twice smooth function of $x$ and $x_1$. You seem to be having trouble allowing the arguments $x$ and $x_1$ to depend on each other. This is an abstraction you have to allow.

I hope the concept becomes clear through examples.

a) Most often, there are no relations between the arguments indeed. Then $\frac{\partial x_i}{\partial x}=0$ and $\frac{\partial x}{\partial x_i}=0$. Then the total derivative coincides with the partial derivative.

b) $\quad x'=\frac{dx}{dx}=\frac{\partial x}{\partial x}=1$. Then $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}$

c) Imagine $x_1$ as a function of $x$, for instance $x_1(x)=x^2$. Now define $f$ by $f(x,x_1)=x+x_1(x)$. This is a smooth function of $x$ and of $x_1$. Also $$\frac{df}{dx}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial x}=1+2x$$

d) Using the above notation consider the restrictions of $x_1$ and of $f$ to the non-negative part of the reals. Then $x=\sqrt x_1$ and $$\frac{df}{dx}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial x}=1+\frac{1}{2\sqrt x_1}$$

e) More generally consider a smooth function $f(x_1,\dots,x_n)$ and a relation $R(x_1,\dots,x_n)=0$ such that the function $R$ is also smooth. Now you may talk about $\frac{\partial x_i}{\partial x_j}$ for you can always solve the relation for $x_i$ as a function of the rest. This is as clear as it can get.

f) Consider also a change of variables.

g) Consider implicit functions.

h) Foremost I recommend reading about constraints and holonomic systems. This makes the concept more sensible... Wikipedia may not be a good source. I recommend E. T. Whittakers Analytical Dynamics and Arnolds Mathematical methods of classical mechanics

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  • $\begingroup$ Sorry for asking this, but what does $\frac{\partial f}{\partial x}$ and $\sum_{i=1}^n\frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial x}$ mean exactly (again, this is just about the notation)? As for the second part as well, when $g(x)$ is defined then we evaluate the function at $x$ but we also differentiate w.r.t. $x$ (= the two meanings i see). $\endgroup$ – Phil-ZXX Nov 24 '13 at 20:10
  • $\begingroup$ You should really take a look at this page first. Let me know if you still have questions. $\endgroup$ – superAnnoyingUser Nov 24 '13 at 20:22
  • $\begingroup$ Okay, I phrased my question poorly. I am just wondering how to understand $\frac{\partial f}{\partial x_i} \frac{\partial x_i}{\partial x}$. I guess the $\partial x_i$ terms do not simply cancel, but how do I differentiate the variable $x_i$ by $x$ anyway? $\endgroup$ – Phil-ZXX Nov 27 '13 at 15:11
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Let $x\in \Bbb R$.

  1. I find the red notations below pedagogically problematic $$\color{red}{\frac{d}{dx}f(x)}=\frac{df}{dx}(x)=\color{red}{\frac{\partial}{\partial x}f(x)}=\frac{\partial f}{\partial x}(x)=f'(x),$$ but they are OK just as long as you don't look at $\color{red}{f(x)}$ as $f$ evaluated at $x$, but rather look at $\dfrac{d}{dx}f(x)$ and $\dfrac{\partial}{\partial x}f(x)$ as the derivative of $f$ evaluated at $x$.
    It should be mentioned that $d$ is usually used when it is a function of one variable and $\partial$ when it is a function of more than one variable.
  2. You're absolutely right. The notation $\dfrac{\partial \log (\varphi(x))}{\partial x}$ is simply wrong. Just look at the notations on item 1. Does it look like any of them? No. Even if you add the notation $\dfrac{\partial f(x)}{\partial x}$ to the list (and this is also a notation I deem bad) and interpret it as the derivative of $f$ at $x$, it still doesn't concurr with $\dfrac{\partial \log (\varphi(x))}{\partial x}$. What would your $f$ be here? If you're consistent, it would be $f=\log$, but then $\dfrac{\partial \log (\varphi(x))}{\partial x}$ would be, according to our interpretation, the derivative of $\log$ evaluated at $\varphi(x)$, i.e., $\dfrac{1}{\varphi(x)}$. What it is intended is $f=\log \circ \varphi$, but then the notation should be $\dfrac{\partial (\log \circ \varphi)(x)}{\partial x}$ which something completly different. With the notation $\dfrac{\partial (\log \circ \varphi)(x)}{\partial x}$ you do get $\dfrac{\partial (\log \circ \varphi)(x)}{\partial x}=\dfrac{\varphi'(x)}{\varphi(x)}$ as wanted.
    Indeed it would be more correct to write $\dfrac{\partial \log (f(y))}{\partial y}\Bigg|_{y=x}$, but not entirely correct because, once again, the notation isn't on the list you gave on 1. After you add notation $\dfrac{\partial f(x)}{\partial x}$ and give it a proper interpretation, then yes, it is correct.
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    $\begingroup$ I refrained from commenting on 3. because I don't know about differentials. $\endgroup$ – Git Gud Nov 24 '13 at 21:02
  • $\begingroup$ Your 1. implicitly suggest that for a map $f:\mathbb{R}\to\mathbb{R}$, we should parse $\frac{df}{dx}(x)$ as $\left(\frac{df}{dx}\right)(x)$ and that hence $\frac{df}{dx}$ by itself denotes the derivative of $f$. But $f$, being a map, "doesn't know anything about the name of its input". Hence the notation $\frac{df}{dx}$ is ill defined. A related discussion was at the nForum What is a variable?. $\endgroup$ – Michael Bächtold Aug 23 '17 at 8:18
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    $\begingroup$ In case you're interested in understanding the history of how we got to these standard abuses of notation I suggest the following two discussion: Who first considered the $f$ in $f(x)$ as an object in itself, and who decided to call it a function? and When did the abuse of notation $y=y(x)$ start?. $\endgroup$ – Michael Bächtold Aug 23 '17 at 8:19
  • $\begingroup$ @MichaelBächtold, I've seen that sentiment expressed elsewhere, such as in Jakub Marian's thesis about single-variable calculus notation, but it is not exactly ubiquitous. What is your take on Student's answer above that that \frac{df(x)}{dx} is actually the empty set? $\endgroup$ – hyperum Feb 5 '18 at 2:42
  • $\begingroup$ @AaronCruz: I haven't seen a student answer that $\frac{df(x)}{dx}$ is the empty set, so I'm not sure what a student who answers that might have been thinking. In case a student says $\frac{df(x)}{dx}$ is zero since f(x) is a constant, then I'd try to explain that f(x) is not a constant. But from a modern set theoretic perspective my explanation wouldn't quite make sense. Since for $x\in \mathbb{R}$ the notions constant and variable are not defined. $\endgroup$ – Michael Bächtold Feb 5 '18 at 9:20

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