# Partial vs Full Derivative

1. Jul 10, 2014

### Hertz

Let's say we have a function $F(\vec{r})=F(s, \phi, z)$. Then (correct me if I'm wrong):
$\frac{dF}{dx}=\frac{\partial F}{\partial s}\frac{ds}{dx}+...$
So then what is $\frac{\partial F}{\partial x}$? Is it zero because $F$ doesn't depend explicitly on x? Is it the same as $\frac{dF}{dx}=\frac{\partial F}{\partial s}\frac{ds}{dx}+...$? Is it $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial s}\frac{\partial s}{\partial x}+...$? Is there even any difference between these last two things?

More generally:
If we have $F(\vec{r})=F(e^1(\vec{r}), e^2(\vec{r}), e^3(\vec{r}))$, then is this correct?:
$\frac{dF}{dx}=\sum_i\frac{\partial F}{\partial e^i}\frac{de^i}{dx}$
and:
$\frac{\partial F}{\partial x}=\sum_i\frac{\partial F}{\partial e^i}\frac{\partial e^i}{\partial x}$

I guess what my question is is what exactly is the difference between a partial and a full derivative? My previous understanding is that you should only take partial derivatives with respect to variables that are explicitly included in the expression, whereas you consider all implicit and explicit dependencies on a variable when you take a full derivative. However, I don't think this understanding of a partial is sufficient anymore. For example, the case above, where we are taking a partial derivative of something which doesn't depend explicitly on the variable we are considering. My past understanding would tell me that the partial derivative would be zero, but I'm fairly certain this is not the case.

2. Jul 10, 2014

### mathman

Partial derivative is used when the function depends on more than one variable. When the function depends on only one variable, the derivative is total.

3. Jul 10, 2014

### Fredrik

Staff Emeritus
There's already an issue here. If $\vec r=(x,y,z)$, and $s,\phi,z$ are cylindrical coordinates with s the distance from $(0,0,0)$, then the F on the left isn't the same function as the F on the right. Notations like $F(x(s,\phi,z),y(s,\phi,z),z)=G(s,\phi,z)$ or $F(x,y,z)=G(s(x,y,z),\phi(x,y,z),z)$ are more appropriate.

It's sometimes useful to use the same notation for both functions. For example, if the outputs of these functions are to be interpreted as the energy of a particle, then you may want to denote both functions by E. But you need to be aware that this is a potentially confusing abuse of notation.

It depends on what you mean by F. We have
$$\frac{d}{dx}F(x,y,z)=\frac{\partial F(x,y,z)}{\partial x},$$ but
$$\frac{d}{dx}G(s(x,y,z),\phi(x,y,z),z) =\frac{\partial G(s(x,y,z),\phi(x,y,z),z)}{\partial s}\frac{\partial s(x,y,z)}{\partial x}+\dots$$

When you use this notation and F is a function defined on a subset of $\mathbb R^3$, then $\partial F/\partial x$ should always denote one of the partial derivatives of $F$, i.e. $D_1F$, $D_2F$ or $D_3F$. Which one it is depends on which of the three variables that you usually denote by $x$. So you want F to be a function that takes three cylindrical coordinates as input (none of which is denoted by x), you may want to avoid the notation $\partial F/\partial x$. It is however OK to write
$$\frac{\partial}{\partial x}G\big(s(x,y,z),\phi(x,y,z),z\big).$$ When there's no function symbol in the numerator of the $\frac{\partial}{\partial x}$ expression, it's convenient to interpret the notation like this:
The value of the first partial derivative of the function $(a,b,c)\mapsto G\big(s(a,b,c),\phi(a,b,c),c\big)$ at the point (x,y,z).​
But you can also interpret it as
The value of the third partial derivative of the function $(a,b,c)\mapsto G\big(s(c,b,a),\phi(c,b,a),a\big)$ at the point (z,y,x).​
That's the same number. If you change the $\partial/\partial x$ to $d/dx$, the interpretation is
The value of the derivative of the function $t\mapsto G\big(s(t,y,z),\phi(t,y,z),z\big)$ at the point x.​
But it's still the same number.

Again, these two F's aren't the same function. Let's say that we have $F(\vec{r})=G(e^1(\vec{r}), e^2(\vec{r}), e^3(\vec{r}))$. Then you'd have to change F to G in your calculation. The $de^i/dx$ in the first line should be written with the partial derivative notation, since the $e_i$ are defined on a subset of $\mathbb R^3$, not a subset of $\mathbb R$. You could argue that since $\frac{d}{dx}e^i(x,y,z)=\frac{\partial}{\partial x}e^i(x,y,z)$, it doesn't really matter, but I think this is one of those times when the partial derivative notation is very appropriate.

I actually don't like either of the left-hand sides, even if you change F to G. The notation $dG/dx$ makes it look like G is defined on a subset of $\mathbb R$, and the notation $\partial G/\partial x$ makes me wonder if we're talking about $D_1G$, $D_2G$ or $D_3G$ when we're in fact not talking about either of them. So I would write the left-hand side as $\frac{\partial}{\partial x}G$.

The distinction between $\frac{\partial G}{\partial x}$ and $\frac{\partial}{\partial x}G$ that I'm recommending is not standard. (If anything is standard, it's to use ambiguous and confusing notation). I'm just describing how I had to think about these things to make sense of them.

Last edited: Jul 10, 2014
4. Jul 11, 2014

### Hertz

Ok, I see.

So then $\frac{\partial s}{\partial x}=\frac{ds}{dx}$ because s depends explicitly on x?
And is $\frac{d}{dx}G(s(x, y, z), \phi(x, y, z), z(x, y, z)=\frac{\partial}{\partial x}G(s(x, y, z), \phi(x, y, z), z(x, y, z)$?
What is a case where $\frac{\partial F}{\partial x}≠\frac{dF}{dx}$?

The reason I'm asking is because I'm trying to understand the gradient in other coordinate systems better. The gradient is defined in terms of partial differentiation operators right? So, if you are given a function that is defined in terms of cylindrical coordinates, you treat the cylindrical coordinates as functions of Cartesian position right? And then you take the partial derivative with respect to a Cartesian coordinate as we discussed above right?

I'm sorry, I'm not very familiar with this notation. What exactly do you mean by #2 and what is the difference between #1 and #3?

Ok.

Last edited: Jul 11, 2014
5. Jul 11, 2014

### Fredrik

Staff Emeritus
I would avoid the latter notation entirely, since it makes it look like the domain of s is a subset of $\mathbb R$ rather than a subset of $\mathbb R^3$. But with the notation I suggested (which makes a distinction between $\frac{d}{dx}f(x)$ and $\frac{df(x)}{dx}$), we do have $\frac{\partial s(x,y,z)}{\partial x}=\frac{d}{dx}s(x,y,z)$.

Let $f$ be an arbitrary differentiable function from $\mathbb R^2$ into $\mathbb R$. Let $g$ be an arbitrary differentiable function from $\mathbb R$ into $\mathbb R$.

$$\frac{d}{dx}f(x,g(x))=D_1f(x,g(x))+D_2f(x,g(x))g'(x)=\frac{\partial f(x,g(x))}{\partial x}+\frac{\partial f(x,g(x))}{\partial y}g'(x).$$ In the notation that I will explain below, the left-hand side is equal to $\big(u\mapsto f(u,g(u))\big)'(x)$ and the first term on the right is equal to $\big(u\mapsto f(u,g(x))\big)'(x)$.

If you don't understand this notation, read the rest of this post, in particular the explanation of the $\mapsto$ notation at the end, and then read the above again.

Yes, that sounds right. If you want to discuss the details of a proof here, you can post the proof or a reference to a book that contains it.

The $\mapsto$ notation gives us a way to define new functions from the ones we're already familiar with, and refer to the new functions without inventing notations for them. For example, suppose that $f:\mathbb R^2\to\mathbb R$. Let $x\in\mathbb R$ be arbitrary. The notation
$$y\mapsto f(x,y)$$ refers to the function $f_x:\mathbb R\to\mathbb R$ defined by $f_x(y)=f(x,y)$ for all $y\in\mathbb R$.

It's actually a bit sloppy to just mention "the function $y\mapsto f(x,y)$". We should also specify the domain of the function. So it's better to say "the function $y\mapsto f(x,y)$ with domain $\mathbb R$".

For all $y\in\mathbb R$, the value of this function at $y$ is $f(x,y)$. Note that in this statement, y can be swapped for any other variable, since it's the target of a "for all". So the statement is equivalent to this: For all $z\in\mathbb R$, the value of this function at $z$ is $f(x,z)$. Because of this, the notation $z\mapsto f(x,z)$ means exactly the same thing as $y\mapsto f(x,y)$.

The value of the third partial derivative of the function $(a,b,c)\mapsto G\big(s(c,b,a),\phi(c,b,a),a\big)$ at the point (z,y,x).​
The explanation goes like this: Define the function $H:\mathbb R^3\to\mathbb R$ by $H(a,b,c)=G\big(s(c,b,a),\phi(c,b,a),a\big)$ for all $a,b,c\in\mathbb R$. Compute its third partial derivative $D_3H$, i.e. the "partial derivative with respect to the third variable". Then see what you get when you plug (z,y,x) into this function. In other words, compute $D_3H(z,y,x)$. Note that
\begin{align}
D_3H(z,y,x)&=\lim_{h\to 0}\frac{H(z,y,x+h)-H(z,y,x)}{h}\\
&=\lim_{h\to 0}\frac{G\big(s(x+h,y,z),\phi(x+h,y,z)+z\big)-G\big(s(x,y,z),\phi(x,y,z),z\big)}{h}\\
&=\frac{\partial}{\partial x}G\big(s(x,y,z),\phi(x,y,z),z\big).
\end{align} You asked about the difference between #1 and #3. They are both the same number, but the former is computed as the value of a function from $\mathbb R^3$ to $\mathbb R$, at the point (x,y,z) in its domain, and the latter is computed as the value of a function from $\mathbb R$ to $\mathbb R$, at the point x in its domain.

Perhaps the following example is the best way to answer your questions both about the $\mapsto$ notation and about the difference between partial derivatives and ordinary derivatives. If $f:\mathbb R^3\to\mathbb R$, then we have
$$D_3f(x,y,z)=\lim_{h\to 0}\frac{f(x,y,z+h)-f(x,y,z)}{h} =\big(u\mapsto f(x,y,u)\big)'(z).$$ The third partial derivative of f is the function $D_3f:\mathbb R^3\to\mathbb R$, but the value of that function at $(x,y,z)$ is equal to the value of the ordinary derivative of the function $u\mapsto f(x,y,u)$ at u.

This can of course be said without the $\mapsto$ notation as well. It's just a bit awkward to have to stop and state a definition like this: For each $x,y\in\mathbb R$, define $f_{x,y}:\mathbb R\to\mathbb R$ by $f_{x,y}(z)=f(x,y,z)$ for all $z\in\mathbb R$. But once we have stated this definition, we can write the above as $D_3f(x,y,z)=f_{x,y}{}'(z)$.

Unfortunately, a lot of people are unfamiliar with the $\mapsto$ notation, so I pretty much always end up having to explain it here at PF. But I find it very useful when I do calculations.

6. Jul 12, 2014

### TSC

Yes, the last notation is the most accurate one. D1, D2... or f1, f2 ... It is the parameter position or place holder location that is important