What is the difference between $\frac{\mathrm{d}}{\mathrm{d}x}$ and $\frac{\partial}{\partial x}$?

Is there not any difference between $\frac{\mathrm{d}}{\mathrm{d}x}$ and $\frac{\partial}{\partial x}$ as long as your function has one variable?

$f(x) = x^3\implies \left\{\begin{align}&\dfrac{\mathrm{d}}{\mathrm{d}x}f = \dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{\mathrm{d}(x\mapsto x^3)}{\mathrm{d}x} = x\mapsto 3x^2&\color{green}{\checkmark}\\&\dfrac{\partial}{\partial x}f = \dfrac{\partial f}{\partial x}= \dfrac{\partial(x\mapsto x^3)}{\partial x} = x\mapsto 3x^2&\color{green}{\checkmark}\end{align}\right.$

And if so, why does this change with two (or more) variables?

$\require{cancel} f(x,y) = yx^3\implies \left\{\begin{align}&\color{grey}{\cancel{\dfrac{\mathrm{d}}{\mathrm{d}x}f = }} \dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{\mathrm{d}((x,y)\mapsto yx^3)}{\mathrm{d}x} \neq x\mapsto 3yx^2&\color{green}{\checkmark}\\&\dfrac{\partial}{\partial x}f = \dfrac{\partial f}{\partial x} =\dfrac{\partial((x,y)\mapsto x^3)}{\partial x} = x\mapsto 3yx^2&\color{red}{\mathcal{X}}\end{align}\right.$

I get that it is supposed to be something like this

$f(x,y) = yx^3\implies \left\{\begin{align}&\color{grey}{\cancel{\dfrac{\mathrm{d}}{\mathrm{d}x}f = }} \dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{\mathrm{d}((x,y)\mapsto yx^3)}{\mathrm{d}x} \neq\\&\cdots\quad (x,y)\mapsto 3y\dfrac{\mathrm{d}\color{red}{(x\mapsto x^3)}}{\mathrm{d}x}+\dfrac{\mathrm{d}\color{red}{(y\mapsto y)}}{\mathrm{d}x}x^3 =\\&\cdots\quad (x,y)\mapsto 3y\color{red}{(x\mapsto x^2)}+\dfrac{\mathrm{d}\color{red}{(y\mapsto y)}}{\mathrm{d}x}x^3&\color{green}{\checkmark}\\&\dfrac{\partial}{\partial x}f = \dfrac{\partial f}{\partial x} = \dfrac{\partial(x,y)\mapsto x^3)}{\partial x} = x\mapsto 3yx^2&\color{green}{\checkmark}\end{align}\right.$

Solution 1:

The notation $\dfrac {\partial}{\partial x}$ indicates that all variables other than $x$ should be treated as constant, whereas $\dfrac{d}{dx}$ would treat the other variables as exactly that: variable.

Thus for $f(x,y)=yx^3$, we have $$ \begin{align} \frac{\partial f}{\partial x}&=3yx^2 \\ \frac {df}{dx}&=3yx^2+ \frac {dy}{dx}x^3 \\ \end{align} $$

Solution 2:

Neither of the answers given so far is correct. The correct answer is somewhat disappointing: we use $\partial$ instead of $\Bbb d$ purely for historical reasons.

Back in the 18th century, mathematicians were not as rigorous as today. French mathematicians, in particular, working on what we call today "partial differential equations", encountered the following problem: imagine that you have a quantity $u$ that depends on position $x$ and on time $t$ (in modern parlance, you're talking about a smooth function $(t,x) \mapsto u(t,x)$);

• you first need a notation meaning "the derivative of $u$ with respect to $t$";

• alternatively, you may evaluate $u$ on some trajectory $t \mapsto x(t)$ and derive this quantity (which in modern parlance is $t \mapsto u(t,x(t))$) with respect to $t$, so you need a notation for "the derivative of $u(t,x(t))$ with respect to $t$.

Where is the problem, then? The problem resides in the fact that back then the concept of "function" did not exist, therefore often times mathematicians used to write $u(t,x)$ instead of $u(t,x(t))$ (i.e. they were using $u(t,x)$ for both $u(t,x)$ and for $u(t,x(t)$). In this case, using the notation $\frac {\Bbb d} {\Bbb d t}$ would have created confusion (you may still encounter this ambiguity in books about mechanics written in the '60s -yes!-, especially in many Soviet ones). Therefore, they decided to come up with the notation $\frac {\partial} {\partial t}$ for the first case above, keeping the old one ($\frac {\Bbb d} {\Bbb d t}$) for the second.

The inventor of this "curly d" was Legendre, who wrote: "Pour éviter toute ambiguité, je répresenterai par $\frac {\partial u} {\partial x}$ le coéfficient de $x$ dans la différence de $u$, & par $\frac {\Bbb d u} {\Bbb d x}$ la différence complète de $u$ divisée par $\Bbb dx$." ("In order to avoid all ambiguity, I shall represent by $\frac {\partial u} {\partial x}$ the coefficient of $x$ in the difference of $u$, & by $\frac {\Bbb d u} {\Bbb d x}$ the complete difference of $u$ divided by $\Bbb dx$.") What Legendre says is that he considers a Taylor expansion of order $1$ of $u$ around some $(x_0,y_0)$, and the coefficient of $x - x_0$ will be called $\frac {\partial u} {\partial x}$, in order to distinguish it from the coefficient of $x-x_0$ in the Taylor expansion of order $1$ of $u(x,y(x))$ - which would be denoted $\frac {\Bbb d u} {\Bbb d x}$. As you can see, everything was meant to resolve an ambiguity in notation, ambiguity that disappeared with the birth of modern mathematics and its new, more rigorous notations.

Why keep it then, anymore? Bluntly put - for historical reasons and laziness. Why change it? This change, if done, should be adopted by every country, and students should be taught both the "old" version (in order to be able to read the literature published so far), and the "new" one. Well, a bit of life wisdom tells you that it's very difficult to make all humans accept one decision - plus that it's not really an important one, and it's nice to carry with us this piece of living history (who doesn't love history and old things?).

Is this the only oddity kept until the present time? No, there are many others. Here are two more: why don't we write the simpler $\dfrac {\partial f} {\partial x_1 ^{i_1} \dots \partial x_n ^{i_n}}$ instead of the more complicated $\dfrac {\partial ^{i_1 + \dots + i_n}f} {\partial x_1 ^{i_1} \dots \partial x_n ^{i_n}}$? And why do we write $\frac {\partial ^2 f} {\partial x^2}$ instead of $\frac {\partial ^2 f} {\partial ^2 x}$ (two things that confuse many students upon first encounter)? Again, for historical reasons, that nobody bothered correcting anymore.