# 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.

• The derivative of $x^3$ is $3 x^2$. – Travis Willse May 31 '16 at 11:46
• It comes down to how $\frac {d}{dx}$ and $\frac{\partial}{\partial x}$ handle expressions with $y$. We don't know a priori that $\frac{dy}{dx} = 0$, but by definition we have $\frac{\partial y}{\partial x} = 0$. – Ben Grossmann May 31 '16 at 11:46
• Read my answer to the following question: math.stackexchange.com/questions/1626028/… – orion May 31 '16 at 11:55
• Have you looked at the limit definitions of both? If you have not it would be instructive. There is no distinction in one dimension. – th0masb May 31 '16 at 11:56
• There is a differenece between partial and total derivatives. But both are the same in the one-dimensional case. Del is used for partial and d is used for the total derivative. Edit: for what ever reason I did not see the answer of @ChristianBlatter. So read this. – Abbraxas May 31 '16 at 19:16

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.

• $\frac{\partial^2 f}{\partial x^2}$ makes some sense for consistency with operator composition. In other words it is meant to be read like $\left ( \frac{\partial}{\partial x} \right )^2 f$. – Ian May 31 '16 at 18:58
• Also, the distinction between $d$ and $\partial$ is still sometimes useful inasmuch as it allows us to fall back on the "old" style. For instance it allows us to use the same letter for $u(t,x)$ as we use for $u(t,x(t))$; in the new style we should really define $u(t,x)$ and $y(t)$ and then introduce $v(t)=u(t,y(t))$ instead. This can be very annoying when what we are really trying to describe is a quantity (like temperature) that we would like to refer to by just one name. – Ian May 31 '16 at 19:01
• This answer, as well as the others, shows why there is still a distinction between total and partial derivatives. To that extent, all the answers are correct. This answer also makes some excellent points about the weakness of the notation, but it would be stronger if it showed specific modern mathematical notation that represents these concepts better. How would you write $\partial u/\partial t$ in your first example? (I might write $u_1$, but maybe you have a better idea.) – David K May 31 '16 at 19:13
• The discussion of the Euler-Lagrange equation $\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=0$ in Sussman and Wisdom's Structure and Interpretation of Classical Mechanics is relevant here. They point out its ambiguity and argue for the functional notation $D(\partial_2 L\circ\Gamma[q]) - \partial_1 L\circ\Gamma[q] = 0$ instead. Personally I find the latter harder to read, but that's at least partially because I'm not as used to it. – user856 May 31 '16 at 20:07
• If one one function is defined on $\mathbb R \times M$ then so is the other. The symbols $t$ and $x$ in $(t,x)\mapsto u(t,x)$ are just placeholders--they don't have any external meaning, and have meaning within the expression only according to where they occur. If the second parameter of $u$ must be of type $M$, then whatever you put in the second position after $u$ will be of type $M$. Now, if you were to write $(x,t)\mapsto u(t,x)$, that would be a different function. – David K Jun 3 '16 at 13:45

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}

• Of course not, nobody said that $y$ is a function of $x$; in both situations, they are independent variables and the OP asks why do we use "a different kind of d" when dealing with several variables. The correct answer is somewhat disappointing: for historical reasons - and nobody bothered changing this anymore. – Alex M. May 31 '16 at 14:47
• @AlexM., your "of course not" suggests that the distinction between $y$ and $y(x)$ is entirely clear, and that it's inconceivable that '$y$' could implicitly be intended to refer to a function of $x$. While this may be technically true if strictly correct notation is always used, in reality, casual use of $y$ to mean $y(x)$ is pretty common, particularly in introductory calculus, so I think the different notation does help make things a lot clearer, even if it's not strictly needed. – jst345 Jun 1 '16 at 11:39

If in a certain situation three variables $x$, $y$, $z$ are identified as "truly" independent, and are agreed on as the variables used for identifying points of the underlying "ground set" $\Omega$ then any function $f:\>\Omega\to{\mathbb R}$ appears as a function $f:\>(x,y,z)\mapsto f(x,y,z)$, and ${\partial f\over\partial x}:\>\Omega\to{\mathbb R}$ is the "partial derivative with respect to $x$" we all are fond of. If, however, in such a situation a certain quantity $u$ depending on $(x,y,z)$ plays a rôle then the expression ${\partial f\over\partial u}$ makes no sense.

To elaborate further: If in such a situation we are given a curve $$t\mapsto\bigl(x(t),y(t),z(t)\bigr)\in \Omega$$ describing the orbit of a spaceship in time then the astronaut in this spaceship feels the temperature $$\hat f(t):=f\bigl(x(t),y(t),z(t)\bigr)\ .$$ It then makes sense to talk about the ("total") derivative $${d\hat f\over dt}={\partial f\over\partial x}\dot x(t)+{\partial f\over\partial y}\dot y(t)+{\partial f\over\partial z}\dot z(t)\ .$$

The first one $$\dfrac{\partial f}{\partial x} \overset{\color{orange}{?}}{=} \dfrac{\partial(\color{orange}{(x,y)}\mapsto x^3)}{\partial x} = x\mapsto 3x^2$$ is wrong since you change the function $f$, which should be $$(x,y)\mapsto yx^3$$

In the second part

\left\{\begin{align}&\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\\&\dfrac{\partial}{\partial x}f = \dfrac{\partial f}{\partial x} \overset{\color{orange}{?}}{=} \dfrac{\partial(\color{orange}{x}\mapsto x^3)}{\partial x} = x\mapsto 3x^2\end{align}\right.

• in the first line, the notation $\frac{d}{dx}$ is incorrectly used since $f$ is a function with two variables
• in the fourth line, you make a mistake again as the very first one, namely, $x\mapsto x^3$ $$x\mapsto yx^3$$

If you want to use $\frac{d}{dx}$ for the function $f(x,y)=yx^3$, a possible way is treating $y$ as a parameter and define $$g_y(x):=yx^3.$$ Then $$\frac{d}{dx}g_y(x)=\frac{d(x\mapsto yx^3)}{dx}=y\cdot\frac{d(x\mapsto x^3)}{dx}=y\cdot(3x^2)$$

• In the line with $\color{orange}{\dfrac{\partial}{\partial x}}$ I was mainly curious if $f$ becomes a one-variable function; is it $\dfrac{\partial((x,y)\mapsto yx^3)}{\partial x}$ or $\dfrac{\partial(x\mapsto yx^3)}{\partial x}$? Given that $\dfrac{\partial}{\partial x}$ treas $y$ as a constant I was assuming the latter. I admit that I may have done the parsing a bit too quick, but whether it's $yx^3$ or $y^2x^5$ is less important. Thanks for the correction anyhow! – Frank Vel May 31 '16 at 15:29
• So $\dfrac{\mathrm{d}}{\mathrm{d}x}f \neq \dfrac{\mathrm{d}f}{\mathrm{d}x}$ when $f$ is a multi-variable function? In which case: how do I treat $\dfrac{\mathrm{d}}{\mathrm{d}x}f$ using the $f = (x_1,x_2,\cdots,x_n) \mapsto f(x_1,x_2,\cdots,x_n)$ definition? – Frank Vel May 31 '16 at 15:32
• When $n>1$, we never use the notation $\frac{d}{dx}f$. – Jack May 31 '16 at 16:35
• You might want to take a look at en.wikipedia.org/wiki/Partial_derivative#Basic_definition – Jack May 31 '16 at 16:36
• I guess I'm just a bit confused about $\dfrac{\partial ((x,y)\mapsto yx^3)}{\partial x} = x \mapsto 3yx^2$, as a variable simply vanishes... the $\mapsto$-notation seems a bit awkward here... Is it even sensible to try using it? – Frank Vel May 31 '16 at 16:45

In the following we consider real-valued functions \begin{align*} &f:\mathbb{R}\rightarrow\mathbb{R}&\text{and}\qquad\quad&g:\mathbb{R}^2\rightarrow\mathbb{R}\\ &x\mapsto f(x)&&(x,y)\mapsto g(x,y) \end{align*}

We denote with $\frac{\partial}{\partial x}$ the partial derivative of a function $f$ with respect to the variable $x$ and with $\frac{d}{dx}$ the total derivative of a function $f$ with respect to the variable $x$.

• In the one-variable case there is no difference between the total derivative and the partial derivative of $f$ with respect to $x$.

• In the multi-variable case there is in general a difference between the total and the partial derivative.



Multivariable case: We consider the total derivative of $g=g(x,y)$ with respect to $x$. It is defined as \begin{align*} \frac{dg}{dx}&=\frac{\partial g}{\partial x}\cdot\frac{dx}{dx}+\frac{\partial g}{\partial y}\cdot\frac{dy}{dx}\\ &=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\cdot\frac{dy}{dx}\tag{1} \end{align*} and this is generally not the same as \begin{align*} \frac{\partial g}{\partial x} \end{align*}

Example: $g(x,y)=yx^3$

We obtain \begin{align*} \frac{d}{dx}g(x,y)&=\frac{\partial }{\partial x}g(x,y)+\frac{\partial }{\partial y}g(x,y)\cdot\frac{dy}{dx}\\ &=\frac{\partial }{\partial x}(yx^3)+\frac{\partial }{\partial y}(yx^3)\cdot\frac{dy}{dx}\\ &=3x^2y+x^3\frac{dy}{dx}\\ \end{align*} whereas \begin{align*} \frac{\partial }{\partial x}g(x,y)&=\frac{\partial }{\partial x}(yx^3)\\ &=3x^2y \end{align*}

We observe the total derivative and the partial derivative are different in general. They are the same in the example above only when \begin{align*} \frac{dy}{dx}\equiv 0 \end{align*}

Single variable case:

In this case the total derivative and the partial derivative are the same since applying (1) in the single variable case is \begin{align*} \frac{df}{dx}&=\frac{\partial f}{\partial x}\cdot\frac{dx}{dx}=\frac{\partial f}{\partial x} \end{align*}