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

…is a question I get surprisingly often when tutoring friends.

**Short version:** The difference is all about dependency!

The “regular d” in $\frac{d}{d x}$ denotes ordinary differentiation: assumes all variables are dependent on $x$ ($\rightarrow$ envoke chain/product rule to treat the other variables as functions of $x$).

The “wobbly d” in $\frac{\partial}{\partial x}$ denotes \partial differentiation and assumes that all variables are independent of $x$.

You can make a straight d from a wobbly d by using a beautiful thing:

$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} + \frac{\partial f}{\partial z}\frac{dz}{dx}$

You can remember it this way: the \partial derivative is \partially a “d”, or the the “wobbly d” is \partial to nemself and bends nir neck down to look at nirs reflection $\partial$.

*Sidenote: Ne/nem/nir/nirs/nemself is a pretty swell gender neutral pronoun set!*

**Longer version: **Before we start – what is a derivative, anyway?

The derivative of a function at a chosen point describes the linear approximation of the function near that input value. Recall the trusty formula: $\Delta y = m\Delta x$, where m is the slope and $\Delta$ represents the change in the variable? For a (real-valued) function of a (real) single variable, the derivative at that point is = tangent line to the graph at that point.

The beauty of math is: $m = \frac{\Delta y}{\Delta x} equiv$ “How much one quantity (the function) is changing in response to changes in another quantity ($x$) at that point (it’s input, assuming $y(x)$).”

So, it makes sense that derivative of any constant is 0, since a constant (by definition) is constant $\rightarrow$ unchanging!

$\frac{d(c)}{dx}=0$, where $c$ is any constant.

What about everything that isn’t a constant?

That means, $\frac{d(x^2)}{dx} = 2x$, since $x^2$ is dependent on $x$. $\frac{d}{dx}$ denotes ordinary differentiation, i.e. all variables are **dependent** on the given variable (in this case, $x$).

But what about $\frac{d(y)}{dx}$? Looking at this equation, we immediately assume $y$ is a function of $x$. Otherwise, it makes no sense. $\frac{d(y)}{dx} equiv \frac{d(y(x))}{dx}$

On the other hand, $\frac{\partial (y)}{\partial x}$ denotes \partial differentiation. In this case, all variables are assumed to be **independent**.

$\frac{\partial (y)}{\partial x} = 0$

Let’s compare them with an example. $f(x,y) = ln(x)sec(y) + y$

$\frac{\partial f}{\partial x} = \frac{sec(y)}{x}$

$\frac{df}{dx}$ implies that $y$ is dependent(a function of) $x$, i.e. $f = (ln(x)sec(y(x)) + y(x))$

$\frac{df}{dx} = \frac{dy}{dx}(ln(x)tan(y(x))sec(y(x)) + 1) + \frac{sec(y(x))}{x}$

As you can see, $\frac{\partial f}{\partial x} \neq \frac{df}{dx}$.

By the way, to use LaTex in Blogger include the following before `</head>`

in your Template (source):

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