# What Does This Wobbly “d” Do?

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 $source$:

<script type="text/x-mathjax-config"> MathJax.Hub.Config${tex2jax: {inlineMath: [['$','$'], ['(','$']]}}); </script>
<script src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML" type="text/javascript"> </script>

Written on October 16, 2013