# What is a derivative, anyway?

#### Prerequisites

There are a lot of different ways of looking at a derivative. Basically, it is a rate of change — we have something ($f$) that depends on something else ($x$).

We can look at the derivative in a number of different ways.  It's useful to know them all and to see how the different views complement each other.

• Derivative as a ratio
• Derivative on a graph
• Derivative as algebra
• Derivative as a stepping rule

## Derivative as a ratio

The basic and most important definition of the derivative for use in science is the derivative as a ratio. It answers the question, "If I make a small change in an independent variable, how much will something that depends on it change?"  An example is the position along a line, $x$, which might change with time, $t$.  So when $t$ changes its value by a little bit, $\Delta t$, what's the change in $x$, $\Delta x$We define the ratio of these two changes to be the derivative when the time change is small -- smaller than anything we care to pay attention to. (If it's big, then we have an "average rate of change" over the big time interval, and we don't call that a derivative.)

$$t \rightarrow t + \Delta t$$

$$x \rightarrow x + \Delta x$$

$$\frac{dx}{dt} \approx \frac{\Delta x}{\Delta t}$$

## Derivative on a graph

We can also look at the derivative as something on a graph. If we pick a particular point, say $t = 0$, as shown by the red arrow on the left, we go to a time a little before that and a little after and find what x's they correspond to. The horizontal red line is the time interval, the vertical red line is the position change during that time interval. The ratio, $\frac{\Delta x}{\Delta t}$, becomes the derivative when $\Delta t$ gets small. We can plot the value of the ratio (the velocity = the derivative) as a function of time.  For the case shown below, the rate of change of position with time is the same for all times shown, so the velocity in that time range is a constant. (Note that the graph on the left uses a little time interval -- creating a triangle -- while the graph on the right shows the calculated velocities at points -- single times. This is because our triangle should really be "smaller than any time interval we care to pay attention to". See more discussion of this on the page instantaneous velocity.)

## Derivative as algebra

We can see how a derivative arises from algebra when we know the functional dependence of the dependent variable on the independent variable.  We can't actually do this when we have quantities that have units without doing some rearranging of our equations (for a more advanced analysis see the discussion of creating dimensionless equations), since it is nonsense to say, for example "that $(\Delta t)^2$ is small compared to $\Delta t$ when $\Delta t$ is small" since $\Delta t$ and $(\Delta t)^2$ are different kinds of quantities and cannot be compared. (Which is why we put a "strikethrough" on the phrase so you don't miss that it's wrong.)

So let's consider a function of a pure number that is itself a pure number -- say $y = f(x)$ -- where both $x$ and $y$ have no units. Let's take an example:

$$y = f(x) = x^3$$

If we want the derivative, we want to see how much $y$ changes when we make a small change in $x$.  So if we let $x \rightarrow x + \Delta x$, then what happens to $y$? Well, we can figure it out. It's a bit messy, but straightforward.

$$y \rightarrow f(x+\Delta x) = (x + \Delta x)^3 = x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3$$

Since $\Delta x$  is a pure number, if it is small (much less than 1) then we can compare it to $(\Delta x)^2$.  The square (and the cube) of $\Delta x$ will be much smaller than $\Delta x$ (try $\Delta x = 0.01$ for example) so we can ignore them, giving:

$$y \rightarrow f(x+\Delta x) = (x + \Delta x)^3 \approx x^3 + 3x^2(\Delta x)$$

This means that the change in y is only the last term so

$$\Delta y = 3x^2\Delta x$$

and the derivative is the ratio of $\Delta y$  to $\Delta x$ when $\Delta x$   is very small; or

$$\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x} = 3x^2$$

which is the familiar rule ($\frac{d}{dx}x^N = Nx^{N-1})$.

## Derivative as a stepping rule

One of the really interesting ways of expressing a derivative is as a "stepping rule".  Since a derivative relates two changes, if we open them up -- say writing a change in $x$  as $\Delta x = x_f - x_i$ where the "f" subscript means "final" and the "i" subscript means "initial" -- then using the definition of derivative in terms of change (we'll use the position-velocity example so as to keep units in the discussion), we can solve for the final position, given the initial velocity, the initial time, and the time interval. Here's the math.

$$v = \frac{dx}{dt} \approx \frac{\Delta x}{\Delta t}$$

$$\Delta x = v\Delta t$$

$$(x_f - x_i) = v(t_f - t_i)$$

$$x_f = x_i + v(t_f - t_i)$$

So if we know about $x$ at an initial time $t_i$, and we know the velocity, then we can figure out what $x$ is going to be a little in the future, at the time $t_f$.

This is really what the derivative is good for. If we know the quantity now and its time derivative now, then we can predict the quantity's value a little time in the future.

There is a slight problem. When we defined the derivative, it was at a particular time. We scrunched the two times needed to construct the velocity down to very close to some time $t$ and called the velocity "the velocity at that time, $v(t)$ as if it were a function defined at a point.  When we open the derivative up into deltas, we have to think about at what time we want to choose $v$ to be. The most reasonable for this discussion is to choose to put in $v$ at the initial time.  Then the info at the initial time gives us a result at the later time.  (When we are solving Newton's second law by stepping rules we will see that there is a better way.)

Julia Gouvea, Kerstin Nordstrom, Joe Redish 9/9/13

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