The derivative

A colorful geometric mural of interlocking cubes painted around the corner of a concrete wall. — Photo by Jan van der Wolf on Pexels.

In the previous lesson, we estimated the sensitivity of a function at one particular input by examining smaller and smaller changes. That process gave us a slope at one point, but it was rather tedious. So what if there were a more precise and more efficient way to find the slope by computing it at every point all at once?

In this lesson, we'll look for patterns that turn those individual slopes into a whole new function. This derivative function will let us find instantaneous rates directly, opening the door to many applied problems.

By the end of this lesson, you should be able to:

Interpret the derivative as a new function that gives an instantaneous rate of change at each input.
Use the Power Rule, Sum and Difference Rules, and Constant Multiple Rule to differentiate polynomial-like functions.

In order to be able to use calculus, we'll need to be able to compute $\displaystyle \dv{y}{x}$ more efficiently than repeatedly calculating $\displaystyle \frac{\Delta y}{\Delta x}$ over smaller and smaller intervals. What we'd really like is a way to be able to take a function $$ y=f(x) $$ , perhaps with an algebraic formula, and be able to figure out $\displaystyle \dv{y}{x}$ immediately just by looking at it.

For a simple example, let's consider the quadratic function $$ x^2 $$ , whose graph is a parabola. If we can find the slope of this curve at any point, then any time an $$ x^2 $$ shows up in something we're studying, we'll know what to do with it without having to do any extra work.

To make it easier to see what's happening, we'll visualize $$ x^2 $$ by imagining it as the area of a whose side length is $$ x $$ .

To start off with something concrete, suppose we have an $3\times 3$ square, whose area is . What happens to the area if we increase the side lengths by a small amount, say, $$ 0.1 $$ ?

Loading the graph…If it does not appear, check the Desmos API configuration.

3\times 3

square whose side lengths have been increased by

\( 0.1 \)

after you drag the slider. Figure not drawn to scale.

Doing so adds three new pieces to the original square: two thin rectangles with an area of each, and one small square with an area of . Adding the four areas together, we find that the area of the new $3.1\times 3.1$ square is . Subtracting the original area of $$ 9 $$ , we get that the change in area is $\Delta A=$ . Finally, to see how the change in area is related to the change in side length, we find the average rate of change $\frac{\Delta A}{\Delta x}\text,$ which comes out to .

What we'd really like to know now is how sensitive the area is to changes in side length, which means we need to find $\displaystyle \dv{A}{x}$ . In our previous lesson we would have done this by redoing the above problem for smaller and smaller values of $\Delta x$ , but let's see if we can find a slicker way to do it.

Instead of starting with a particular size for our square, let's just use $$ x $$ for the length of the original square, and let $\Delta x$ be how much that length changes.

Loading the graph…If it does not appear, check the Desmos API configuration.

x\times x

square whose side lengths have been increased by

\Delta x

after you drag the slider.

The original square has an area of . When we change the side length by $\Delta x$ , we get three new pieces: two thin rectangles with an area of each, and one small square with an area of . Adding the four areas together, we find that the area of the new square is

$\begin{aligned}A\tsub{new}&=(x+\Delta x)^2\\ &=\mathhlred{x^2}+\mathhlblue{2x\,\Delta x}+\mathhlviolet{(\Delta x)^2}\text{.}\end{aligned}$

To find how much the area changed, we subtract the old area from the new area:

$\begin{aligned}\Delta A &= A\tsub{new}-A\tsub{old}\\ &= \qty(x^2+2x\,\Delta x+(\Delta x)^2) - x^2\\ &= 2x\,\Delta x+(\Delta x)^2\text.\end{aligned}$ Finally, to see how the change in area is related to the change in side length, we find the average rate of change:

$\begin{aligned}\frac{\Delta A}{\Delta x} &= \frac{2x\,\Delta x+(\Delta x)^2}{\Delta x}\end{aligned}$

Simplifying this expression, we find that $\displaystyle \frac{\Delta A}{\Delta x}=$ .

Now, if we imagine that $\Delta x$ becomes infinitely small, then the term will be much larger than the term. Hence if we treat that tiny change as negligible, we end up with $\dv{A}{x}=2x\text.$

This is big.

What we've done is taken a function we wanted to know about, and derived a brand-new function that instantly tells us the slope of the original function at any point.

For example, when $$ x=3 $$ , we have $\displaystyle \eval{\dv{A}{x}}_{x=3}=\)$\displaystyle \text.$ And if you zoom in close enough to the graph of $\( y=x^2 $$ around $$ x=3 $$ to the point where the graph looks linear, it does indeed seem to have a slope of about $$ 6 $$ .

Loading the graph…If it does not appear, check the Desmos API configuration.

Drag the slider below to zoom in on the point

\( (3,9) \)

; you should notice that as the graph starts to look more like a straight line, the slope of that line appears to approach

\( 6 \)

This new function $$ 2x $$ is called the derivative of $$ x^2 $$ , because we derived it from $$ x^2 $$ by reasoning about infinitesimal quantities. The derivative will be our main object of study for the next two chapters.

From individual slopes to a slope function

In the previous lesson, we learned how to find an instantaneous rate of change $\displaystyle \dv{y}{x}$ at one particular input by repeatedly computing the average rate of change $\displaystyle \frac{\Delta y}{\Delta x}$ over smaller and smaller intervals. However, we can make this process much more efficient by finding general patterns in those rates of change.

The function $$ f' $$ is derived from $$ f $$ , which is why we call it the "derivative." The process of finding the derivative is called differentiation, since the slope we find can be thought of as ratio of differentials.

Notations for derivatives and differentials

We now have seen two types of notation for the derivative:

We can use differential notation, writing $\displaystyle \dv{f}{x}$ or $\displaystyle \dv{y}{x}$ .
We can use prime notation, writing $$ f'(x) $$ or $$ y' $$ .

While these notations are largely interchangeable, you'll sometimes find that one is easier to use than the other depending on the situation. We can also relate them together: if $$ y=f(x) $$ , then we can say $\dv{y}{x}=f'(x)\text.$ You can think of this as saying that the slope of $$ f $$ (the ratio between infinitesimal changes $\dd{y}$ and $\dd{x}$ ) is given by this new derivative function $$ f' $$ .

One more shorthand that often comes in handy is the notation $\displaystyle \dv{}{x}$ , which you can read as "the derivative with respect to $$ x $$ ." (The phrase "with respect to $$ x $$ " just means that we're specifying our independent variable is $$ x $$ .) For example, we can write:

$\dv{x}\qty(x^2)=2x\text.$

This can be read out loud as "the derivative of $$ x^2 $$ (with respect to $$ x $$ ) is $$ 2x $$ ."

We call $\displaystyle \dv{x}$ an operator: it does something to the expression to its right, namely, it takes the derivative. If this notation seems a bit weird, don't worry — we'll see exactly where it comes from when we talk a bit more about differentials later in the section.

Finding derivatives of well-known functions

Let's look at the derivatives of some often-used functions, starting with linear functions.

A constant function is a function of the form $$ f(x)=c $$ (for some number $$ c $$ ) that outputs the same value ( $$ c $$ ) no matter what the input is. This can be thought of as a linear function with a slope of $$ 0 $$ .

Next let's consider powers of $$ x $$ . We already saw the derivative of the quadratic function $$ x^2 $$ in the Explore activity:

Theorem

Derivative of $$ x^2 $$

$\dv{x}(x^2)=2x\text.$

Proof

Imagine a square with side length $$ x $$ and area $$ A=x^2 $$ . Increasing both dimensions by $\Delta x$ adds two thin rectangles with area $x\,\Delta x$ and a small square with area $(\Delta x)^2$ .

Loading the graph…If it does not appear, check the Desmos API configuration.

The enlarged square consists of the original area together with two rectangular strips and a small square.

Subtracting the old area from the new area gives

$\begin{aligned}\Delta A&=(x+\Delta x)^2-x^2\\ &=2x\,\Delta x+(\Delta x)^2\text{.}\end{aligned}$

Dividing by $\Delta x$ , we get

$\begin{aligned}\frac{\Delta A}{\Delta x}&=\frac{2x\,\Delta x+(\Delta x)^2}{\Delta x}\\ &=2x+\Delta x\text{.}\end{aligned}$

Finally, if we imagine that $\Delta x$ becomes infinitesimal, the $$ 2x $$ term will be much larger than the remaining $\Delta x$ term, so we treat that term as negligible. Hence $\displaystyle \dv{A}{x}=2x$ .

We can apply this same logic to find the derivative of the cubic function $$ x^3 $$ .

Theorem

Derivative of $$ x^3 $$

$\dv{x}(x^3)=3x^2\text.$

Proof

Imagine a cube with side length $$ x $$ and volume $$ V=x^3 $$ . Increasing every dimension by $\Delta x$ adds three thin slabs with volume $x^2\,\Delta x$ , three thin bars with volume $x(\Delta x)^2$ , and a small cube with volume $(\Delta x)^3$ .

Loading the graph…If it does not appear, check the Desmos API configuration.

Drag the figure to rotate it.

We can subtract the two volumes to find how much it changes:

$\begin{aligned}\Delta V &= V\tsub{new}-V\tsub{old}\\ &=(x+\Delta x)^3-x^3\\ &=\qty(x^3+3x^2\,\Delta x+3x(\Delta x)^2+(\Delta x)^3)-x^3\\ &=3x^2\,\Delta x+3x(\Delta x)^2+(\Delta x)^3\text{.}\end{aligned}$

Now we can divide by $\Delta x$ :

$\begin{aligned}\frac{\Delta V}{\Delta x} &= \frac{3x^2\,\Delta x+3x(\Delta x)^2+(\Delta x)^3}{\Delta x}\\ &= \frac{3x^2\,\Delta x}{\Delta x} + \frac{3x(\Delta x)^2}{\Delta x} + \frac{(\Delta x)^3}{\Delta x}\\ &= 3x^2+3x\,\Delta x+(\Delta x)^2\text{.}\end{aligned}$

Finally, if we imagine that the change $\Delta x$ becomes infinitesimal, the $$ 3x^2 $$ term will be much larger than the terms involving $\Delta x$ , so we treat those terms as negligible. Hence $\displaystyle \dv{V}{x}=3x^2$ .

Now step back and take a look at what we've found for the derivatives of $$ x $$ , $$ x^2 $$ , and $$ x^3 $$ :

Function	Derivative
$$ x^1 $$	$$ 1x^0 $$
$$ x^2 $$	$$ 2x^1 $$
$$ x^3 $$	$$ 3x^2 $$

Notice that in each case, the exponent 'moves' to the front as a coefficient, and the new exponent is one less than the original. This pattern seems to suggest the following rule:

Note that we've actually only justified this rule geometrically for $$ n=1,2,3 $$ , but it turns out that the same pattern works for larger powers, negative powers, fractional powers, and even arbitrary real number powers. We'll go ahead and start using the Power Rule for all of these cases, but later sections will give us new ways to explain why this pattern continues to work.

Sums, differences, and constant multiples

Derivatives preserve several of the familiar properties of ordinary changes. These results allow us to break a complicated function into simpler pieces and differentiate those pieces separately.

Theorem

Sum and Difference Rules

If $$ f $$ and $$ g $$ are functions, then

$\begin{aligned}\dv{x}\bigl(f(x)+g(x)\bigr)&=f'(x)+g'(x),\\ \dv{x}\bigl(f(x)-g(x)\bigr)&=f'(x)-g'(x)\text{.}\end{aligned}$

Proof

Suppose $$ f $$ and $$ g $$ both change as $$ x $$ changes from $$ a $$ to $$ b $$ :

$\begin{aligned}\Delta f&=f(b)-f(a)\\ \Delta g&=g(b)-g(a)\text{.}\end{aligned}$

Then we can also find the change of the sum $$ f+g $$ :

$\begin{aligned}\Delta(f+g)&=(f+g)(b)-(f+g)(a)\\ &=\qty(f(b)+g(b))-\qty(f(a)+g(a))\\ &=f(b)-f(a)+g(b)-g(a)\\ &=\Delta f+\Delta g\text{.}\end{aligned}$

A similar argument shows that $\Delta(f-g)=\Delta f-\Delta g$ . If we expect these properties to remain true when the changes are infinitesimal, then the derivative of a sum or difference is the corresponding sum or difference of the derivatives.

Together with the Power Rule, these properties now allow us to differentiate a polynomial one term at a time.

Derivatives and differentials

Starting with the equation $\displaystyle f'(x)=\dv{y}{x}$ , we can multiply by $\dd{x}$ to solve for the corresponding differential in the output:

$\dd{y}=f'(x)\dd{x}\text{.}$

This form emphasizes what the derivative does: near the input $$ x $$ , it converts a small nudge $\dd{x}$ into the corresponding small nudge $\dd{y}$ .

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The parabola

\( y=x^2 \)

near

\( (1,1) \)

Loading the graph…If it does not appear, check the Desmos API configuration.

The straight line that matches the parabola infinitely close to

\( (1,1) \)

For $$ y=x^2 $$ , we have $\dd{y}=2x\dd{x}$ . At $$ x=1 $$ , this becomes $\dd{y}=2\dd{x}$ : along the matching line, the infinitesimal vertical nudge is twice the infinitesimal horizontal nudge.

We can write any of our derivative rules in differential form by multiplying both sides by the differential of the independent variable. For example, the derivative rule

$\dv{x}(x^2)=2x$

can be rewritten as

$\dd(x^2)=2x\dd{x}\text{.}$

These equations express the same relationship in two different ways. The derivative form gives the ratio between the changes; the differential form solves for the change in the output produced by an infinitesimal change $\dd{x}$ .

This also explains the operator notation $\displaystyle \dv{}{x}$ : first we take the differential of an expression, then we divide by $\dd{x}$ . For instance,

$\begin{aligned}\dv{x}(x^2)&=\frac{\dd(x^2)}{\dd{x}}\\ &=\frac{2x\dd{x}}{\dd{x}}\\ &=2x\text{.}\end{aligned}$

Thus $\displaystyle \dv{}{x}$ can be read not only as “the derivative with respect to $$ x $$ ,” but also as an instruction: take the differential, then divide by $\dd{x}$ .

With the Power Rule and various other properties in hand, we can now find derivatives for a wide variety of functions that appear across many fields of study. In the exercises, we'll practice carrying out these calculations and interpreting what the resulting derivatives mean in context.

Function	Derivative
$\( x^1 \)$	$\( 1x^0 \)$
$\( x^2 \)$	$\( 2x^1 \)$
$\( x^3 \)$	$\( 3x^2 \)$