**Overview of the derivative **

In the previous lesson, we discuss how when \(f'(x)\) is a constant, we can go from \(f(x)\) to \(f'(x)\) and from \(f'(x)\) to \(f(x)\) by finding the slope and area, respectively. The "calculation procedures" we used to find the slope and area were pretty simple. When \(f'(x)\) is changing (that is, when \(f'(x)\) is *not *a constant), we still have to find the slope and area to go back and forth between between \(f(x)\) and \(f'(x)\); but we'll see that we'll have to use a fairly sophisticated calculation procedure to get the slope and the area. For about the first half of our studies of calculus, we're going to be studying differential calculus: how to go from \(f(x)\) to \(f'(x)\). In this lesson we're going to discuss something called a derivative. The derivative of \(f(x)\) is written as either \(f'(x)\) or \(df/dx\). To go from \(f'(x)\) (or \(df/dx\)) to \(f(x)\), we must take the integral of \(f'(x)\) to get the area underneath \(f'(x)\). This integral is written as

$$\int{f'(x)}dx=f(x).$$

We're going to learn later that both of these notations involving the '\(d\)' letters and the elongated s-shape are very appropriate. You can already see, albeit on a superficial level, that \(df/dx\) looks kind of similar to \(Δf/Δx\) and \(\sum{f'(x)Δx}\) looks similar to \(\int{f'(x)}dx\)—but we'll get in all of that later.

In Figure 1, I have drawn a graph of some arbitrary function \(g(x)\) which could be, pretty much, anything.\(^1\) Before discussing how to calculate \(g'(x)\) given \(g(x)\), let's start out by discussing what a derivative is. In a nutshell: the derivative \(g'(x)\) is just the slope of \(g(x)\) at each point. We can see, qualitatively, that the steepness of \(g(x)\) within the whereabouts of the point \((x_1,g(x_1))\) looks pretty flat whereas the steepness around the point \((x_2,g(x_2))\) is quite steep.

In order to calculate \(g'(x)\) from \(g(x)\) at each point along the curve, we'll have to introduce a new concept known as a limit which is perhaps the most important idea in calculus since it makes taking derivatives and integrals possible.

**Basic example of a limit**

Below, I have drawn a graph of the function \(f(x)=x^2\) (see Figure 2). As the value of \(x\) gets closer and closer to \(2\) (which we write as \(x→2\)) and as \(Δx\) gets closer and closer to zero (or, a shorter way of saying that: as \(Δx→0\)), what does the value of \(f(x)=x^2\) get closer and closer to equaling? That entire sentence can be rewritten as

$$\lim_{Δx\to0}x^2=?\tag{1}$$

Equation (1) is read as follows: "the limit of \(x^2\) as \(Δx\) approaches \(0\) is equal to \(?\)" Equation (1) should be interpreted like this: as we keep picking values of \(Δx\) which get closer and closer to equaling zero (which, to restate in another way to make sure that we're on the same page; as we keep picking values of \(x\) that get closer and closer to equaling \(2\)), what are the values of \(x^2\) getting closer and closer to equaling? What goes on the right-hand side of Equation (1) is the thing that \(x^2\) is getting closer and closer to equaling (I just put a \(?\) mark on the right-hand side since we're not sure yet what that thing is).

This is fairly straightforward and can be solved through repetition by calculating Equation (1) for several different values of \(x\) which get successively closer and closer to equaling \(2\). Since we want to choose values of \(Δx\) that are close to zero, let's choose \(Δx=0.1\). This means that \(x=2.1\) and \(x^2=4.41\). To find the answer to Equation (1), we want to keep choosing values of \(Δx\) that keep getting closer to zero. Let's pick \(Δx=0.01\). In this case, we have \(x=2.01\) and \(x^2=4.0401\). Ok, let's now pick a value of \(Δx\) that's even closer to being zero; let's say \(Δx=0.001\). Then, \(x=2.001\) and \(x^2=4.004001\). For convenience, I have listed all of the calculations we just did in Table 1 on the right. I'll restate Equation (1): as we choose values of \(Δx\) that get closer and closer to zero, what value (the thing that goes on the right-hand side of Equation (1)) does \(x^2\) get closer and closer to? Hopefully you can see the pattern in Table 1. As we choose values of \(Δx\) that get closer and closer to zero, \(x^2\) is getting closer and closer to being \(4\). Thus, we've found the answer to the problem and we can write

$$\lim_{Δx→0}x^2=4\tag{2}$$

**Using a limit to define the derivative **

This problem might have seemed trivial, but the importance of the notion of a limit is that it can be applied to any\(^1\) arbitrary function \(f(x)\) (not just \(x^2\)) and we can let \(Δx\) approach anything (not just zero). One very important function from algebra to apply the limit to is the function \(\frac{f(x+ Δx)-f(x)}{(x+Δx)-x}\) which we'll just call \(h(x)\) to make life easier. Let's take the limit of this function as \(Δx→0\) to get

$$\lim_{Δx→0}h(x)=?\tag{3}$$

To figure out what \(h(x)\) gets closer and closer to as \(Δx\) approaches zero, we'll do the same steps that we did to solve Equation (1). Let's start off with \(Δx=Δx_1\); then we have \(h(x)=\frac{f(x+ Δx_1)-f(x)}{(x+Δx_1)-x}\). As you can see from looking at Figure 3, \(h(x)=\frac{f(x+ Δx_1)-f(x)}{(x+Δx_1)-x}\) gives us the slope of the line \(AB_1\). Let's choose a smaller value of \(Δx\). Let's choose \(Δx=Δx_3\). Then, \(h(x)=\frac{f(x+ Δx_3)-f(x)}{(x+Δx_3)-x}\). You can see from Figure 3 that this gives the slope of the line \(AB_3\). If we choose an even smaller value of \(Δx\) (say \(Δx=Δx_4\)), then \(h(x)=\frac{f(x+ Δx_4)-f(x)}{(x+Δx_4)-x}\) which gives us the slope of the brown line in Figure 3. The question is: as \(Δx\) approaches zero, what does the function \(h(x)\) get closer and closer to equaling? Well, you can see that as \(Δx→0\), the function \(h(x)\) gives us the slopes of lines which resemble more and more closely the dashed grey line in Figure 3. We'll give this slope a name and call the slope of the dashed grey line the derivative of \(f(x)\) at the point \(x)\). We write the derivative of \(f(x)\) at \(x\) like this: \(f'(x)\) or \(\frac{df}{dx}\).

To solve the problem we initially started with, we can see that as the values of \(Δx\) get closer and closer to zero, the values of \(h(x)\) are getting closer and closer to the slope of the dashed grey line (which we called \(f'(x)\)). Thus, the solution to Equation (3) is

$$\lim_{Δx→0}\frac{f(x+ Δx)-f(x)}{(x+Δx)-x}=f'(x).\tag{3}$$

Equation (3) is the definition of the derivative function.

This article is licensed under a CC BY-NC-SA 4.0 license.

**References**

1. Khan Academy. "Introduction to limits 2 | Limits | Precalculus | Khan Academy". Online video clip. YouTube. YouTube, 30 September 2007. Web. 19 June 2017.

2. Khan Academy. "Derivative as a concept". Online video clip. YouTube. YouTube, 19 July 2017. Web. 07 November 2017.

**Notes**

1. \(g(x)\) can be any arbitrary function so long that it is smooth and continuous.