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Section 2.1 Introduction to Derivatives

We’ll start this off by thinking about slopes. Before we begin, you should be able to answer the following questions:
  • What is a slope? How could you describe it?
  • How do you calculate the slope of a line between two points?
  • If we have a function \(f(x)\) and we pick two points on the curve of the function, what does the slope of a straight line connecting the two points tell us? What kind of behavior about \(f(x)\) does this slope describe?

Subsection Defining the Derivative

Activity 2.1.1. Thinking about Slopes.

We’re going to calculate and make some conjectures about slopes of lines between points, where the points are on the graph of a function. Let’s define the following function:
\begin{equation*} f(x)=\frac{1}{x+2}\text{.} \end{equation*}
(a)
We’re going to calculate a lot of slopes! Calculate the slope of the line connecting each pair of points on the curve of \(f(x)\text{:}\)
  • \((-1, f(-1))\) and \((0,f(0))\)
  • \((-0.5, f(-0.5))\) and \((0,f(0))\)
  • \((-0.1, f(-0.1))\) and \((0,f(0))\)
  • \((-0.001, f(-.001))\) and \((0,f(0))\)
(b)
Let’s calculate another group of slopes. Find the slope of the lines connecting these pairs of points:
  • \((0,f(0))\) and \((1,f(1))\)
  • \((0,f(0))\) and \((0.5,f(0.5))\)
  • \((0,f(0))\) and \((0.1,f(0.1))\)
  • \((0,f(0))\) and \((0.001,f(0.001))\)
(c)
Just to make it clear what we’ve done, lay out your slopes in this table:
Between \((0,f(0))\) and... Slope
\((1,f(1))\)
\((0.5,f(0.5))\)
\((0.1,f(0.1))\)
\((0.01,f(0.01))\)
\((-0.01,f(-0.01))\)
\((-0.1,f(-0.1))\)
\((-0.5,f(-0.5))\)
\((-1,f(-1))\)
(d)
Now imagine a line that is tangent to the graph of \(f(x)\) at \(x=0\text{.}\) We are thinking of a line that touches the graph at \(x=0\text{,}\) but runs alongside of the curve at that point instead of through it.
Make a conjecture about the slope of this line, using what we’ve seen above.
(e)
Can you represent the slope you’re thinking of in the table above with a limit? What limit are we approximating in the slope calculations above? Set up the limit and evaluate it, confirming your conjecture.

Activity 2.1.2. Finding a Tangent Line.

Let’s think about a new function, \(g(x) = \sqrt{2-x}\text{.}\) We’re going to think about this function around the point at \(x=1\text{.}\)
(a)
Ok, we are going to think about this function at this point, so let’s find the coordinates of the point first. What’s the \(y\)-value on our curve at \(x=1\text{?}\)
(b)
Use a limit similar to the one you constructed in Activity 2.1.1 to find the slope of the line tangent to the graph of \(g(x)\) at \(x=1\text{.}\)
(c)
Now that you have a slope of this line, and the coordinates of a point that the line passes through, can you find the equation of the line?

Definition 2.1.1. Derivative at a Point.

For a function \(f(x)\text{,}\) we say that the derivative of \(f(x)\) at \(x=a\) is:
\begin{equation*} f'(a) = \lim_{x\to a} \left(\frac{f(x)-f(a)}{x-a}\right) \end{equation*}
provided that the limit exists.
If \(f'(a)\) exists, then we say \(f(x)\) is differentiable at \(a\text{.}\)
We can investigate this definition visually. Consider the function \(f(x)\) plotted below, where we will look at the point \((-1, f(-1))\text{.}\) In the definition of the limit, we’ll let \(a=-1\text{,}\) and so consider:
\begin{equation*} \lim_{x\to-1}\left(\dfrac{f(x)-f(-1)}{x-(-1)}\right)\text{.} \end{equation*}
Can you estimate the limit of the slope of the tangent line as \(x\to -1\text{?}\)
Does it look like the limit of the slope between \((-1,f(-1))\) and \((x,f(x))\) exists as \(x\to -1\text{?}\) What do you think it is?

Subsection Calculating a Bunch of Slopes at Once

Activity 2.1.3. Calculating a Bunch of Slopes.

Let’s do this all again, but this time we’ll calculate the slope at a bunch of different points on the same function.
Let’s use \(j(x)=x^2-4\text{.}\)
(b)
Stop calculating the above derivatives when you get tired/bored of it. How many did you get through?
(c)
Notice how repetitive this is: on one hand, we have to set up a completely different limit each time (since we’re looking at a different point on the function each time). On the other hand, you might have noticed that the work is all the same: you factor and cancel over and over. These limits are all ones that we covered in Section 1.3 First Indeterminate Forms, and so it’s no surprise that we keep using the same algebra manipulations over and over again to evaluate these limits.
Do you notice any patterns, any connections between the \(x\)-value you used for each point and the slope you calculated at that point? You might need to go back and do some more.
(d)
Try to evaluate this limit in general:
\begin{align*} j'(a)\amp= \lim_{x\to a}\left(\frac{j(x)-j(a)}{x-a}\right) \\ \amp = \lim_{x\to a} \left(\frac{(x^2-4)-(a^2-4)}{x-a}\right)\text{.} \end{align*}
Remember, you know how this goes! You’re going to do the same type of algebra that you did earlier!
What is the formula, the pattern, the way of finding the slope on the \(j(x)\) function at any \(x\)-value, \(x=a\text{?}\)
(e)
Confirm this by using your new formula to recalculate the following derivatives:
We’re going to try to think about the derivative as something that can be calculated in general, as well as something that can be calculated at a point. We’ll define a new way of calculating it, still a limit of slopes, that will be a bit more general.

Definition 2.1.2. The Derivative Function.

For a function \(f(x)\text{,}\) the derivative of \(f(x)\text{,}\) denoted \(f'(x)\text{,}\) is:
\begin{equation*} f'(x) = \lim_{\Delta x\to 0} \left( \frac{f(x+\Delta x) - f(x)}{\Delta x} \right) \end{equation*}
for \(x\)-values in the domain of \(f(x)\) where this limit exists.
This definition feels pretty different, but we hopefully can notice that this is really just calculating a slope. Notice in the following plot there is a significant difference. In the visualization of the Derivative at a Point, the first point was fixed into place and the second point was the one that we moved and changed. It was the one with the variable \(x\)-value.
Notice in the following visualization that the first point is the one that is moveable while the second point is defined based on the first one (and the horizontal difference between the points, \(\Delta x\)). This means that we don’t need to define one specific point, and can find the slope of the line tangent to \(f(x)\) at some changing \(x\)-value.

Practice Problems Practice Problems

1.

We define a derivative of\(f(x)\) at \(x=a\) as\(f'(x) = \displaystyle \lim_{x\to a} \left(\dfrac{f(x)-f(a)}{x-a}\right)\text{.}\) What is this limit? What does it represent in terms of\(f(x)\text{?}\) Feel free to sketch some pictures in order to discuss it visually.

2.

For some function\(f(x)\text{,}\) we have that \(f'(3)=-4\text{.}\) What does this mean? Interpret this value in the context of\(f(x)\text{.}\)

3.

For each of the following functions, use the definition of the derivative at a point to find the slope of the line tangent to the function’s curve at the specified point.
(a)
\(f(x) = 6x+1\) at \((1, 7)\)
(b)
\(g(x) = x^2-5\) at \((-1, -4)\)
(c)
\(j(x) = \sqrt{x}\) at \((4, 2)\)
(d)
\(f(x) = \dfrac{1}{4-x}\) at \((3,1)\)
(e)
\(g(x) = 4+3x-2x^2\) at \((2,2)\)
(f)
\(j(x) = \dfrac{1}{x^2}\) at \((1,1)\)

4.

We have talked about two limit-based definitions of derivatives:
\begin{align*} f'(a) \amp = \lim_{x\to a} \left(\dfrac{f(x)-f(a)}{x-a}\right)\\ f'(x) \amp = \lim_{\Delta x \to 0} \left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right) \end{align*}
(a)
What are some similarities between these definitions? It might help to compare visual depictions of each.
(b)
What are some differences between these definitions? Focus on differences in how they are built as well as how we use them.
(c)
Pick a situation where it might make sense to use each definition. Explain your choice.

5.

For each of the functions below, find the derivative function \(f'(x)\text{,}\) and evaluate that function at \(x=1\text{,}\) \(x=2\text{,}\) \(x=3\text{,}\) and \(x=4\text{.}\) Interpret these values.

6.

Find the derivative function for each of the following functions.
(b)
\(f(x) = {\dfrac{5}{x-2}}\)
(e)
\(f(x) = {\dfrac{1}{x^2+1}}\)