Lesson 6 MA 16010 Nick Egbert

Overview

From kindergarten we all know how to ﬁnd the slope of a line: “rise over run,” or “change in y over change in x.” We want to be able to determine slopes of functions which are not lines. To do this we use the idea of the derivative, which is the “instantaneous rate of change” at a single point.

Lesson Secant and tangent lines Before we figure out how to formally define the derivative we first need the notion of a secant line and a tangent line. Consider the graph of y = f(x) below. y

y = f(x) x a b

The red line which passes through the points (a, f(a)) and (b, f(b)) is called a secant line. In general any line which intersects the curve at two points we call a secant line. The green line above is called a tangent line at x = a. In other words, a tangent line touches (but does not cross) the curve locally. The word “locally” here just means that it’s possible for the line to intersect the curve at multiple points. If this happens we just restrict our attention to a smaller interval. For example, if we extended the green line to the left, we would eventually cross the curve at some negative y-value. We are eventually after the slope of the tangent line at x = a, which is what we will call the derivative of f at x = a. This is no easy task, so we start with an easier problem: ﬁnding the slope of the secant line between x = a and x = b. As we mentioned before, we already know how to do this from kindergarten.

The slope of the secant line between the points (a, f(a)) and (b, f(b)) is given by

∆y f(b) − f(a) slope = = . ∆x b − a

So how are we going to ﬁnd the slope of the tangent line at x = a? Using the notion of limits from the previous few lessons of course! If we imagine moving b closer and closer to a, then eventually our secant line will move into the tangent line. To visualize this, we’ll show a progression of a few graphs that demonstrate this idea.

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y y

y = f(x) y = f(x) x x

y y

y = f(x) y = f(x) x x

The diﬀerence quotient Now that we know where we’re headed, let us again consider the original graph with the secant line. y

y = f(x) x a b

We will denote h = b − a; that is, h is the distance along the x-axis from x = a to x = b. Some people will also denote this ∆x. This means that we can rewrite b = a + h, and then we can write the slope of the secant line as f(a + h) − f(a) slope = (a + h) − a Notice that we can simplify the denominator by canceling out the a and −a. We call this the diﬀerence quotient, which we often denote

f(a + h) − f(a) DQ = h

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The important takeaway from the diﬀerence quotient is that it is just the slope of the secant line between a and b = a + h. Recall that we wanted to move the point (b, f(b)) closer and closer to the point (a, f(a)). Mathematically this means we want to take the limit at h goes to 0. Since the derivative is also a function of x (meaning that the point a that we picked wasn’t special), we’ll replace a with x, and we have the following deﬁnition.

The derivative of f(x) at x, denoted f 0(x) is

f(x + h) − f(x) f 0(x) = lim . h→0 h

Remark. The definition in the box above is often referred to as the “formal definition of the derivative.” Alternatively, finding the derivative in this manner is referred to as “using the limit process to find the slope of the tangent line.” Important. In subsequent lessons we will learn more efficient methods for computing the derivative. However, throughout this lesson and whenever specified, you must compute the derivative in this manner.

Examples Example 1. Use the limit process to ﬁnd the slope of the tangent line to the graph of

f(x) = 8 − 3x at x. Solution. Step one in these problems is always to ﬁnd the diﬀerence quotient, DQ. Here,

f(x + h) = 8 − 3(x + h) = 8 − 3x − 3h.

So f(x + h) − f(x) DQ = h 8 − 3x − 3h − (8 − 3x) = h 8 − 3x − 3h − 8 + 3x = h −3h¡ = h¡ = −3.

Then by deﬁnition,

f 0(x) = lim −3 = −3. h→0

This shouldn’t be surprising since f(x) is a line whose slope is −3.

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Note. Recall that f(x + h) means that we want to plug in “(x + h)00 for every instance of x that we see. That is, we are evaluating f at x + h. Example 2. Use the limit process to ﬁnd the slope of the tangent line to the graph of

f(x) = 6x2 + 8 at x. Solution. In this case,

f(x + h) = 6(x + h)2 + 8 = 6(x2 + 2xh + h2) + 8 = 6x2 + 12xh + 6h2 + 8.

Now, f(x + h) − f(x) DQ = h 6x2 + 12xh + 6h2 + 8 − (6x2 + 8) = h 6x2 + 12xh + 6h2 + 8 − 6x2 − 8 = h 12xh + 6h2 = h h¡(12x + 6h) = h¡ = 12x + 6h.

Finally,

f 0(x) = lim (12x + 6h) h→0 = 12x.

Remark. Simplifying the diﬀerence quotient completely is generally the key to being able to take the limit as h → 0. Notice that has we tried just evaluating at h = 0 from the beginning, 0 we would have gotten 0 . Example 3. Find the equation of the tangent line to the graph of 7 f(x) = 6x2 at x = 1. Remark. In order to ﬁnd the equation of a line, we need to pieces of information: the slope and a point. From there we will be able to use point-slope form to determine the equation of the line:

y = m(x − x1) + y1 where m denotes the slope of the line and (x1, y1) is a point on that line.

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Solution to Example 3. By the above remark we need the slope and a point. The only point we know for sure that will be on the tangent line is at x = 1. To find the y-value, we simply 7 plug x = 1 into f(x). Doing so, we get f(1) = 6 . So (1, 7/6) is a point on the tangent line. Now we need to find the slope. As you’ll recall f 0(1) is the slope of the tangent line at x = 1. So we need to find the derivative evaluated at x = 1. We start by computing the difference quotient. Observe first that 7 f(x + h) = 6(x + h)2

so that f(x + h) − f(x) DQ = h 7 7 2 − 2 = 6(x+h) 6x h 7x2 7(x+h)2 2 2 − 2 2 = 6x (x+h) 6x (x+h) h 7x2−7(x+h)2 2 2 = 6x (x+h) h 7x2 − 7(x + h)2 = 6hx2(x + h)2 7x2 − 7(x2 + 2xh + h2) = 6hx2(x + h)2 7x2 − 7x2 − 14xh − 7h2 = 6hx2(x + h)2 −14xh − 7h2 = 6hx2(x + h)2 h¡(−14x − 7h) = 6hx¡ 2(x + h)2 (−14x − 7h) = . 6x2(x + h)2

And ﬁnally we are ready to take the limit as h → 0.

0 (−14x − 7>h) f 0(x) = lim h→0 0 6x2(x + h¡¡)2 −14x = 6x2 · x2 14 = − 6x3 7 = − . 3x3 But unfortunately we’re still not ﬁnished. We need to ﬁnd the equation of the tangent line

5 Lesson 6 MA 16010 Nick Egbert at x = 1. As mentioned above, the slope of this line is simply f 0(1). Well, 7 7 f 0(1) = − = − . 3(1)3 3

Now all that’s left to do is to plug this into point-slope form: 7 7 y = − (x − 1) + . 3 6 Remark. The given answer is a ﬁne and complete ﬁnal answer. Point-slope form is an equation of a line. If you prefer slope-intercept form or if you are asked to give an answer in slope-intercept form, we could also write this as 7 7 y = − (x − 1) + 3 6 7 7 7 = − x + + 3 3 6 7 14 7 = − x + + 3 6 6 7 21 = − x + 3 6 7 7 = − x + 3 2 In particular, on Loncapa you may leave your answer in point-slope form.