# 2.2 Derivative Graphically And Numericallyap Calculus

So you may have memorized all of the derivative rules. You might be able to derive f '(x) from f(x) no matter how complicated the function is. But how are you at estimating derivatives directly from the graph?

MATH 12002 - CALCULUS I x2.2: Di erentiability, Graphs, and Higher Derivatives Professor Donald L. White Department of Mathematical Sciences Kent State University. Graph the Holling type I equation, given latexa=0.5/latex. Determine the first derivative of the Holling type I equation and explain physically what the derivative implies. Determine the second derivative of the Holling type I equation and explain physically what the derivative implies. Home Calendar Q1 Q2. 2.2 Derivative Graphically and Numerically Notes 2.2 Key. C) 2 /; 2; /2 2 2 2 2 2 2 V H r h r r h V r V rh r hr = + + + + = = +π π π π πH h 6. If the sides of lengths a an b in a triangle form an angle θ, use the Law of Cosines to find the third side c of the triangle. What are the rates of change of c with respect to a, b, and θ? Ans: c =a2 +b2 −2ab cos θ., 2 2 2 2 cos cos; 2 cos 2 cos a b. Here is a set of practice problems to accompany the The Shape of a Graph, Part I section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University.

As we’ll see in this review article, it’s all about slope!

## The Derivative Measures Slope

Let’s begin with the fundamental connection between derivatives and graphs of functions.

The derivative value f '(a) equals the slope of the tangent line to the graph of y = f(x) at x = a.

I recommend brushing up on the idea of tangent lines first. Here are a few resources that might help.

### Example — Estimating Derivatives using Tangent Lines

Trend micro titanium internet security 2012 trial download windows 10. Use the information in the graph of f(x) below to estimate the value of f '(1).

Graph of a parabola with a tangent line attached at (1, 1).

#### Solution

Remember, derivative values are slopes! So f '(1) is equal to the slope of the tangent line attached to the graph at x = 1.

All it takes is two points on a line to determine slope. One point is easy to spot because it’s also on the graph of f itself: (1, 1). Next we look along the tangent line until we find another point whose coordinates are easy to estimate. Try to find a point that crosses an “intersection,” because then it will have integer coordinates. For example, (2, 3), or (3, 5), or (0, -1), etc.

I’m going to pick (3, 5) as my second point. However, if you pick any other point, as long as it’s on the tangent line, then your answer should be equal (or very close) to mine.

Next, use the slope formula (RISE over RUN) to compute the slope of the tangent line.

Therefore f '(1) = 2.

### Increasing, Decreasing, and Turning Around

Ok, so the first example may have been pretty easy. How hard can it get? Sometimes we have to estimate all of the derivative values! In other words, given the graph of a function f(x), it should be possible to sketch a graph of f '(x).

There are three things to keep in mind for differentiable functions.

• If f is increasing on an interval, then f ' > 0 (above the x-axis) in that interval.
• If f is decreasing on an interval, then f ' < 0 (below the x-axis) in that interval.
• If f turns around smoothly at a point x = a, then f '(a) = 0 (intersecting the x-axis).

### Example — Estimating the Graph of a Derivative

Sketch the graph of the derivative of the function whose graph is shown below.

#### Solution

First identify the two turnaround points: at x = -2 and 0. This means that f '(-2) = f '(0) = 0.

Then, identify the intervals on which the graph increases and decreases. When f is increasing, we have f ' > 0. When f is decreasing, we have f ' < 0.

The graph of a function gives information about its derivative… if you know how to analyze it.

The graph below shows the original in black and a sketch of its derivative in blue.

Notice how the blue curve fits the description of f '.

• The blue curve is above the x-axis whenever f increases.
• The blue curve is below the x-axis whenever f decreases.
• The blue curve crosses the x-axis whenever f has a turnaround point.

## Non-Differentiable Points

The methods for estimating derivatives so far have ignored an essential issue. What happens when the function fails to have a derivative value at a given point?

Any point x = a at which f '(a) does not exist is called a point of non-differentiability.

If a is such a point, then there will either be a hole or break in the graph of f ' at x = a.

Three things could cause such behavior. 1. The original function is undefined or discontinuous.
2. There is a corner point in the original function’s graph.
3. The tangent line is vertical.

Let’s explore the three situations in the following example.

### Example — Estimating Derivatives With Non-Differentiable Points

Sketch the graph of the derivative of the following function.

#### Solution

There is a lot going on in this graph!

• There’s a vertical asymptote at x = -5. Because f is undefined at this point, we know that the derivative value f '(-5) does not exist.
• The graph comes to a sharp corner at x = 5. Derivatives do not exist at corner points.
• There is a cusp at x = 8. The derivative value becomes infinite at a cusp.

Aside from these important landmarks, there’s also one turnaround point, at x = 0. Let’s analyze what happens in the intervals between the special points.

But what exactly happens near x = -5, 5, and 8?

At x = -5, the original graph follows a vertical asymptote. By definition, the function values are approaching ∞ or -∞ the closer x gets to -5. As a result, the function gets infinitely steep as x → -5. Infinite steepness means infinite slope values, so f ' must also have a vertical asymptote at x = -5.

Next, the corner point at x = 5 represents a very sudden change in direction. Rather than turning around smoothly, the function instantly changes course. That means that there will be a jump in the derivative value when crossing x = 5.

(For more about jump discontinuities and related topics, check out: AP Calculus Review: Discontinuities.) Finally, there is a cusp at x = 8. At a cusp, the graph’s tangent line becomes so steep that it’s effectively vertical. That means that the slope is infinite, and again there will be a vertical asymptote in the graph of f '.

Let’s put it all together now. The blue graph represents just a sketch of the derivative curve (not 100% accurate, but close enough for our purposes).

Notice not only the odd behavior near each point of discontinuity, but also how the derivative values are above the x-axis when f is increasing and below the axis when f is decreasing.

Sketch of the derivative of a complicated function. Original in black; derivative in blue.

## Conclusion

It’s important to know how to identify the derivative of a function based only on its graph. Fortunately, the AP Calculus exams will not require you to sketch the derivative curve itself, but may ask you to pick which answer choice best matches it.

Use smooth turnaround points as landmarks. Make sure you understand the strange behavior at non-differentiable points. And fill in the details by analyzing where f increases and decreases.

### More from Magoosh

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!

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When you start looking at graphs of derivatives, you can easily lapse into thinking of them as regular functions — but they’re not. Fortunately, you can learn a lot about functions and their derivatives by looking at their graphs side by side and comparing their important features. For example, take the function, f (x) = 3x5 – 20x3.

f (x) = 3x5 – 20x3 and its first derivative”/>

You’re now going to travel along f from left to right, pausing to note its points of interest and also observing what’s happening to the graph of

at the same points. But first, check out the following (long) warning.

This is NOT the function! As you look at the graph of

in the figure, or the graph of any other derivative, you may need to slap yourself in the face every minute or so to remind yourself that “This is the derivative I’m looking at, not the function!” It’s easy to mistake graphs of derivatives for regular functions. You might, for instance, look at an interval that’s going up on the graph of a derivative and mistakenly conclude that the original function must also be going up in the same interval — an understandable mistake.

You know the first derivative is the same thing as slope. So when you see the graph of the first derivative going up, you may think, “Oh, the first derivative (the slope) is going up, and when the slope goes up that’s like going up a hill, so the original function must be rising.” This sounds reasonable because, loosely speaking, you can describe the front side of a hill as a slope that’s going up, increasing. But mathematically speaking, the front side of a hill has a positive slope, not necessarily an increasing slope. So, where a function is increasing, the graph of its derivative will be positive, but that derivative graph might be going up or down.

Say you’re going up a hill. As you approach the top of the hill, you’re still going up, but, in general, the slope (the steepness) is going down. It might be 3, then 2, then 1, and then, at the top of the hill, the slope is zero. So the slope is getting smaller or decreasing, even as you’re climbing the hill or increasing. In such an interval, the graph of the function is increasing, but the graph of its derivative is decreasing. Got that?

Okay, let’s get back to the f and its derivative in the figure. Beginning on the left and traveling toward the right, f increases until the local max at (–2, 64). It’s going up, so its slope is positive, but f is getting less and less steep so its slope is decreasing — the slope decreases until it becomes zero at the peak. This corresponds to the graph of

(the slope) which is positive (because it’s above the x-axis) but decreasing as it goes down to the point (2, 0). Let’s summarize your entire trip along f and

with the following list of rules.

• An increasing interval on a function corresponds to an interval on the graph of its derivative that’s positive (or zero for a single point if the function has a horizontal inflection point). In other words, a function’s increasing interval corresponds to a part of the derivative graph that’s above the x-axis (or that touches the axis for a single point in the case of a horizontal inflection point). See intervals A and F in the figure.

• A local max on the graph of a function (like (–2, 64) corresponds to a zero (an x-intercept) on an interval of the graph of its derivative that crosses the x-axis going down (like at (2, 0)).

On a derivative graph, youve got an m-axis. When you’re looking at various points on the derivative graph, don’t forget that the y-coordinate of a point, like (2, 0), on a graph of a first derivative tells you the slope of the original function, not its height. Think of the y-axis on the first derivative graph as the slope-axis or the m-axis; you could think of general points on the first derivative graph as having coordinates (x, m).

• A decreasing interval on a function corresponds to a negative interval on the graph of the derivative (or zero for a single point if the function has a horizontal inflection point). The negative interval on the derivative graph is below the x-axis (or in the case of a horizontal inflection point, the derivative graph touches the x-axis at a single point). See intervals B, C, D, and E in the figure (but consider them as a single section), where f goes down all the way from the local max at (–2, 64) to the local min at (2, –64) and where

is negative between (–2, 0) and (2, 0) except for at the point (0, 0) on

which corresponds to the horizontal inflection point on f.

• A local min on the graph of a function corresponds to a zero (an x-intercept) on an interval of the graph of its derivative that crosses the x-axis going up (like at (2, 0)).

Now let’s take a second trip along f to consider its intervals of concavity and its inflection points. First, consider intervals A and B in the figure. The graph of f is concave down — which means the same thing as a decreasing slope — until it gets to the inflection point at about (–1.4, 39.6).

So, the graph of

decreases until it bottoms out at about (–1.4, –60). These coordinates tell you that the inflection point at –1.4 on f has a slope of –60. Note that the inflection point on f at (–1.4, 39.6) is the steepest point on that stretch of the function, but it has the smallest slope because its slope is a larger negative than the slope at any other nearby point.

Between (–1.4, 39.6) and the next inflection point at (0, 0), f is concave up, which means the same thing as an increasing slope. So the graph of

increases from about –1.4 to where it hits a local max at (0, 0). See interval C in the figure. Let’s take a break this trip for some more rules.

• A concave down interval on the graph of a function corresponds to a decreasing interval on the graph of its derivative (intervals A, B, and D in the figure). And a concave up interval on the function corresponds to an increasing interval on the derivative (intervals C, E, and F).

• An inflection point on a function (except for a vertical inflection point where the derivative is undefined) corresponds to a local extremum on the graph of its derivative. An inflection point of minimum slope (in its neighborhood) corresponds to a local min on the derivative graph; an inflection point of maximum slope (in its neighborhood) corresponds to a local max on the derivative graph.

Resuming your trip, after (0, 0), f is concave down till the inflection point at about (–1.4, 39.6) — this corresponds to the decreasing section of

from (0, 0) to its min at (1.4, –60) (interval D in the figure). Finally, f is concave up the rest of the way, which corresponds to the increasing section of

beginning at (1.4, –60) (intervals E and F in the figure).

Well, that pretty much brings you to the end of the road. Going back and forth between the graphs of a function and its derivative can be very trying at first. If your head starts to spin, take a break and come back to this stuff later.

Now, look again at the graph of the derivative,

in the figure and also at the sign graph for

in the next figure.

f(x

) = 3

## 2.2 Derivative Graphically And Numericallyap Calculus Formulas

x5 − 20x
A second derivative sign graph for f(x) = 3x5 − 20x3.

That sign graph, because it’s a second derivative sign graph, bears exactly (well, almost exactly) the same relationship to the graph of

as a first derivative sign graph bears to the graph of a regular function. In other words, negative intervals on the sign graph in the figure

show you where the graph of

is decreasing; positive intervals on the sign graph

show you where

is increasing. And points where the signs switch from positive to negative or vice versa

show you where

has local extrema.