A very typical calculus problem is given the equation of a function, to find information about it (extreme values, concavity, increasing, decreasing, etc., etc.). This is usually done by computing and analyzing the first derivative and the second derivative. All the textbooks show how to do this with copious examples and exercises. I have nothing to add to that. One of the “tools” of this approach is to draw a number line and mark the information about the function and the derivative on it.

A very typical AP Calculus exam problem is given the graph of the derivative of a function, but not the equation of either the derivative or the function, to find all the same information about the function. For some reason, student find this difficult even though the two-dimensional graph of the derivative gives all the same information as the number line graph and, in fact, a lot more.

Looking at the graph of the derivative in the x,y-plane it is easy to very determine the important information. Here is a summary relating the features of the graph of the derivative with the graph of the function.

 Feature the function ${y}'$> 0 is increasing ${y}'$ < 0 is decreasing ${y}'$ changes  – to + has a local minimum ${y}'$changes + to – has a local maximum ${y}'$ increasing is concave up ${y}'$ decreasing is concave down ${y}'$ extreme value has a point of inflection

Here’s a typical graph of a derivative with the first derivative features marked.

Here is the same graph with the second derivative features marked.

The AP Calculus Exams also ask students to “Justify Your Answer.” The table above, with the columns switched does that. The justifications must be related to the given derivative, so a typical justification might read, “The function has a relative maximum at x-2 because its derivative changes from positive to negative at x = -2.”

 Conclusion Justification y is increasing ${y}'$> 0 y is decreasing ${y}'$< 0 y has a local minimum ${y}'$changes  – to + y has a local maximum ${y}'$changes + to – y is concave up ${y}'$increasing y is concave down ${y}'$decreasing y has a point of inflection ${y}'$extreme values

For notes on vertical asymptotes see

For notes on horizontal asymptotes see Other Asymptotes

17 thoughts on “Reading the Derivative’s Graph”

1. hello! what does it eman if the graph of the y’ has an asymptote or has an undefined part? Thanks!

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• Thanks for this great question. I will try to answer it more fully in a new blog post that I hope to have ready for next Tuesday November 5, 2019.
The short answer is that if there is a vertical asymptote on the derivative, then there is a vertical asymptote or a cusp on the function on the function. Please check next week for more details.
Again thanks for the suggestion; I’m always looking for more topics to write about.

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2. Pingback: Graph Analysis (Type 3) | MATHMANMCQ

3. Alex Epperly says:

I’ve never had this information provided so simply and easy to understand, thank you so much!

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4. Peter says:

That’s great, but where’s the original graph: f(x) ? Not seeing the whole picture at the same time is hurting my understanding. Thanks

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• Peter;
Yes, I can see that would help, however I don’t have an equation for the derivative, so I cannot write the equation for the function and graph it. it was just a sketch from an old AP exam. Its equation is something like y = (x+2)(x-4)(x-1)^2. So its antiderivative is
f(x) =$\displaystyle \frac{1}{6}{{x}^{6}}-\frac{2}{5}{{x}^{5}}-\frac{11}{4}{{x}^{4}}+\frac{8}{3}{{x}^{3}}+10{{x}^{2}}-16x+C$

Here’s what you can try: On a calculator or Desmos or other graphing utility, graph a function and its derivative. Then you can compare the two. Starting with the functions see how its features show up in the derivative; starting with the derivative see how its features tell you about the function. Try several different functions. They do not have to be complicated.

Here is two links to Desmos graphs that may help https://www.desmos.com/calculator/ovtzrjunz9 and https://www.desmos.com/calculator/tk241itk7v You can change the function (first line), leave the other parts alone.

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5. siya says:

Thanks so much! This is extremely useful! Beautiful explanation, demystified a week’s worth of lessons!

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6. R says:

Thank you very much.

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7. Angie says:

Also, note that the justifications must refer to calculus on the AP exam. For example, stating “concave up” is not OK for justification; stating that f ” > 0 or f ‘ increasing is OK.

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8. f'(1)=0. You have it in the section labeled “decreasing”. I tell my kids at f'(x) = 0 f is neither increasing nor decreasing. There may be an extrema or not. Am I wrong?

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• Barbara

Good question. Thank you. This is a very common misunderstanding; the question comes up a lot.

There is a theorem that says, if the derivative is positive on an interval, then the function is increasing on the interval. This theorem does not cover the case where the derivative is zero. Also, the converse is false. Consider the function $f\left( x \right)={{x}^{3}}$ which is increasing on any interval you choose, even though the derivative is zero at the origin. It is increasing because on every interval every point is higher (i.e. has a greater y-value) than all of those to its left.

For more on this see my post Open or Closed? and the link at the end.

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• R says:

Im only a student but yes you are correct

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9. k says:

Hey, this has been REALLY helpfully, outlining all the rules. I have always been confused with it, but not now. Thank you.

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