We can use derivatives to help when graphing a function. But wait, don't we have calculators and computers for that? Yes, we do. But calculators are very imprecise when graphing and really just give us a feel for what the function looks like. And, although computer graphing is more precise than calculators, they still don't usually give us exact values and are really just used to show generally what is going on. Once you learn how to use derivatives to describe a graph and how to interpret the results, you will have a much better idea of what is going on and how to use the graph for your application.
First and Second Derivatives
The first derivative and the second derivative give you information about the shape of the graph. Each derivative tells you different things but they do parallel one another, i.e. the pattern with the first derivative is repeated in the second derivative, as shown in this table. These may not make sense initially but, after you have studied each, they will.
1. increasing/decreasing intervals 
2. critical points \(f'(x)=0\) 
3. first derivative test 
4. concavity 
5. inflection points \(f''(x)=0\) 
6. second derivative test 
For the discussion of a topic, select a link above, starting with the the first derivative, increasing and decreasing intervals. The topics need to be discussed in order, since they build on one another. Once you have studied each of these topics, come back to this page and we will help you put all your newlylearned knowledge together.
Asymptotes
We touched on asymptotes on the continuity page. After you are comfortable with that material, come back here and watch this next video. It has lots of good examples of specific types of vertical asymptotes.
video by Krista King Math 

Putting It All Together
Okay, at this point you should have read all six topics listed in the table above and worked those practice problems. These practice problems put all those concepts together to build a picture of what a graph should look like and how it behaves. In addition, you will need to have covered specifics about domain and range.
Practice
Unless otherwise instructed, use the first and second derivative as well as other concepts linked to on this page to sketch these functions. Show your work by building tables and label your graph carefully and completely.
Basic 

\(\displaystyle{y=\frac{x1}{x^2}}\)
Problem Statement 

Use the first and second derivative to sketch the function \(\displaystyle{y=\frac{x1}{x^2}}\). Show your work by building tables and label your graph carefully and completely.
Solution 

This problem is solved in two videos.
video by PatrickJMT 

video by PatrickJMT 

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\(\displaystyle{f(x)=\frac{x}{x+4}}\)
Problem Statement 

Use the first and second derivative to sketch the function \(\displaystyle{f(x)=\frac{x}{x+4}}\). Show your work by building tables and label your graph carefully and completely.
Solution 

video by PatrickJMT 

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\(f(x)=3x^26x+5\)
Problem Statement 

Use the first and second derivative to sketch the function \(f(x)=3x^26x+5\). Show your work by building tables and label your graph carefully and completely.
Solution 

video by Krista King Math 

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Intermediate 

\(\displaystyle{f(x)=\frac{x}{\sqrt{x^2+1}}}\)
Problem Statement 

Use the first and second derivative to sketch the function \(\displaystyle{f(x)=\frac{x}{\sqrt{x^2+1}}}\). Show your work by building tables and label your graph carefully and completely.
Solution 

This problem is solved in four videos.
video by PatrickJMT 

video by PatrickJMT 

video by PatrickJMT 

video by PatrickJMT 

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You CAN Ace Calculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, use the first and second derivative as well as other concepts linked to on this page to sketch these functions. Show your work by building tables and label your graph carefully and completely.