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This page consists of some basic topics to get you started with derivatives including the limit definition of the derivative and a few basic techniques.

Derivative Definition and Meaning

\(\displaystyle{ f~'(x) = }\) \(\displaystyle{ \lim_{\Delta x \to 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}} }\)

derivative = slope

you may also see the derivative referred to as the rate of change or instantaneous rate of change

There are two very important things to remember about the derivative, the definition and what it means.
First, the definition of the derivative is from a limit. When you were studying limits, you may have run across this limit but not known what it meant. There are at least three different forms that you might see. These are sometimes called difference quotients.

Three Ways to Write the Limit Definition of the Derivative

\(\displaystyle{ f~'(x) = \lim_{\Delta x \to 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}} }\)

original form; used the most since it contains the most information

\(\displaystyle{ f~'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} }\)

alternative form of the first equation; some students find this form easier to use

\(\displaystyle{ f~'(a) = \lim_{x \to a}{\frac{f(x) - f(a)}{x-a}} }\)

this form is usually used when finding the derivative at a point \(x=a\); used the least since it is specific to one point

The second thing you need to really know is that the derivative is the slope of the graph. Every time you hear the word 'derivative' you should think 'slope' and, when a problem asks for the slope (or the equation of the tangent line, since it involves slope) your first thought should be 'derivative'.

Okay, so what do all these equations actually mean and how are you supposed to understand them? Here is a fun video, using the secant line, which gives you an intuitive understanding of what the derivative actually is.

Krista King Math - Secant line [2min-34secs]

video by Krista King Math

Before we discuss some details about the derivative, let's talk about notation.

Derivative Notation

Why Notation is Important in Calculus

Using correct notation is extremely important in calculus. If you truly understand calculus, you will use correct notation. Take a few extra minutes to notice and understand notation whenever you run across a new concept. Start using correct notation from the very first.

You may not think this is important. However, if your current (or previous) teacher doesn't require correct notation, learn it on your own. You may (and probably will) get a teacher in the future that WILL require correct notation and this will cause you problems if you don't learn it now. It is much easier to learn it correctly the first time than to have to correct your notation later, after you have been doing it incorrectly for a while.

Not only is it important in class to use correct notation, when you use math ( or any other subject that uses special symbols ) in your career, you will need to be able to communicate what you mean. Without correct notation, your ideas could be misunderstood. It's a lot like speaking English ( or whatever language you use regularly ) or speaking a variant that has meaning only to you. You will be misunderstood and it may even affect your ability to keep a job.

So, just decide to start using correct notation now. It's not that hard and it will pay off in the long run.

You will see and use various types of notation to write this derivative. Most books just tell what to write without explaining why. Let me try to explain. In the following examples, I will assume \( y = f(x) \).

Various Notation Methods to Write the Derivative

\( f~'(x) \)

Used quite often and the small quote symbol indicates the derivative. Make sure that this symbol appears clearly in your work.

\( y' \)

Also used a lot. The disadvantage of this notation is that you need to pick up from the context what the variable of differentiation is, since it is not explictly stated in the expression, like it is in the previous example.

\( \displaystyle{\frac{d}{dx}[f(x)]} \)

Probably the clearest way to write the derivative but is also longer than the previous two ways. Use of this notation is usually limited to situations when it is important to clearly show the function itself and the variable of differentiation.

\( \displaystyle{\frac{dy}{dx}} \)

Also quite common. This is shorthand for \( \displaystyle{\frac{d[y]}{dx} }\) and \( \displaystyle{ \frac{d}{dx}[y]} \). [ see note below ]

\( D_x[y] \) and \( D_x[f(x)] \)

Not used much in beginning calculus. However, they are important when you study partial derivatives.

Note - - The notation \( dy/dx \) does NOT mean dy divided by dx. It means the derivative of y with respect to x. Right before you begin studying integration, you will probably cover a section about differentials. This section will prove that you can write \(\displaystyle{\frac{dy}{dx} = f~'(x)} \) in differential form as \( dy = [ f~'(x) ] dx\). This can propagate the illusion of division but don't let yourself be fooled. It takes quite a bit to prove the differential form.

Using The Limit Definition

Okay, let's watch some videos related to the derivative and how to use the limit definition to get our heads around this concept.

This first video contains a great explanation of how to think about this limit definition and what it means. It also contains a good example of how to calculate and interpret the limit.

Krista King Math - Derivatives [7min-48secs]

video by Krista King Math

After watching that first video, if the ideas are still not clear, here is another good video for you to watch explaining the same concepts.

PatrickJMT - Understanding the Definition of the Derivative [12min-8secs

video by PatrickJMT

Here is video that compares the graph of a function to the graph it's derivative. It is a great follow-up to the previous video.

PatrickJMT - Sketching the Derivative of a Function [14min-49secs]

video by PatrickJMT

Fortunately, we are not going to have to use this limit definition to calculate derivatives all the time. There are some basic rules, based on the limit definition, that will very quickly replace the use the limit definition to calculate derivatives. Let's look at a few here.

Constant Rule

The constant rule is the simplest and most easily understood rule. The derivative calculates the slope, right? So, if you are given a horizontal line, what is the slope? Right! The slope is zero. That's it. That's the slope of every horizontal line. We can write the equation of a horizontal line as \(f(x)=c\) where \(c\) is a real number. Since these are always horizontal lines, the slope is zero. Therefore, the derivative of all constant functions (horizontal lines) is zero. We can derive this idea from the limit definition as follows. If \(f(x)=c\)

\(\displaystyle{ f~'(x) = }\) \(\displaystyle{ \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} = }\) \(\displaystyle{ \lim_{h \to 0}{\frac{c - c}{h}} = \lim_{h \to 0}{0} = 0 }\)

Notice that c is gone from the final answer, \(f~'(x)=0\), so this holds for all horizontal lines. Thinking about this, it makes sense intuitively, right?

Constant Multiple Rule

This rule works as you would expect. Mathematically, it looks like this.

\(\displaystyle{ \frac{d}{dx}[cf(x)] = c \frac{d}{dx}[f(x)] }\)

Nothing surprising, just pull out the constant and take the derivative of the function. This is discussed in more detail with examples on the power rule page.

Addition and Subtraction Rules

When you have two functions that are added or subtracted, you just take the derivative of each individually. Mathematically, it looks like this.

\(\displaystyle{ \frac{d}{dx}[f(x) \pm g(x)] = }\) \(\displaystyle{ \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)] }\)

Nothing surprising or tricky here. It works just as you would expect.
[ However, you will find out soon that this idea does NOT hold for multiplication and division. We have some special rules for those called the product rule and quotient rule. ]

Coming Up

Here is a great video with quick explanations of the next 4 or 5 rules that you will learn to calculate derivatives. This video is especially helpful if you are refreshing your calculus skills or just need a review before an exam. If you are first learning derivatives, it is better to skip this video for now and come back to it right before your first exam on derivatives.

Krista King Math - Solving Derivatives [9min-0secs]

video by Krista King Math

After working some practice problems, your next logical step is the power rule.

 
power rule →

faq: Which derivative rule is used first?

This section is best understood after you have learned some of the other, more involved rules, including the product rule, quotient rule and chain rule.

A common question we get a lot when we teach differentiation is, which rule is used first, the product rule, quotient rule or chain rule?

The answer is, there is no rule that tells you which one to use first in all cases. You have to decide depending on the problem. We teach our students to take derivatives starting from the outside and working their way in. Here are two examples that might help. We will show you how to get started and then show the complete solutions on the chain rule page.

Example 1:
Evaluate \( \displaystyle{\frac{d}{dx}\left[ (2x)(x^3 +7 )^{10} \right]} \).
If we start on the outside we use the product rule first where \( f(x) = 2x \) and \( g(x) = (x^3 + 7)^{10} \).
Then we use the chain rule on \( g(x) \). [ complete solution on chain rule page ]

Example 2:
Evaluate \( \displaystyle{\frac{d}{dx}\left[ (2x)(x^3 +7) \right]^5} \).
If we start on the outside again, we need to use the chain rule first and then use the product rule. The first step is \( \displaystyle{ 5 \left[ (2x)(x^3+7) \right]^4 \frac{d}{dx}\left[ (2x)(x^3+7)\right] } \). [ complete solution on chain rule page ]

So you can see that there is no rule telling you which to use first. You need to decide by looking at the configuration of the expression.

By the way, the chain rule videos show the instructor starting on the inside and working his way out. We find that more difficult than starting on the outside. We recommend that you try both and see which makes more sense for you unless your instructor requires you to use a specific technique.

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

Derivatives - Practice Problems Conversion

[A01-899] - [A02-901] - [A03-905] - [A04-906]

[B01-898] - [B02-900] - [B03-902] - [B04-903] - [B05-907] - [B06-908] - [B07-1312] - [C01-904]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, use the limit definition to calculate the derivative in these problems and then use the constant rule, constant multiple rule and addition and subtraction rules to check your answers.
Give your answers in exact terms, completely factored.

Basic Problems

\( f(x) = 12 + 7x \)

Problem Statement

Use the limit definition to calculate the derivative of \( f(x) = 12 + 7x \).

Final Answer

\(f'(x) = 7\)

Problem Statement

Use the limit definition to calculate the derivative of \( f(x) = 12 + 7x \).

Solution

From the problem statement, we are given \(f(x)=12+7x\)
We will use the equation \(\displaystyle{ f'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} }\) but the equation \(\displaystyle{ f'(x) = \lim_{\Delta x \to 0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}} }\) will also work.
In our equation of choice, we have two main factors, \(f(x)\) and \(f(x+h)\). We are given \(f(x)=12+7x\). To get \(f(x+h)\) we replace x with x+h everywhere in \(f(x)\). This gives us \(f(x+h) = 12+7(x+h)\). Notice that there is only one x here, so we just need to do the substitution once. If there was more than one place that an x appears, we would need to replace EVERY x with x+h.
Okay, so now we are ready to set up the limit.

\(\displaystyle{ f'(x) = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} = \lim_{h\to0}\frac{[12+7(x+h)]-(12+7x)}{h}}\)

Now we just use algebra to simplify and solve.

\(\displaystyle{ f'(x) = \lim_{h\to0}\frac{12+7x +7h-12-7x}{h} }\)

\(\displaystyle{ f'(x) = \lim_{h\to0}\frac{7h}{h} }\)

\(\displaystyle{ f'(x) = \lim_{h\to0}7 = 7 }\)

Final Answer

\(f'(x) = 7\)

close solution

\( f(x) = x^2 + x - 3 \)

Problem Statement

Use the limit definition to calculate the derivative of \( f(x) = x^2 + x - 3 \)

Final Answer

\(f'(x) = 2x + 1\)

Problem Statement

Use the limit definition to calculate the derivative of \( f(x) = x^2 + x - 3 \)

Solution

901 solution video

video by PatrickJMT

Final Answer

\(f'(x) = 2x + 1\)

close solution

\( f(x) = x^2 + 5 \)

Problem Statement

Use the limit definition to calculate the derivative of \( f(x) = x^2 + 5 \).

Final Answer

\(f'(x) = 2x \)

Problem Statement

Use the limit definition to calculate the derivative of \( f(x) = x^2 + 5 \).

Solution

905 solution video

video by Krista King Math

Final Answer

\(f'(x) = 2x \)

close solution

\(\displaystyle{f(x)=\frac{\sqrt{x}}{8}}\)

Problem Statement

Use the limit definition to calculate the derivative of \(\displaystyle{f(x)=\frac{\sqrt{x}}{8}}\)

Final Answer

\(\displaystyle{ f'(x)=\frac{1}{16\sqrt{x}}}\)

Problem Statement

Use the limit definition to calculate the derivative of \(\displaystyle{f(x)=\frac{\sqrt{x}}{8}}\)

Solution

In the video solution below, he uses a shortcut that you are not usually given to solve this problem. To solve this problem without the shortcut, you need to use the concept of rationalizing that we discuss on the finite limits page.

\(\displaystyle{\frac{1}{8}\lim_{h \to 0}{\left[\frac{\sqrt{x+h}-\sqrt{h}}{h}\right] }}\) \(\displaystyle{ \frac{\sqrt{x+h}+\sqrt{h}}{\sqrt{x+h}+\sqrt{h}}}\)

\(\displaystyle{ \frac{1}{8}\lim_{h \to 0}{ \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} } = \frac{1}{8} \left[ \frac{1}{2\sqrt{x}} \right] = \frac{1}{16\sqrt{x}}}\)

906 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ f'(x)=\frac{1}{16\sqrt{x}}}\)

close solution

Intermediate Problems

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

Problem Statement

Use the limit definition to calculate the derivative of \(\displaystyle{ f(x)=\frac{1}{1+x^2} }\)

Solution

898 solution video

video by MIT OCW

close solution

\(\displaystyle{ G(t) = \frac{4t}{t+1} }\)

Problem Statement

Use the limit definition to calculate the derivative of \(\displaystyle{ G(t) = \frac{4t}{t+1} }\).

Solution

900 solution video

video by PatrickJMT

close solution

\(\displaystyle{f(x)=\frac{x}{1-2x}}\)

Problem Statement

Use the limit definition to calculate the derivative of \(\displaystyle{ f(x) = \frac{x}{1-2x} }\)

Solution

907 solution video

video by Krista King Math

close solution

Use the limit definition to calculate \(f'(x)\) for \(f(x)= 3/(4x)\) and find the slope of the tangent line at \(x=2\).

Problem Statement

Use the limit definition to calculate \(f'(x)\) for \(f(x)= 3/(4x)\) and find the slope of the tangent line at \(x=2\).

Solution

902 solution video

video by PatrickJMT

close solution

Use the limit definition to calculate \(f'(x)\) for \(\displaystyle{f(x)=\sqrt{2x+1}}\) and the slope of the tangent line at \(x=4\).

Problem Statement

Use the limit definition to calculate \(f'(x)\) for \(\displaystyle{f(x)=\sqrt{2x+1}}\) and the slope of the tangent line at \(x=4\).

Solution

903 solution video

video by PatrickJMT

close solution

Use the limit definition to calculate the slope and equation of the tangent line to the function \(f(x)=\sqrt{x+2}\) at the point \((7,3)\)

Problem Statement

Use the limit definition to calculate the slope and equation of the tangent line to the function \(f(x)=\sqrt{x+2}\) at the point \((7,3)\)

Solution

1312 solution video

video by PatrickJMT

close solution

\(\displaystyle{ f(x) = \frac{x}{x+3} }\)

Problem Statement

Use the limit definition to calculate the derivative of \(\displaystyle{ f(x) = \frac{x}{x+3} }\).

Solution

908 solution video

video by Krista King Math

close solution

Find the equation of the tangent line to the graph of \(f(x)=x^2-3\) at \((3,6)\) using the limit definition of the derivative.

Problem Statement

Find the equation of the tangent line to the graph of \(f(x)=x^2-3\) at \((3,6)\) using the limit definition of the derivative.

Solution

2202 solution video

video by PatrickJMT

close solution

Advanced Problems

Use the limit definition to determine the point on the graph of \(\displaystyle{y=1/x+2}\) where the slope is equal to \(-2\).

Problem Statement

Use the limit definition to determine the point on the graph of \(\displaystyle{y=1/x+2}\) where the slope is equal to \(-2\).

Solution

If you were confused by the equation in the question whether it meant \((1/x)+2\) or \(1/(x+2)\), you need to review order of operations in the precalculus section.

904 solution video

video by PatrickJMT

close solution
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