Calculus I
Famous Math Professor Quote"...BUT WITH CALCULUS, WE CAN DO BETTER!"

Special Thanks to Eric Gumtow at SWT University for allowing me to use most of his code and content for my Calculus I page. His Calculus site is great! To go there, Click HERE.



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The Vital Points Of Calculus


Delta-Epsilon Proof

Definition Of A Derivative

Rules Of Differentiation

Trig Function Derivatives

The Chain Rule

Implicit Differentiation


Indefinite Integrals

Definite/Riemann Integrals


Key: Do they exist or not?


The concept of the limit of a function is central to the study of calculus. It will be used to define just about everything else of any importance throuout your strudy of the subject. Whether or not a function is continuous or not, and to define the derivative and definite integral of a function to mention a few.

The limit describes the behavior of a function close to a particular value for x.

means if you have a continuous function f(x) such as (x + 3) then the limit as x approaches a is the function evaluated at a. Substitute a in wherever you see an x.

This means that if you picture the graph of the funcion (a 45 degree line with y intercept at 3) and you start to approach x=5 from the right or left, the value of y will approach 8.

Sometimes direct substitution doesn't yield a sensible answer and more work is necessary.


direct substitution yields

which is called an indeterminate form. For the first step, factor the expression out and see what you get.

Now divide out the (x - 2) in the numerator and denominator and you have 1 / (x + 4). Now direct substitution yields

Limits can also be one-sided. To evaluate them, you evaluate x approaching c from the right or left only. So, for example if you want to find the limit of Öx as x approaches 0 from the right, you get 0, but from the left it doesnt exist because as x approaches 0 from the left it is always negative and you end up with the square-root of a negative number.

INFINITE LIMITS exist when the value of f(x) increases or decreases without bound near certain values for x.


The sign of the infinity symbol is determined by the sign of the quotient at values close to the number that x approches.

LIMITS AT INFINITY describe the behavior of a function as x increases or decreases without bound.


The limit does not exist because f(x)doesnt get close to any limiting value as x decreases.

Graph of the above example

What is: "a continuous function?"

A function is continuous at a point (c, f(c)) if:

1) f(c) exists

2) the limit of f(x), with x approaching c exists, and

3) the limit of f(x) with x approching c is equal to f(c)
This means that there are no gaps, splits, or missing points for f(x) at c, and you could take your pencil and move it along the graph through (c,f(c)) without taking the point of the pencil off the graph




Delta-Epsilon Proof

Key: be able to construct a picture in your head

You really don't need to learn this to be able to do Calculus, but it is necessary if you wish to understand what Calculus is doing.
Formal definition of a limit:

For e > 0 
There is a d > 0 
Such that if 0 < | x - a | <d
Then | f(x) - A | <

This means you are looking for a point at (a, A) (note that A may also be denoted as f(a) or L for "limit"). Since that point might not exist, you need a way to know where it is without "seeing" it. One way to do this is to put a box around where you expect to find the point. This is what d (the Greek letter delta) is for. You know the point should be at (a, A), so you can box it in on the left side with a -d and on the right side by a +d This works well for the X component of the point, but the Y component is still left unchecked. With that thought, we need to evaluate the function at a - d to get A - e (the lowercase Greek letter epsilon), and evaluate the function at a +d to get A +e Now we have a box drawn around (a, A). If we make d very small, as close to zero as we can without it being zero, we have a minuscule box that contains only one point -- the point we are looking for, which is the point (a, A), where A is f(a) (the limit of the function).


The Definition Of A Derivative
Key: be able to combine algebra with limits as h®0
When speaking of "the definition of a derivative," you are referring to the formula:

This is nothing more than your old high school algebra formula for finding the slope of a line. Take the length of one side of the triangle and divide it by the other side of the triangle (rise over run). This gives you the slope of the line. Since a function will not always be a straight line, we have to construct our own triangle. Instead of using an exact number, such as 4 or 10, we need to use variables. We'll pick x as one variable, and h as the second. The variable x is any number along the X axis. The variable h is a distance (the distance from x). It follows that the distance from x to (x + h) is h. That is the run of our triangle. The rise can be found by evaluating the function at x, then evaluating it again, this time at (x + h), and subtracting one from the other. In short, the distance between f(x) and f(x + h) is the rise of our triangle. The slope of the segment that we used to form our triangle is the rise divided by the run. It can be seen in the picture that we are indeed only finding the slope of a segment (denoted by the green brace). It is easy to see that if we made smaller triangles, we would get a more precise measurement of the slope. This is why we make h approach 0.


1) The Derivative of a function is the slope of the line tangent to the curve (original function) at a given point.

2) The Derivative is also the instantaneous velocity of a function which represents the position of a particle along a line at time t, where y=s(t).

3) The Derivative is the instantaneous rate of change of a function at a point.



Find the instrantaneous velocity of s(t)=1/(t+2) at the time t=3. s(t)=1/(t+2) is called the position function.

First find the derivative of the position function which works out to be: s'(t)=-1/(t+2)2.

Probable Mistake: Follow this example closely and review it often. the common error is to do it backwards. DO NOT plug f(x), in this case 2x+x, into x in the formula.


To evaluate the function f(x)=2x+x at (x + h):
Wherever you see an x in the function, substitute (x + h). So if f(x) = 2x + x, then it would become 2(x + h) + (x + h), which simplifies to 3x + 3h. To complete the derivative using the definition, substitute it in the first expression of the numerator in the formula, then plug in f(x) in the second expression:

Factor out the 3

The first derivative of (2x+x) is 3




The Rules Of Differentiation

Many differentiation rules can be proven using the limit definition of the derivativeand are useful in finding the derivatives of corresponding functions. We use these formulas tjo eliminate the need for using the combersome formal definition every time we want to find a derivative.

  • Constant Rule:
  • Power Rule: 
  • Constant Multiple Rule:
  • Sum Rule:
  • Product Rule:
  • Quotient Rule: 

Your professor will expect you to be able to do the proof for all the above rules. I'll show you the Product Rule Proof. They are all about the same. Basically you plug the problem in the definition (click here if you still don't know what the definition is).

To have a product you have two have two things being multiplied together. We have f(x) and g(x). Plug that into the definition.

Now do some algebra. First we'll use the properties a - a = -a + a = 0 and b + 0 + c = b + c.

Combine like terms ab + ac = a * (b + c).

Now use the property lim (a + b) = lim a + lim b.

Since the limit as h ®0 of f(x + h) is f(x), we can use the property lim [a * b(x)] = a * lim b(x).

The above looks just like f * d/dx g + g * d/dx f (or f * g' + f ' * g).

Question: what is d/dx?
d/dx is short for "the derivative of". Calculus professors thought "the derivative of" would be too easy to understand, so they started using d/dx to make it look confusing. If you see dy/dx that means "the derivative of y with respect to x."

Good example: d/dx (4 * un + u * v / w + 1) would simplify to
4 * d/dx un + (w * (u * d/dx v + v * d/dx u) - u * v * d/dx w) / v2 + d/dx 1.
It looks confusing, but it is simpler than using the definition to figure it out. (This particular example uses all the rules).




Derivatives of Trigonometric Functions

Key: memorize these
d/dx sin x = cos x 
d/dx cos x = -sin x
d/dx tan x = sec2x
d/dx cot x = -csc2x
d/dx sec x = sec x * tan x
d/dx csc x = -csc x*cot x


Figure them out by plugging them into the definition and doing some algebra (like we did in the previous section). Take sin x, for example. You would have to evaluate

to find the derivative of sin x. It would be easier to memorize them, However, if your professor asks for the proof of the derivative of sin x, the above is what you would have to evaluate.

There are also a couple properties that would be nice to know.


Question: how do I differentiate sin2x?
sin2x can also be written as sin x * sin x. Taking the derivative of that requires the use of the product rule.




The Chain Rule

Learn it and don't forget it!
f(g(x)) = f(u) if u = g(x)

Take the derivative of the outside function times the derivative of the inside function. The chain rule provides the technique for finding derrivatives of composite functions. The number of functions that make up the composition determines how many differentiation steps are necessary.


look at sin(2x) 
d/dx sin(2x) = 2cos(2x) 
sin(u) is f(u) and 2x is g(x)


Look at (2x + x2)3
d/dx (2x + x2)3 = 3 * (2x + x2)2 * (2 + 2x)
You've got f(u), which is (u)3 and g(x), which is 2x + x2.

There are a few keys to the chain rule:

1) Identify your two functions (f and g) 
2) Write f as a function of u [eg. sin(u) or u3
3) Take the derivative of f with respect to u, and g with respect to x
4) Plug g back in wherever there is a u (not du) left in df/du [after differentiating f]
5) Multiply step 4 by dg/dx.


Implicit Differentiation

Key: remember to use the chain and product rules

By now you should know how to differentiate using all the rules. So your professor will throw something new at you. What if you have something that is not a "nice" function, like y3 = x2y? How would you take the derivative of that? Simply take the derivative of both variables, x and y. Remember that when you differentiate y, you get dy/dx. When you

differentiate y3 you get 3y2 * dy/dx. Since you are taking the derivative of y with respect x, you have an extra dy/dx term left over after differentiation. Now for x2y. You need to use the product rule for this. The derivative of x2 is 2x, and the derivative of y is dy/dx. So the product rule yields d/dx (x2y) = 2x * y + x 2 * dy/dx.

  • 3y2 * dy/dx = 2x * y + x2 * dy/dx 
  • Move all terms that contain dy/dx onto one side
  • 3y2 * dy/dx - x2 * dy/dx = 2x * y 
  • Factor out dy/dx from all the terms
  • dy/dx * (3y2 - x2) = 2x * y 

The equation of a circle with radius 2 is x2 + y2 = 22. You should be able to see how that could tie into implicit differentiation. Also, you might have something like "tan(xy) = 2xy + cos(x), find the derivative" on your test. Just remember to use the chain rule. In implicit differentiation, the chain rule leads to the product rule, because to find the derivative of tan(xy) you need to find the derivative of xy..

The following is a shortcut for implicit differentiation. Your professor will probably count it wrong on a test, but it is a quick way to check your answer. You will learn this in Calculus II.

Take the derivative of the function with respect to x, treating all the y terms as a constant.
Solve for dx.
Take the derivative of the function with respect to y, treating all the x terms as a constant.
Solve for dy.
Divide the first (dx) by the second (dy), and put a negative sign in front.



Key: Be able to think backwards and check your result by going through it frontwards again.


Now you are no doubt able to amaze family and freinds with your ability to find the derivative of any function they can throw at you. Well now it's time to "put things back the way they were" so to speak, or at least get as close as you can to putting them back to their original states. Its time to learn about antidifferentiation, and that means just what it sounds like it means.

The "inverse" process for differentation is antidifferentiation. Given a function, f, we are looking for a function, F whose derivative is is the given function , f. In other words,

F is called the antiderivative for f.


Antidifferentiation requires the ability to think backwards through the process of differentiation.

Example: F(x)=x2 is an antiderivative of 2x because F'(x)=2x=f(x).

Use these steps fir finding an antiderevative of f(x)=cos(x):


So, sin(x) is the antiderivative of cos(x) BECAUSE:

d/dx sin(x)=cos(x).

Notice that the derivatives of sin(x)+2, sin(x)-12, sin(x)+67 are also cos(x). There are an infinite number of antiderrivatives for any function and they differ only by a constant. So we can see that a better solution to the problem above would be to say that the antiderrivative of cos(x)=sin(x)+C, where C is un unknown constant. The constant in the original function could have been any number, including zero.

Now let's look at what we are trying to do geometrically. Remember that the differentiation process gives us the formula for the slope of the line that is tangent to the curve at any point. Now we are beginning with a slope function and we are asking "what function has this slope at each x?" The graphs below show that differentiating the function x2-3x+3 gives the slope function 2X-3, as shown.

Differentiating the parabolic function produces a slope.

Notice that the three tangents to the graph I have illustrated as occurring at x1, x2 and x3 in the graph of the parabola are represented as points along the graph of the function's derivative, on the right. That is, the big picture of this, simply put, at x1 the slope is negative, at x2, the slope of the tangent is positive but small and at x3, the slope of the tangent is positive but getting larger.

The figures below illustrate the problem with "working our way back to the original function (the antiderrivative) from a given function which was the derivative of the first function! Observe:

Slope field showing slopes of tangents to the antiderrivative of the function on the left.

So, you see the antiderrivative of our slope function, 2x-3, when represented on a graph, gives us what looks like an infinite number of parabolas stacked one on top of the other. That is precisely what the graphs of x2-3x+C (the antiderrivative of 2x-3) will give us! So, we can differentiate the original function and come up with a unique new function, but when we antidifferentiate the new function the best we can do is come up not with the original function again, but a whole family fo functions with an infinite number of members, differing ONLY by a constant, ONE of which is the original function with C=3.


The Indefinite Integral


The indefinite integral of a function, f, on an interval I, is the family of all antiderivatives for f on I and is denoted by:

That is:

Where F'(x)=f(x) and C is any constant.


All of the properties for antiderivatives are given throughout my page on Calculus II. Just remember it's the inverse process of differentiation. I will introduce the "power rule" of antidifferentiation here and then you can study the other methods of finding an antiderrivative in my Calculus II page. On that page look at the methods of integration by substitution, integration by parts, integration by partial fraction decomposition, etc., as well as stated antiderrivatives for log and exponential functions and trigonometric functions. By the way, an antiderrivative is called an indefinite integral, because it does not yield a number as does a definite integral, but a family of functions....we're getting a little ahead of ourselves here. Indefinite integrals or antiderrivatives can be foud by reversing the formulas for differentiation as shown above.

Power Rule for finding inefinite Integrals (antiderrivatives)

The following example illustrates the fact that:


Let f(x)=x3+sin(x)+cos(x). Then,

Here are some basic formulas for finding antiderrivatives. For more, go to my Calculus II page HERE


Now go to the Definite Integral Page or Go straight to Calculus II.

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