Introduction to Calculus Tutorial

ThePerfectHacker · #1 $Old$ 12-27-2006, 09:30 PM

This tutorial will be limited to polynomial functions in single and multi-variables. But note these theorems work for the standrard functions in general too. Thus, this will help you familiarize thyself with the most important rules in Calculus.

Definition: A polynomial function (real) can be expressed in the form:
$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ .
Here are some explames,
$f(x)=a_0$ is a constant function (it looks like a horizontal line).
$f(x)=a_1x+a_0$ is a linear function (it looks like a slanted line when $a_1\not = 0$ ).
$f(x)=a_2x^2+a_1x+a_0$ is a quadradic (it looks like a parabola when $a_2\not =0$ ).
Thus, the largest non-vanishing exponent (when it is not zero) is the degree of the polynomial.

I will not introduce multi-varaible functions just yet. I will do that when I reach partial differenciation.

I am sure you understand the basic notion, that is "What does the function approach as the value gets closer to some specific number".

I am also sure you are familar with the geometric meaning of the derivative. That is, we draw a secant line and make it closer and closer to a point. (Since I do not have aninamation you should find one somewhere on the internet und see their animation. Klicken heir.).

The last statement tells us (about the derivative) that say we have a point on a polynomial $f(x)$ which is $(c,f(c))$ .
How do we find the derivative at the point? We choose a point nearby, let us use $\Delta x$ to represent the small increase in the domain (input value). Then the nearby point is $(c+\Delta x, f(c+\Delta x))$ . Now we find the slope of these two points,
$\frac{f(c+\Delta x)-f(c)}{c+\Delta x-c}=\frac{f(c+\Delta x)-f(c)}{\Delta x}$ . And we take the limit as $\Delta x\to 0$ .
Thus, the derivative at the point is,
$\boxed{ \lim_{\Delta x \to 0}\frac{f(c+\Delta x)-f(c)}{\Delta x}}$ .
Note, we could have also done this "backward". Meaning instead of $\Delta x$ being an increase it could have been a decrease, but that is of no importance because it leads to the same result.
There is another way how to get the derivative at a point. Again let the point on the polynomial curve be $(c,f(c))$ then let a nearby point be $(x,f(x))$ thus the slope (together with the limit):
$\boxed{\lim_{ x\to 0}\frac{f(x)-f(c)}{x-c}}$ .

That is the derivative of a polynomial at a point. The derivative of a polynomial is a function whose input values (domain) are the points on the curve and whose output values (the function value) are the derivatives (the slope of tangent line) at the point.

Here is a classic example.

Example 1: Consider the curve $y=x^2$ what is the derivative at $(1,1)$ ?
We need to find,
$\lim_{\Delta x\to 0}\frac{(1+\Delta x)^2-1^2}{\Delta x}=\lim_{\Delta x \to 0}\frac{1+2\Delta x+(\Delta x)^2-1}{\Delta x}$
Combine,
$\lim_{\Delta x\to 0}\frac{(2\Delta x)+(\Delta x)^2}{\Delta x}=\lim_{\Delta x\to 0}2+\Delta x=2$
The final limit is true because when you add the very small number to 2 you get almost the same result back. Thus, when you approach it to zero the final result is 2+0=2.

Example 2: But what is the derivative of $y=x^2$ ? It means a formula that enables us to calculate the derivative at a point (a function that produces the derivative at a point). If the point is $x$ then the derivative at the point is,
$\lim_{\Delta x\to 0}\frac{(x+\Delta x)^2-\Delta x}{\Delta x}$
Skipping some steps (similar as before) we arrive at,
$\lim_{\Delta x\to 0}2x+\Delta x=2x$ .
Thus, the derivative of $y=x^2$ is a new curve $y=2x$ .
Now if we go back to the problem just before it asks to find the derivative (or slope of tangent line) at $x=1$ .
Just substitute that for derivative function $y=2(1)=2$ . Thus the derivative is 2 at that point.

We need to know one important thing about derivatives: "The derivative of a function is itself a function".

Another thing about derivatives that we will pay no attention to is the concept of differenciability. It means that the derivative exists (that means the limit that we take exists, because as you know not everything does exists as a limit). The reason why we will ignore that because we will use polynomials for which we can always take derivatives of. The reason why I am mentioning this is because this is like division by zero it leads to faulty reasoning. It is also another feature that divides immortals (the mathematicians) from the mortals (scientists) who do not pay attention to differenciability. Thus, if you ever play around with derivatives and get some strange results remember, it is probably what I said.

Time to introduce some notation. If $y=f(x)$ is our function. Then the derivative can be expressed as:
$\frac{dy}{dx}$ , $y'$ , $f'(x)$ .
In the first notation we cannot cancel the d's. They are not numbers rather represent some operation, that is, derivative.
I myself am not in favor of this notation, because it has a purpose to it. We can "magically" think of this as a fraction and split the $dy$ and $dx$ which is a favorite in physics. And sometimes it leads to faulty conclusions because it is non-mathematical (but it looks cool). To add some history it is called "Leibniz" notation.

Example 3: If $y=x^2$ then $y'=2x$ .

Example 4: If $y=2x^3$ then to find the derivative we need to find,
$f(x+\Delta x)=2(x+\Delta x)^3$
Substract $f(x)=2x^3$
Then divide by $\Delta x$
And then take the limit.
I will write the limit in the end because it takes less space.
$2(x+\Delta x)^3=2x^3+6x^2\Delta x+6x(\Delta x)^2+2(\Delta x)^3$ .
Next from this we subtract,
$2x^3$
Thus we have,
$6x^2\Delta x+6x(\Delta x)^2+2(\Delta x)^3$
Divide through by $\Delta x$
Thus,
$6x^2+6x\Delta x+2(\Delta x)^2$
Take the limit,
$6x^2$ .
Thus,
$y'=6x^2$ .

There got to be a better way! Does it mean we need to do always this long mess? No! There are rules. In fact I will make you develope them.
~~~
Excerises

1)Find $y'$ for $y=x^2+x$ .

2)Find equation of tangent line at $(1,2)$ in problem above.

3)Use the limit definition of derivative and find,
$y'$ for $y=\frac{1}{x+1}$ .

4)If $y=f(x)$ and $y'=f'(x)$ .
What does you think happens with,
$y=k\cdot f(x)$ then $y'=?$

5)Find derivative for.....
$y=1=x^0$
$y=x=x^1$
$y=x^2$
$y=x^3$
$y=x^4$
And guess what the pattern is.

*6)Prove the pattern always holds.
(Hint use the binomial theorem:
$(x+y)^n=\sum_{k=0}^n {n\choose k}x^{n-k}y^k, n\geq 0$
Where,
${n\choose k}=\frac{n!}{k!(n-k)!}$ are called "binomial coefficients").

ThePerfectHacker · #2 $Old$ 12-28-2006, 03:18 PM

Okay, since we know what derivatives mean.
We can develope a few rules that will help us.

1)Derivative of Constant Function: The derivative of a function $f(x)=k$ (some number) is zero, that is, $f'(x)=0$ . One way is is to think of this as the slope of a vertical line (which is zero). Another way is through the limit:
$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x\to 0} \frac{k-k}{\Delta x}=\lim_{\Delta x\to 0}0=0$

2)Derivative of a Sum: You might have expected
$(y_1+y_2)'=y_1'+y_2'$ where $y_1,y_2$ are some functions. Again, this is easy to show through the limit.
Let $y_1=f(x)$ and $y_2=g(x)$ .
That means,
$y_1'=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
$y_2'=\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}$
And,
$(y_1+y_2)'=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)+g(x+\Delta x)-f(x)-g(x)}{\Delta x}=$ $\lim_{\Delta x\to 0}\underbrace{\frac{f(x+\Delta x)-f(x)}{\Delta x}}_{y_1'}+ \underbrace{\frac{g(x+\Delta x)-g(x)}{\Delta x}}_{y_2'}=y_1'+y_2'$

3)Derivative of a Difference: The same thing as addition. That is,
$(y_1-y_2)'=y_1'-y_2'$

Note: If you want to sound cool

and impress your teachers you can say rules #1,2,3 are true because "Differenciation is a linear transformation for the vector space of differenciable functions over the field of reals". Or can you can "Differenciation is a homomorphism". Basically "derivative" has the property that the derivative of a sum is the sum of the derivatives. You will found many operators during you math learning that have this propert, pay attention to them.

4)Derivative of a Product: The rule says,
$(y_1y_2)'=y_1'y_2+y_1y_2'$ .
The derivation is a bit strange, but it relys on a trick mathematicians love to use.
Again using the same convention for $y_1,y_2,y_1',y_2'$ as above we can write,
$(y_1y_2)'=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}$
Add and subtract $f(x+\Delta x)g(x)$ (thus no change in the expression).
Thus,
$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)\overbrace{-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)}^{\mbox{ This is zero }}-f(x)g(x)}{\Delta x}$
Thus, some factoring,
$\lim_{\Delta x\to 0}f(x+\Delta x)\cdot \frac{g(x+\Delta x)-g(x)}{\Delta x}+g(x)\cdot \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Note,
$\lim_{\Delta x\to 0}f(x+\Delta x)=f(x+0)=f(x)$
Thus,
$y_1y_2'+y_2'y_1$

Note the only questionable step was,
$\lim_{\Delta x\to 0}f(x+\Delta x)=f(x)$ .
Because, in the other post I spoke about limits and I mentioned you can only substitute the value when the function is continous at the point. It happens to be true. Because a differenciable function at a point is continous at a point, this is extremely useful in theoretical demonstrations.

5)Derivative of a Quoteint: If $y_2\not = 0$ then,
$(y_1/y_2)'=\frac{y_1'y_2-y_1y_2'}{y_2^2}$
This time we add and subtract $f(x)g(x)$
$\lim_{\Delta x\to 0}\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)} }{\Delta x}$
Thus,
$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{\Delta xg(x)g(x+\Delta x)}$
Thus,
$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{\Delta xg(x)g(x+\Delta x)}$
Thus,
$\lim_{\Delta x\to 0}\frac{g(x)\cdot \frac{f(x+\Delta x)-f(x)}{\Delta x } - f(x)\cdot \frac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x)g(x+\Delta x)}$
Thus,
$\frac{y_1'y_2-y_1y_2'}{y_2^2}$

6)The Power Rule: The power rule says that for any in positive integer $n$ we have,
$y=x^n$
$y'=nx^{n-1}$ .
I am going to present two proofs.
One the standard way.
The second which is my way.
The standard way is by the binomial expansion. The only important thing to know is that,
$(x+y)^n=x^n+nx^{n-1}y+A_{n-2}x^{n-2}y^2+A_{n-3}x^{n-3}y^3+...+A_1xy^{n-1}+A_0y^n$
Where, $A_{n-2},....,A_0$ are some numbers which we really do not care about.
That means,
$(x+\Delta x)^n=x^n+nx^{n-1}\Delta x+....$
From this we subtract the orginal function $x^n$ .
Thus,
$nx^{n-1}\Delta x+A_{n-2}x^{n-2}(\Delta x)^2+...$
Then we divide through by $\Delta x$ ,
$nx^{n-1}+A_{n-2}x^{n-2}\Delta x+...$
But everything to the right of $nx^{n-1}$ is zero because a $\Delta x$ exists.
Thus,
$nx^{n-1}$ .

Here is my way....
Note, by product rule,
$(y_1y_2y_3)'=((y_1y_2)y_3)'=(y_1y_2)'y_3+(y_1y_2)y_3'$
Thus, by product rule again,
$(y_1'y_2+y_1y_2')y_3+y_1y_2y_3'$
Thus,
$y_1'y_2y_3+y_1y_2'y_3+y_1y_2y_3'$ .
In fact this pattern hold for some positive integer of functions,
$(y_1y_2...y_n)'=y_1'y_2...y_n+y_1y_2'...y_n+...+y_1y_2...y_n'$
Thus,
$x^n=\underbrace{x\cdot x\cdot ... \cdot x}_n$
Thus, the derivative by Hacker's product rule above is,
$\underbrace{(x)'x\cdot ...\cdot x+x\cdot (x)'\cdot ... \cdot x+...+x\cdot x\cdot ... \cdot (x)'}_n$
But, $x'=1$ .
Thus,
$\underbrace{\overbrace{x\cdot ... \cdot x}^{n-1}+...+\overbrace{x\cdot ... \cdot x}^{n-1}}_n$
Thus,
$nx^{n-1}$

The easy way to remember this rule is to bring down the exponent in front of the varaible and reduce exponent by 1.
The nice thing is that though we proved it only for positive integers it hold for any power.
Thus, we can find,
$y'$ for $y=\sqrt{x}=x^{1/2}$
Time for examples.
But I am sure you do not need them, you are good with algebra.

Example 5: $y=x^3+x^2+x$ .
By, the sum and power rule we have,
$y'=3x^2+2x+1$

Example 6: $y=(x+1)(x+2)$
By the product rule,
$y=(x+1)'(x+2)+(x+1)(x+2)'$
But the power and sum rules,
$y=1(x+2)+(x+1)1=2x+3$
We could have also multiplied out,
$y=(x+1)(x+2)=x^2+3x+2$
By, power and sum rules,
$y'=2x+3$ .
Same result.

Example 7: $y=\frac{1}{x}$ .
We can write,
$y=x^{-1}$ .
Though it is not a positive integer the rule still holds,
$y'=(-1)x^{-2}=-\frac{1}{x^2}$ .

Now we reach the most important rule. The "chain rule".
I will state it, and explain why it is called chain rule.

First we need to know what a composite of functions is. If we have a functions $f(x)$ and $g(x)$ . The composition of these functions is a new function,
$f(g(x))$ where $g(x)$ is the "inner" and $f(x)$ is the "outer".
Note, $f(g(x))\not = g(f(x))$ , (always)meaning it is not commutative.

Example 8: If $f(x)=2x$ and $g(x)=x+1$ .
Then,
$f(g(x))=f(x+1)=2(x+1)=2x+2$ .

7)Limit of a Composition: The rule says,
$[f(g(x))]'=g'(x)f'(g(x))$ .
Take the derivative of the inside, multiply it by the composition of the derivative of outer function.
There is an easy to do it.
$y=f(g(x))=f(u)$
Where $u=g(x)$ .
Then, the composition,
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ .
As if we can cancel the $du$ 's.

Example 9: $y=(x+1)^{10}$
We can write,
$y=u^{10}$ and $u=x+1$
Thus,
$\frac{dy}{du}=10u^9$
$\frac{du}{dx}=1$
Thus,
$\frac{dy}{du}\cdot \frac{du}{dx}=\frac{dy}{dx}=10u^9=10(x+1)^9$

Example 10: $y=((2x+1)^9+1)^8$
We can write,
$y=(u^9+1)^8$ and $u=2x+1$
But we can write more,
$y=v^8$ , $v=u^9+1$ , $u=2x+1$ .
Thus,
$\frac{dy}{dv}=8v^7=8(u^9+1)^7=8((2x+1)^9+1)^7$
$\frac{dv}{du}=9u^8=9(2x+1)^8$
$\frac{du}{dx}=2$
Thus,
$\frac{dy}{dx}=\frac{dy}{dv}\cdot \frac{dv}{du}\cdot \frac{du}{dx} = 8((2x+1)^9+1)^7(9(2x+1)^8)(2)$
Hence the name "chain rule".

I will not prove the chain rule. The proof is difficult. But there is a nice trick, a weaker result, that makes it an easy prove. I do not want to post it because that will be the next problem of the week.

Thus, now you know the most important rule about derivatives!
---
Excersices.
Find $y'$ for the following functions:

1) $y=x^9+x^2+1$

2) $y=x^6-x^2-\sqrt[3]{x}$

3) $y=\frac{1}{1+x^2}$

4) $y=(x+1)(3x-1)^2$

5) $y=\sqrt{0}+1^2-6$

*6)Prove the chain rule for polynomial functions.
That is given,
$f(x)=a_nx^n+...+a_1x+a_0$
$g(x)=b_mx^m+...+b_1x+b_0$
Find their composition and then compute the derivative.
And compute the derivative via chain rule.
And then show the results match.

TD! · #3 $Old$ 12-28-2006, 04:10 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

Note: If you want to sound cool

and impress your teachers you can say rules #1,2,3 are true because "Differenciation is a linear transformation for the vector space of differenciable functions over the field of reals".

If you want to sound really cool, I'd say that the fact that the derivative is a linear operator means that it converts a linear combination as argument, to a linear combination of its image. The fact that a constant is mapped to zero, can be seen as a result of a more general requirement. Being linear means that D(f(x)+g(x)) = D(f(x))+D(g(x)), but also that D(c.f(x) = c.D(f(x)), with c a scalar (an element of the field, in your case the reals). Combined: linearity holds iff D(a.f(x)+b.g(x)) = a.D(f(x)) + b.D(g(x)).

ThePerfectHacker · #4 $Old$ 01-01-2007, 12:51 PM

I was debating what I should lecture on. Finally after many hours of thought I decided to show some applications that the derivative is used for. The first important thing to know about the derivative is that it represents the instantenous rate of change. If $f(t)$ is some function based on time, which can represent: the distance, the amout, the population,.... Then $f(t+\Delta t)-f(t)$ is the change small change in the function. To find the average rate of change we divide through by the time passed $(f(t+\Delta t)-f(t))/ \Delta t$ . Note for a small increase $\Delta t$ the average rate of change of the function is almost its instantenous rate of change (the rate of change at that point). The smaller $\Delta t$ is the more accurate this expression. Thus we need to consider the limit $\Delta t\to 0$ . In that case we have the derivative. Thus, the rate of change at some moment in time is the derivative at that point.

Origin of Differencial Equations:
I think it is a good time to mention what a differencial equation is. To show that we will consider the following problem: "A tank is filled with 10 gallons of pure water. There are two pipes. One taking the water in and one taking the water out. The flow rate is the same for both at 3 gallons/min. The tube that takes the water in contains 1 gallon of salt. Find a function that represents the amount of salt at any given time"
We need to understand the difficultly in this problem. The difficultly is that the flow out tube also takes out salt with it also of the mixture of salt and water. Thus, this is really not an easy question to answer. We will not answer is question but rather set up a differencial equation. Let $Q(t)$ be the amount of salt at any given time passed $t\geq 0$ . Then, as mentioned before, the derivative, $dQ/dt$ is the rate of change of this amount. One way we can find the rate of change is to note that:
$\frac{dQ}{dt}=\mbox{ rate in }-\mbox{ rate out }$ .
Where "rate in, rate out" represent the rate at which salt is entering and leaving the tank.
Every minute 3 gallons of mixure enter the tank with 1 gallon of salt. Thus the rate in is constant at 1 gallon/min.
The rate out is a bit tricker, but not so bad. The rate out is the concetration of salt in the mixture multiplied by the amout leaving (that is 3 gallons). The concetration is the amount of salt at that time which is $Q$ divided by the total volume which is fixed at 10 gallons.
Then we need to multily this result by 3 because 3 gallons are leaving. Thus,
$\frac{dQ}{dt}=1-\frac{3Q}{10}$
$\frac{dQ}{dt}+\frac{3Q}{10}=1$
This is a differencial equation.
We need to find a function that makes this statement true.
Unlike an algebraic equation where we need to find a number that makes a statement true, here we need to find a function. And there is a way to solve for that, I am just not going to do that. The interesting thing is that there are infinitely many solutions to that equation! How do we know which one is it? We use the important fact that at $t=0$ we have pure water, thus, $Q(0)=0$ . And with this condition (called intial condition) the differencial equation will have a unique solution. And that solution will describe the amount of salt in the tank.

The Motion Problem
The motion problem deals with the motion of some object. To understand the technique used we need to be familar with the meanings of: distance, speed and acceleration.
Distance we already know that is means, the function $s(t)$ will represent the total distance traveled for some time $t$ . Speed is distance traveled per time. Thus, $(s(t+\Delta t)-s(t))/\Delta t$ is the average speed per some small amount of time. Again taking the limit we find that $s'(t)=v(t)$ meaning, the derivative of distance is speed. Accelleration is the speed changed by a certain time. Meaning, an accelleration 5 miles per hour per second means each second the speed (5 miles per hour) increases by 5 miles per hour. Again by similar reasoning $a'(t)=v(t)$ . The derivative of acceleration is speed.

Example 11: The distance a particle travels can be expressed as a function: $s(t)=12t^3$ . Find the accelleration of the particle after 1 second. We will actually need the second derivative (meaning just take the derivative twice). Because $a(t)=v'(t)=(s'(t))'=s''(t)$ . Taking 2 derivative of the distance function we find that:
$s'(t)=36t^2$ and $s''(t)=72t$
At $t=1$ we have $s''(1)=72$ units per some time per second.

We can use the motion problem to solve the free falling problem. The free falling problem concerns an object falling under the effect of gravity. Let us assume that $v_0$ is the initial speed which an object is thrown at (positive for up and negative for down). And $s_0$ is the initial height at which we are standing. What function represents the height of the object as a function based on time? To solve this famous problem (historically I think Galileo solved it first, then Newton explained why it works). We need to be familar with an important property that all falling object posses. They accellerate downwards with a constant rate (why that is, we do not know).
The downwards accelleration is based on the gravitational force acting on the object. On Earth the acceleration is 32 feet per second per second.
Thus, we are looking for a function $s(t)$ that represents the height of the object from the ground.
What we do know is that the second derivative is the acceleration,
$s''(t)=-g$
Where $g$ is the acceleration from the force of gravity. It is negative because by our sign convention, down is negative (fallings objects go down) and upwards is positive.
This is actually a differencial equation (the most basic type).
To make it easier to follow we can think of the derivative as,
$(s'(t))'=-g$
What function $s'(t)$ has its derivative equal to $-g$ . Think about this.... You should come up with that $s'(t)=-gt$ . But wait, any constant that we attach in the end will dissappear thus,
$s'(t)=-gt+C$
(What we just did is called taking the integral, or antiderivative. It turns out that all antiderivatives that satisfy this throughout the interval must differ by a constant).
If we subsitute $t=0$ we have,
$s'(0)=v(0)=v_0=-g(0)+C$
Thus, $C=v_0$ (initial velocity).
Thus, $s'(t)=-gt+v_0$ .
What function has its derivative equal to $-gt+v_0$ . Some thought should produce,
$s(t)=-(1/2)gt^2+v_0t+C$
Substitute $t=0$ .
$s(0)=s_0=-(1/2)g(0)^2+v_0(0)+C$ .
Thus, $C=s_0$ .
Thus,
$\boxed{ s(t)=-\frac{1}{2}gt^2+v_0t+s_0}$ .

Example 12: An angry husband throws up his wife with the initial velocity at 96 feet per second. They live on a cliff with the altitude of 960 feet. Find how much time will pass until his wife reach maximum hieght. And find the amount of time until he hits the ground and dies.
The function is,
$s(t)=-16t^2+96t+960$ .
Maximum height is when velocity is zero.
Thus,
$s'(t)=v(t)=-32t+96=0$
The amount of time that passes is when the height is zero when she comes crashing down.