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ThePerfectHacker · #1 December 27th, 2006, 08:30 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):
$\boxed{\lim_{ x\to c}\frac{f(x)-f(c)}{x-c}}$ .

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").

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