More of the Chain rule, Natural Logarithms, e! Sugoi!

Previously on AP Calculus 2008...
(Monday, November 10th)

We have 4 rules for derivatives so far:

The Power Rule: d/dx [x^(n)] nx^(n-1)
The Product Rule: d/dx [f(x) * g(x)] = f(x) * g'(x) + g(x) * f'(x)
The Quotient Rule: d/dx [f(x) / g(x)] = [g(x) * f'(x) - g'(x) * f(x)] / [g(x)]^2
The Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g(x)

And now the continuation...

So, we basically continued working with these rules. The topic of the day was The Leibniz notation for the Chain Rule.

(All this is in the book, pages 220 to 222)

We have two differentiable functions, defined thus:

y = f(u) and u = g(x)

Therefore y = f(g(x))

Then we know: dy/du = f'(u) and du/dx = g'(x). If we apply the chain rule...

dy/dx = f'(u) * g'(x) = dy/du * du/dx

Simply put: dy/dx = dy/du * du/dx

So if we put this into an example:

y(u) = (sin3x)^6

u = sin3x

We can apply the chain rule and power rule here to find dy/dx y(u):

dy/du = (6sin3x)^5

dy/du = 6u^5

Now here we start with the "layers" Dr. Eviatar talked about. Theoretically, your function could be so long that there would be functions within functions (within functions, within functions...!), thus having many "layers" of functions. The usefulness of the chain rule is that we can use it to seperate the layers and find the derivative.

So in this example...

dy/du * du/dx = dy/dx 

The thing is, finding the derivative of sin3x (du/dx) isnt simple. So we'll just seperate it into "layers" with the chain rule, again. Let's have another variable, v.

du/dx = dv/dx * du/dv

And have v = 3x

So now, we have an equation within an equation.

u = sinv

du/dv = cosv

dv/dx = 3

du/dx = cosv * 3 = 3cosv = 3cos3x

And finally find the derivative:

dy/dx = 6sin3x^5 * 3cos3x

Arent you glad there were just two layers?

After that, we looked over the question on page 221 about the oil slick. This question is the same as any other derivative question we've looked at, but with different variables.

dr/dt = 3ft/hr

dA/dr = 10ft^2/ft

A = f(r) where r = g(t)  therefore A = f(g(t)). Thus we can apply the chain rule:

dA/dt  = dA/dr * dr/dt = 10ft^2/ft * 3ft/hr = 30ft^2/hr

Simple, right?

And finally, we started into the mysterious and awe inspiring power of e and the natural logarithm.

e^(lnx) = x

We find the derivative of both sides:

d/dx [e^lnx] = d/dx [x]

The derivative of x is 1 because x is a constant. We apply the chain rule to find the derivative of e^lnx:

e^lnx * d/dx(lnx) = 1 

Remember, we stated earlier on that e^lnx = x...

x * d/dx(lnx) = 1

So finally, the derivative of lnx must be:

d/dx [lnx] = 1/x

So now that we know this...

And that concludes yesterdays episode of AP Calculus 2008. Tune in tomorrow for more exciting mathematical developments!

Oh and we had some questions to solve...

Find the derivative of the following and which rule(s) would you use to solve it:

Function | Rule

---------------

ln2x | Chain Rule

xlnx | Product Rule

(lnx)^3 | Chain + Power Rule

And some homework I'm sure you all did. So because you all did it, I dont have to write down the question numbers you were supposed to do. Because you already did them.

The next scribe is "Hi Im Justu"

Whoever the heck that is.

October 21, 2008

Wow, I'm scribe again! Haha, thanks Lawrence.
Today in class we had our test on Derivative Functions, it was quite tough, because there were questions in there that we hadn`t reviewed before, but next unit should be easier.
That is all we did in class today, now you know why I wanted to be scribe.
The next scribe will be Joseph!

The Pre-Test

Hope you've all been studying, because I know for a fact that some people were as confused as I was on that pre-test.

Today we just had a pre-test and went over the answers. We were all in class, so I don't think I have to go over the answers in detail.

Answers:
1) A
2)D
3)C
4)D
5)D
6)
a)-2.25
b)f(x) = -2.25 (x-5.2) + 8.8
c)8.665
d)Negative

So by request, Francis asked to be scribe, for the same reasons everyone would want to be.
Therefore, Francis will be tomorrow's scribe.

Friday's class of a couple Theroms.

Today in class we learnt about a couple new theorems. The first one is the "Intermediate Value Theorem". This theorem pretty much states that according to slide 1, found on the graph, that if y= f(x) is continuous on an interval [a, b] and y = k in between points y=f(b) and y=f(a) then there is an x-value of c where f(c) = k. If a discontinuity is found, remove it. I believe that what Mrs. E means is if there is a discontinuity between the chosen interval of [a,b] then chose a new interval by moving it to a place on the graph where there is no discontinuity. Confusing? Yeah, just look at the graph and read the notes a couple times. That's what I did.

We also looked at how to find zeros on a continuous function without our calculators. We were given an example on slide 2. f(x) = x³-3x+1. To find the zero of this function, we chose points [0,1] for this example. We then found out where the x-values were either positive or negative. On the example the point 0 was input into the function and equaled to 1. Which was positive ( f(0) = 0-3(0) +1 = 1). At the point of 1, the function was a equal to a negative number ( f(1) = 1 -3(1) + 1 = -1). Now we know that the zero of the function will be in between those two points. Now we want to hone in on a smaller region, so we should cut that interval of 0 to 1, in half. So we find the f(1/2) and see if it's positive or negative, so we know where to look next. The function of f(1/2) turned out to be negative, as found on the slide. We then know to look between points [0 and 1/2] because the zero will be in between a negative and positive value. It's best to divide the section by half to cover greater amount of space between the points. The next point we used was 1/4. We input this into the function and found it to be positive. Now we can look between points [1/4 and 1/2]. We then used the point of 3/8 in our function, which is in between our given points. It turned out this was negative, so now we know our zero is between 1/4 and 3/8 which is small enough of an interval where our zero is found, so we stopped there.

The next theorem we used was entitled the "Extreme Value Theorem". This states that on a continuous function, it will always have a maximum and minimum and will be an extremum value from where all points on the function will either larger or smaller depending if its the maximum or minimum extremum. Lets say that on a continuous function on interval [a,b] there exists numbers c and d where all x-values in [a,b] f(c) ≤ f(x) ≤ f(d). d will be the maximum and all x-values will be less that or equal to d. c is the minimum and all x-values will be greater that c or equal.

That's all we did in today's Calculus class. If anything is wrong, talk to me and I'll be sure to change it. The next scribe will be Lawrence!

Continuity, and some more Limits

Alrighty guys, my name is Justus and I will be your scribe for today. Lets get started.



This image is basically a modified version of the first slide. Now this example was used to explain the concept of continuity. If you look at the image, I've drawn 3 sets of lines, the magenta (purpley) ones, the maroony ones, and the aqua ones. If you try to travel from one aqua line, to the other, you can do it easily, and without any kind of sudden changes. Therefore the graph is said to be continuous along that interval. The same goes for the magenta. However, when trying to travel from one maroon line to another, difficulty arises. At x = 0, there is a sudden change in direction, the line is no longer continuous, and there is now said to be a discontinuity. It is called a discontinuity because there is no way to get from one maroon line to the other in a continuous fashion. The next point we got into, were 3 defining rules of continuity, which I shall type out
here for everyone to see.

Continuity

A function is continuous at x = a if;
1.) lim f(x) exists
x->a
2.) f is defined at a
3.) lim f(x) = f(a)

So after going over these rules, we got some examples on the board, so we might get a better understanding of what each means.

For the first we came up with the following

1.) Example: lim f(x) = 1/x
x->a
Counterexample: (look at slide)

So as seen in this slide image here, the example graph produces a nice, rule following reciprocal graph (reciprocal of the graph f(x) = x), whereas the counterexample gives a piecewise function, which has that huge gap/jump/aka. Discontinuity at x = 10 approximately. Alright, moving on.

2.) Example: f(x) = x/2 at x = 1
Counterexample: f(x) = 1/x at x = 0

In the second example there, it is pretty obvious to see what the problem is. In the counter example, the function is NOT defined at x = 0 because you are then dividing by zero, and we all know how much fun that is. In the example however, all is good and well, and f is defined at a, as required to be continuous.

3.) Example: lim 1/x = 1 = f(1)
x->1
Counterexample: (check slide)


Okay, so the counterexample in this one, might seem a little confusing but I'll try to explain it as best as I can. Basically, its saying that f(x) = x² when x does not equal 2; When x DOES equal 2, then f(x) = 5. Thus you get the graph with the weird point floating in the air. This point also leads to that magical thing called a discontinuity again.

Now it was at this time Dr. E said to checkout Pg's 140-141.

The final little tidbit in the lesson involved continuity on an interval, which basically says that if function is continuous from points a, and b, it is continuous on that open interval a to b. This is shown in more detail on the following slide

The lim f(x) = f(a) basically means that starting above a and moving towards it, the function
x->a⁺
is continuous. The opposite is true for the lim f(x) = f(b) which says that starting below b and x->b^-
moving towards it the function is continuous.



Now thats about it for my scribe post for today. I know we did a couple other things, (two slides), but I forgot my textbook at the school, and I didn't really understand them at all, so if someone could help out with that (ie by talking about it a little bit in their post, that would help me tremendously, and I could then go and fix mine.) Either way, like I said, I will need to get my text to help me explain those two slides.

So I think thats everything, which only leaves choosing the next scribe. I've decided it will be francis mkay. kay.

OH And I almost forget, theres alittle note about pg 143 here in my notes, which means it's probably a good idea to check that page out. As far as I remember there weren't any questions assigned for homework for tonight though.

Anyways I'm off, because I have lots of other things to do now. Hopefully my post made enough sense, and I apoligize for any discontinuities in it; feel free to point those out btw, so I may fix them.

ciao

OCTOBER 8: INFLECTION POINTS, CRITICAL POINTS CONCAVITY AND DERIVS REVIEW

Hello blog. It's me Jamie again with today's scribe post or yesterday's really. I'm starting it up again, after it abruptly stopped for that day. At the moment the class is basically at the peak of chapter 2. It's still a bit vague for me but having class with Mrs. Karras and her famous notations like "fcn" helps.

First we spent the first seven minutes or so settling in and having a few warm up discussions, relating to math of course. Most of it. Time is golden, right? We were deciding amongst ourselves whether we were audio/visual learners who absolutely adore diagrams [and possibly lectures] and those theoretic learners who love...well, theorems and of course, there are the extraordinary who are both. But let's get down to business.

We started the class reviewing what we should know... and I guess emphasizing that knowledge on derivatives. We began with the FIRST DERIVATIVE TEST, although it was not called that in the text. Of course, this is one way to find the relative maxima or minima of a function, without being too tedious.

Like on the slides, there are the notes describing the function itself. But to carve this inside your head, I guess, I'll repeat. If this derivative test is to be used, it follows these conditions:

• f is a continuous function located on a closed interval [a, b]
• f is differentiable on an open interval (a, b)

Along with those two conditions, are indicators of which direction the intervals would be, which are:

1. If f '(x) is pos^ve for the x values in the open interval, then f '(x) is INCREASING on the closed interval, [a, b]
2. If f '(x) is neg^ve for the x values in the open interval, then f '(x) is DECREASING on the closed interval.
   

Secondly, to make sure we were all on the same train of thought, we defined EXTREMA. This the term describing either the local minimum or maximum of a differentiable fcn. 

Extrema occur in a fcn where f is on an interval (I) and c (a critical point; more info on critical points here) is a number in I. They also follow these conditions:

• f (c) is the max value of the function on the interval given that f '(x) is less than or equal to f (c) or vice versa, which is shown in the next pt. [LOCAL MAX]
• f (c) is the min value of the fcn on the interval given that f '(x) is greater than or equal to f (c). [LOCAL MIN]
  

But if f (c) is constant, then the function is BOTH a max and min value for EVERY real number c, of course considering that not every function has both a minimum and/or a maximum.

[more on...] CRITICAL NUMBERS 

This next section was a further explanation of critical numbers. But basically, if a function has a local extrema then either f '(c) equals zero or is undefined.

CONCAVITY and the 2nd DERIV TEST

This was one moment in class where I was quite intrigued. It was one of those moments where you learn a concept through a different perspective, in this case, I learned a new way to describe what concavity is. Concave UP was when the function was curving upwards above the tangent line and concave DOWN was if it was curving downwards below the tangent line.

**NOTE VOCABULARY: INFLECTION POINT [click on the word for a link]

THE SECOND DERIVATIVE TEST

This occurs when a function is differentiable on an open interval containing "c" and f '(c) = 0. Once again, this indicates where there is a local extrema. If f "(c) is negative, then f has a local MAX at "c". If f "(c) is positive, then f has a local MINIMUM at "c".

With taking notes aside, we were given a warm up question that applied some of the things we've learned so far:

If f (x) = 12 + 2x² - x⁴ , use the 2nd derivative test to find the local max and min for f . Discuss concavity, find points of inflection and sketch the graph.

THE SOLUTION:

Use the power law to differentiate the first and second derivative. The power rule is also known as: f ' (x) = nx^{n - 1} 

f ' (x) = 4x - 4x³  first deriv

f " (x) = 4 - 12x^{2 =} 4x (1 - x)(1 + x) second deriv

To find critical numbers, we must know that critical numbers exist when f '(x) [or the 2nd deriv] = 0, meaning that the critical numbers were {-1, 0, 1}.

Then to continue this problem, finding the inflected points makes finding the concavity easier since we can find whether the function increases or decreases by separating the intervals accordingly by using the critical numbers and plugging numbers that were in b/w intervals into the first derivative. But then the bell rang and we are to conclude this problem on Tuesday. Also, Mrs. Karras will be back to teach us about limits. 

Later that week though, just a reminder, when Dr. Eviatar comes back, we will have our pre-test for this unit and the test the following day. My post today wasn't quite exciting, but at least there was no major advertizing... [or is there?] But anyhow... I'm famished-- so are my fingers typing, so I'm done for tonight. The next scribe derived from randomosity is Richard.

Oct.6th/08

Hi, it's me YiNan! I just found out that I'm the scribe for today!!
Well, we did section 2.5 Critical Numbers; Relative Maximum and Minimum Points.

We start off with the local maximum and the local minimum.



A local minimum, also called a relative minimum, is a minimum within some neighborhood that need not be (but may be) a global minimum.

A local maximum, also called a relative maximum, is a maximum within some neighborhood that need not be (but may be) a global maximum.

Global minimum is the smallest overall value of a set, function, etc., over its entire range. It is impossible to construct an algorithm that will find a global minimum for an arbitrary function.

Global Maximum is the largest overall value of a set, function, etc., over its entire range. It is impossible to construct an algorithm that will find a global maximum for an arbitrary function.

As x^2 is increasing, 2x is also increasing.

Critical Point is where f'(c)+0 or does not exist.



Then we did Example 1 on page 115.

Homework for today is 1,7,9,11,15,19,22,23,25 on Exercises 2.5

I'm done! next scribe is Yi Cong ( good luck! ^_^)

Slides October 6th

Here they are ...

AP Calculus October 6th, 2008

View SlideShare presentation or Upload your own. (tags: apcalc2008 calculus)

Me and My TI-83.

In Fridays class, we explored the different ways of finding derivatives with our handy calculator.

On slide 2, we used the change in y / change in x to find the derivative. That method of course is where we find the slope of two points that are very close to the point we want which is where x =2.

On slide 3, the method we used is the quotient one. All we did with this question was to plug in 3 in x and chose 0.01 as our interval. From there we simplified and got the answer of 0.33. This is my favourite method for some reason. We also saw that we'd get the same answer with using the change in y / change in x way. After plugging in numbers like 2, 3, and 4 we noticed there was a pattern. The derivative for x=2 was .5, x=3 was .33, and x=4 was .25. Therefore, the derivative for ln(x) was 1/x.

From what we learned from Mrs. Karras, not all points are differentiable. The points that aren't are where the graph is....pointy! Like on the slide, the absolute graph is not differentiable at x=0. This is because the tangent slopes are -1 from the one side and 1 on the other side.

The other methods we learned were the dy/dx thing in our calculator and the nderive function in our calcs.

That's pretty muuuch it. *looks at scribe list.* The next scribe will be Yi Nan i suppose. Have fun scribing tomorrow. Now is time for me to enjoy the last hours of my weekend :(

Slides October 3rd

Here they are ... one of these days I'll make pretty slides like Mr. K. For now, I'm glad to be putting them on the blog on a regular basis!

AP Calculus October 3rd, 2008

View SlideShare presentation or Upload your own. (tags: math apcalc2008)

The Power Rule...

Okay this is Benofschool and going for another attempt to explain the Power Rule...
The following image shows the equation work that was used to derive the rule. I will explain each line after the image.

Line 1: Okay lets find the derivative of x to the power of n at x

Line 2: So we just throw it into the derivative formula like usual

Line 3: We have to expand the binomial in the brackets. There are 3 things to realize. First the first term will be x to the power of n regardless to what n is. Secondly, the next term will always have n as the coefficient and n-1 as the exponent and also multiplied by one h. Finally, every other term will have a unknown coefficient (this coefficient will become irrelevant and you will see later) and the exponent of h will increase by 1. So in the third term the power of h will be 2 and in the fourth it is 3 and so on

Line 4: Now we can get rid of x to the power of n because it is subtracted to zero by the other term.

Line 5: Now we factor out an h. Notice that the first term in Line 4 no long has an h variable in the term and every other term afterward still has an h variable remaining. This is the key to seeing the rule

Line 6: Now the h reduces and you are left with what was in the brackets in Line 5.

Line 7: Now we apply the limit and substitute all the h values with zero since there is no h in the brackets anymore. This will cause all terms containing the h variable to become zero. Since the first term does not contain an h variable, it won't be affected by the limit.

Line 8: This leaves us with the rule because all of the other terms are turned into zero thus not affecting the first term at all.

I hope this helped. If there is any more help required, please talk to Dr. Eviatar or myself for help...

Okay well I tried =P

The Derivative Function (Cont'd)

OVERVIEW:

• Alternative Notations for the Derivative
• y = nx^(n-1)
• Derive f(x) = x^3
• Derive y = -1/x

We finally finished our lessons with the topics covered in 2.3 of our Calculus: Concepts and Calculators textbook. According to yesterday's substitute's substitute, the textbook we use is the worst textbook she has ever seen in her life because it's convoluted and because its explanations can be explained better too, but that is why we have the BLOG, our online interactive textbook to help better our understanding.


Alternative Notations for the Derivative

f'(a) is the symbol used by Lagrange. E.g. If f(x) = x^2, its derivative is f'(x) = 2x.
dy/dx is the symbol used by Leibniz. E.g. If f(x) = x^2, its derivative is dy/dx = 2x.


y = nx^(n-1)

Then the Benchmeister took over the class and starting teaching us. He taught us a shortcut in finding the formula of a derivative by using a formula. We were so amazed at his genius work that probably only Dr. Eviatar and Bench can understand it, because I know I didn't. Mr. K says, "If you don't understand something in class, you can bet that six other people in that class probably didn't get it as well." So I'm counting on other people to fill in the blanks that I've left in this blog post. Also, the formula that Bench gave us is something to be taught some time in the future. Anyways, we ended up getting the formula, y = nx^(n-1).


Derive f(x) = x^3.



Derive y = -1/x.Using the y = nx^(n-1) formula,
y = nx^(n-1)
y = (-1)x^[(-1)-1]
y = -x^-2
y = -1/x^2


END NOTES:

• Homework: 2.3 The Derivative Function #1, 5, 7, 9, 11, 12, 21, 27
• Note: The principal says we won't get our laptops until he talks to Mr. K (assuming that Mr. K gets back in action at our school). Well, guess what? WE ARE THE RADICAL CALCULUS GROUP. WE WANT OUR LAPTOPS, AND WE WANT OUR LAPTOPS NOW!
  
• Next scribe is Joyce.

Differentiables

Hey guise, its me Not Paul making my blog post for today because I'm Not* Scribe.
*Truthiness may vary

Anyway, onto actual Mathematics, those darn derivative deriving derivates and what in the heck a differentiable function is. I cannot however tell you with absolute certainty what a differential is, but I have a good idea we're going to find out soon enough.

The basic definition of a differentiable is a function where the derivative can be found for any point of a defined segment of that function.

Which is why f(x) = x^2 is differentiable, because we can find the derivative for every point in the function or any portion of the function.

If you get a function with "kinks" or very pointy points when you graph it, it won't have derivatives at those pointy points and thus is not differentiable for the entire function.

However, if you define a segment of that function that doesn't have pointy points then you can say that the function is differentiable for this segment of the function.

The teacher's definition is as follows:

"A function (fcn) is said to be differentiable at variable a if at variable a f '(a) exists. It is called differentiable on an interval if it is differentiable for every number in the interval."


Here's the example from yesterday that we did today [Ex 1] (which we also did yesterday), and another example we did [Ex 2]. Both of these are differentiable, because the derivative f '(x) exists at x = 0. (in Example 1 replace variable a with x)

Ex 1)


(Click to view larger)

Ex 2)


(Click to view larger)

This is an example where f '(0) doesn't exist, therefore it is not differentiable.

Ex 3)



(Click to view larger)

As shown in the image, the function f(x) = |x| is not differentiable because when x = 0, its derivative does not exist/cannot be determined tangentially. If graphed, y = |x| would show it has a "kink" or pointy point at x=0.

At the end of class we started talking about the different people who invented Calculus, and importantly Joseph Louis Lagrange's contribution of f '(x) = dy/dx, which will come up later in the course I'm sure of it, because of this Wikipedia article.

The next scribe shall be...

Zeff (zeph)

Sorry this post is uncredibly late, you can blame Wired and NeoGAF for their unfairly up-to-date news. I started working on this post at 9pm and finished 4 hours later. Of course atleast one of those hours was spent trying to figure out OpenOffice...

Working with Derivatives

Hello again everyone, tis I, Kristina, back for another scribe post! Today we had the wonderful Mrs. Karras as our sub for today, along for the next 3 days.

Anywho, we mostly worked with the formula of a derivative (slope of a tangent line). If you've forgotten what that formula was, here it is again:

While Mrs. Karras finished writing that up on the smartboard, someone asked to re-explore what a limit was. She reviewed it to us briefly, explaining that a limit basically means that something is approaching another without touching it. An example she gave us was that she could get as close to Paul and invade his personal space as much as she wanted but she couldn't technically sit right on top of him since there was basically no room to do so and such, also, he may have exploded =(

She also bestowed us the gift of notes as to what the derivative of a function is in more detail. Due to her not being able to post the slides with these notes on it till Thursday, I shall kindly post them here for anyone that missed it or was not quick enough to take them down.

"The derivative of a fcn (a gift she bestowed upon us, it is short for function) at a chosen input value describes the best linear approximation of the fcn near the x-value under consideration."

With the help of this graph (I forgot the arrows at the ends of the parabola xD), we can see that in this case "a" is the inputted value mentioned in the above note and the ..faded...red line is the linear approximation of the fcn near that inputted value, "a".

"For a fcn with a single real variable, the derivative at that point equals the slope of the tangent line of the graph."

Meaning that we can input that value into the derivative formula since the definition of a derivative is that it is the slope of a tangent line.


Wahoo, now for some examples she gave us!

In the first example, she gave us the function f(x) = 2x² - 5x + 6. We were told to find f'(4). Solving this is as easy as bringing paper plates and cups for Pi Day (Yes, I'm thinking about it already). Back on track, all you have to do is input the value 4 into the derivative formula shown at the top of this post. If you're still lost, here is what it should look like as you work through the problem:

Tada! The derivative is 11 since as "h" approaches 0, it is shown that the value is close to 11 thus the answer. Simple algebra, right?! And since some people may make mistakes in their algebra during a question like this, here's a song to help you (I was bored so I decided to search "math" on imeem TEEHEE)

Kay, so it doesn't exactly help with the algebra...sorta, but I thought it was kinda neat so HA!

Back on track! The second example we got asked us to find the derivative of f(x) = x² - 3x for a number "a". How we solve this is the exact same as the last example, except we are working with a variable, "a". Once solved, we should find out that the answer is:

Now, if we were given a question where it asks us to find the derivative f'(1) for the same function given in the last example, you can do it the long and tedious way (but not hard!) by using the same procedure as example 1, but you don't have to! This is because we've already found the derivative for f(x) = x² - 3x earlier, which is 2a - 3. We can simply think of that as a function and then input 1 into that and voila! Simple and easy, also elegant.

For our last example, we were given the function f(x) = x² and were asked to estimate the derivative at the points {-2, -1, 0, 1, 2}. First we must graph this. After you have graphed it, draw the tangent line to those points and then just take an estimate at what the slope of that tangent line is. To check your answer, just input the points into the derivative equation and that's that!



Now that that's out of the way, the next scribe is.. Not Paul. Thanks for the confusing name, Paul :(. Oh and starting from page 100 up to 105 is review on some of the stuff I've talked about. Also, homework is on page 105. There are 8 questions but just do enough of them till you understand it. Anyways, I'm done for tonight. Have a nice day :)

Deriving Derivatives Using a Derived Derivative

Today, a monday, a tiring monday, a monday that wasn't the same as last monday. Sorry this has nothing to do with Calculus. Okay back to class...

Remember when we were working with derivatives last week? We found the derivative of f(x) = x² at the point where x=1. that could be found using the derivative definition found on previous slides. There are many ways to use this definition, some are more tedious than others but they should all arrive at the same answer. It is like Mr. K's analogy for using more force than necessary.

When you are going to hang up a painting. You need a nail in the wall, should you use the little hammer or should you use the sledge hammer. Common sense would tell us to use the little hammer, because the sledge hammer would blast a hole through the wall. So use the way that would save you time and effort for more tougher questions.

The easier way to find this is to plug the function and x = 1 into the definition and you get:

Picture courtesy of Sitmo

Now all you have to do is to simply the function. What you have to do is to get rid of the h in the denominator or else when you apply the limit, the equation will explode or be undefined. The answer will be 2 + h. so when you apply the limit, h will be so small towards zero that it will be seen as negligible and so the derivative will be 2.

After that little review we learned about locally linear points. Points can be called locally linear when you zoom into a point on your calculator and eventually it appears as a straight line. Take the graph of f(x)=|x²+1| for example:

Picture courtesy of Fooplot.com

If you zoom into the point (2, 3) you will eventually find a straight line meaning that the point is locally linear. But if you zoom into any point where this function has a zero you will never find a line. You will only see a corner or a spiky spot. It will never be a line so it is not locally linear.

So that is what we did today in class. Homework is exercise 2.2 question numbers: 1, 2, 3, 5, 10a, 11, 12, 14, 15, 17. But remember if you do not feel comfortable with the subject just practice more. Everybody is more than welcome to do the entire exercise.

So the next scibe will be Kristina.
Good night...

Slides September 29th

Here they are ....

AP Calculus September 29th 2008

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September 26th

Hi! I'm Yi Nan. ^_^

On Friday, we continued on the unit of derivative function.(2.2)
First, we start with reviewing tangent lines.

Then we did the questions on page 84
instantaneous velocity (at t+a) :





average velocity= distance traveled over time elapsed:



I didn't really get it so I don't really know how to explain it. Sorry!!!*_*

I think that's all we have done! So the next scribe will be Hai Yan!!!
Have a nice day!!

Slides September 26th

Here they are. Sorry about the long slide - I couldn't really see a good way to split it up. I'm open to suggestions ;-).

AP Calculus Slides September 26th 2008

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Slides September 24th

Here they are ....

Calculus AP Slides September 24 2008

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