Scribe Post 10/4/12

In today’s class, we started off with a forty minute follow-up test on Unit 1. Once the forty minutes were up, OB called the class together and asked everyone to turn in their tests. Then, he moved on to the ever-exciting topic of MATH TEAM! 

He called everyone’s attention to the smart board which displayed the rankings of each high school in the “Central Math Team.” Two years ago, we didn’t do so hot. Out of 96 schools, Camden Hills only placed 93rd! This was an outrage, especially since even Rockland beat us! As if we couldn’t have done any worse, OB then showed us the stats from last year, where the only school we beat in the state was Wiscasset. That all changed this year, however. Wednesday’s math meet, narrated by the captivating voice of Mr. McKenzie, had two rooms filled with mathletes. The meet concluded with Camden Hills finishing second behind Lincoln Academy, an outstanding improvement from the past two years. We finished 30th in the state in our first meet. Had we taken the top ten scores of our students at this meet, we would have finished 17th in the state. O’Brien was then quick to congratulate one of our own, Lexi, as being one of the top 10 mathletes from Wednesday’s meet! GO Lexi! Eddie McClusky was the highest scorer of OB’s students, while Alden Parker was the highest scorer for the school, finishing 39th in the state. OB then called out both Margaux and Weston on their absence from Wednesday’s meet and questioned them on their commitment to math team. Clearly, OB is really cracking down the whip on this one. The next math meet was announced to be on October 31st (a.k.a. Halloween). The reason for this scheduling was questioned by Lexi and received a response from OB: “Everyone will come to the math meet in costume!” It’s safe to say, everyone will be attending... 

Now onto the productive work we did in class today. This class, we went over problems 3, 10, and 11 from IW #2. OB refuses to post the answers up to IW#2 because he wants us to keep working on the and use Geogebra to help us. OB claims that he’ll have the answers up before next Friday, but we’ll see how that goes.. 

Question 3 and 10 are both proofs involving derivatives of functions. To help us get a better understanding, or a visualization of what were were being asked to find, OB drew two pictures up on the board that can be applied to both problems. The only difference between the two pictures is that in one, x is approaching c and in the other, (x+h) is approaching x:

OB told us to think of the function f and think of the derivative as the slope of the tangent line to f. As an example, we can use the first picture to find the derivative at the value c. The tangent line to f at the point c is shown in red. We can then take a second point (other than c) and create a secant line between the two points. As the second point, which in the picture is x, travels towards the first point, c, the slope of the secant line (shown in green) will get closer and closer to c until it becomes the tangent line. This make sense because the tangent is the limit of the secant as the two points come together. The reason we need two points to find the derivative, or slope of the tangent line, is because you can’t have a slope with a single point.

Once we had this nice visual, it was time to tackle #3 of IW#2. The first part of question 3 asks us to think about what the derivative of f(x)=k would be at any value of x. Because f(x)=k is a constant function, it is simply a horizontal line. What’s the slope of a horizontal line? Zero! The problem get’s a bit trickier, however, when we’re asked to prove algebraically that the derivative of the constant function is indeed zero. To do this, we have to use our handy definition of derivatives that OB keeps telling us to memorize (especially in the shower):

The proof is completed as follows:

If:

Then prove:

By plugging k in for f(x) and f(c), our handy definition of derivatives at a point c becomes:

This can be simplified to:

Because zero divided by anything is zero, it follows that:

The second part of question 3 asked us what the derivative of a linear function would be at any value of x. The equation, as everyone knows, for linear functions is:

"What's the slope of this function?" asks OB. "m" responds a unanimous class! Now that we know that the derivative of a linear function is m, it's time for us to prove it algebraically. 

If:

Prove:

The proof is as follows:

Question 10 wants us to prove algebraically the derivative of the sum of two functions f and g, the relationship of which we were given at the end of class on Tuesday:

To do this, we have to use the other definition of a derivative as h approaches zero:

In the next equation of what we're trying to prove, s(x) is used as a symbol for the sum of x. The proof is as followed: 

If:

Prove that: 

Using the definition of derivatives as h approaches zero, we can apply our given function to that definition:

In terms of f and g, this becomes:

Getting rid of the brackets and distributing the negative sign, the limit simplifies to:

Here comes the really cool part! Using the commutative property, we can group the f(x)s together and the g(x)s together and create two separate fractions, each with their own limit!

If these two fractions look familiar to you, they should! They're the definitions of derivatives as h approaches zero! Therefore, this can be simplified to:

And just like that, we've proven the sum rule of derivatives!

As cool as all this is, it's time to prove something REALLY useful, as OB says: the Power Theorem. This theorem states that for any real number of n, the derivative of x raised to the nth power is equal to n multiplied by x raised to the n-1 power:

To prove this, we let:

Using the same derivative rule we used in question 10, we can start to prove the limit as follows:

The question now becomes: "how do we foil a quantity raised to the n power? The answer to this question is with the Binomial Theorem and Pascal's Triangle (that is, until we learn about factorials). A great video explaining both the binomial theorem and pascal's triangle can be found below:

With an understanding of the binomial theorem and Pascal's triangle, we can continue our proof (note the triangle in the equation is just some unknown coefficient):

The reason I didn't continue on with the above equation is because it simply doesn't matter because all terms will be multiplied by values of h. This is a good thing, great actually, because the x's raised to the nth power will cancel and leave only terms that can be divided by h. This means that we can simplify to:

Almost done! If we were to now sub in 0 for h, the only part that we are left with is:

Quod erat demonstrandum!

IW for next class: 

finish IW#2

IW#3: pg 124/ 1, 5, 11, 25, 29, 32, 37, 39, 53, 54 and pg 126/ 1, 2, and 4

UPDATE!!!
We now, using the chain rule and power rule, know that the derivative of:

Is equal to:

To proof is as follows:

This can be rewritten as:

Then, using the quotient rule...

This is simplified to:

Using the division of exponents property, this becomes: 

And just like that, we now have a way to find the derivative of a function raised to the negative nth power!