Scribe Post 10/10

Today, we began class by going over the answers to the super correction follow-up test. Mr. O’Brien pointed out a few things to remember and pay attention to:
3. Calculator section of the test, much more efficient to use technology when you can rather solving algebraically.
4. a. change in y over change in x problem
b. choose points just above and just below 1 then use slope formula. Don’t forget to round to three decimal places!
5. Intermediate Value Theorem
7. Non calculator side of the test. Algebra!
a. We were asked to find the limit of:


This can be found algebraically, but Mr. OB pointed out that this limit looks very much like a derivative:



Which means that there is a second, clever way of finding this limit. Looking at our equation and the derivative we can see that


This is a power function and can also be written as



In this form the Power Rule (aka Damian trick) can be used to find the derivative.



Now that we have a derivative of the function we can find the derivative when x=5.


Because the limit in 7a was a derivative we were able to use this method to find the limit.

UPDATE

IW 2 Worksheet
8. Asked us to use nDeriv to find the derivative of at x=0 and whether or not the calculator told the truth.
nDeriv found that the derivative of f(x) was 0. To answer the second part of the question we need to understand how the calculator calculates nDeriv. nDeriv is calculated by taking a point just above and a point just below the given x value and using those points to find the slope of the tangent. However when we graph this function we see that the limit does not exist because as you zoom in on the graph it never flat lines, it is a corner.
The limit does exist. It is the derivative that does not exist. The derivative doesn't exist because two different slopes are approaching the same point.

IW 2 Book Problems
40. If function f is differentiable at x=a, then it is continuous. 
We want to show that the graph is continuous at x=a.
Continuous at x=a:

Equivalent to saying:

We need to show that the LHS=RHS
PROOF:
LHS= 
Multiply numerator and denominator by (x-a), a Fufoo.

The first limit in that equation is the derivative and the second, when a is substituted in for x equals 0.


0 = RHS
Therefore, if a function is differentiable at a point, then the function is continuous.

41. If a function is continuous it must be differentiable.
If we think about the graph of the absolute value of x, which we know is a continuous graph, but we also know has a corner then we can conclude that this statement is false. Its false because functions may still be continuous, like the absolute value of x, but have a corner, cusp, or vertical tangent, which we know makes a function non differentiable.

IW 3 pg. 124 and 126
There were four questions posted that we went over as a class.
39. At what points are the tangents of  parallel to the x-axis. 
First we found the derivative of the function because if the tangents are parallel to the x-axis then we are able to set the derivative of the equation equal to 0.
From here we solved for x and found that x=2,-1. Since the question asked for points we had to go back to the original equation, not the derivative, and plug in the two values of x.  We got the points (2, 0) and (-1, 27). To fully appreciate what we had just found we opened geogebra and graphed the equation and plotted the two points:



We found the the minimum and maximum points of the function using the derivative instead of the min max on our calculator!

53) There was a question 53 posted that asked why was constant. Its constant because = 31.00627668... and the derivative is still just 0.

2) A question was posted asking what a normal line was.
A normal line to a function f at a point A is the line perpendicular to the tangent of f at point A.

To find the equation of a normal line, first find the derivative of the function at point A, which is the slope of the tangent line. The normal line is perpendicular to the tangent line which means that the slope of the normal line is the opposite reciprocal of the slope of the tangent line. Then using point slope form you can find the equation of the normal line.

32) For this problem someone asked how to apply the Power Rule to .
The first step to applying this power rule is to get the equation into a form where the power rule can be used, so rewrite the square roots as fractional exponents.
  Then we are able to apply the power rule and find the derivative:
OB pointed out that on an exam this form of the answer would not appear as one of the choices, even though it is correct, because if a question has radicals then the answer usually does too. So we can rewrite the derivative with radicals like this:




After we finished going over all IW questions we moved on to some new stuff!
Derivatives of Products and Quotients 
aka the Product and Quotient Rules.
Product Rule:
 
Instead of giving this to us first, OB gave us an example and had us try to figure out the rule. In the example instead of using g(x) and f(x) we used u and v because it was easier to write and look at.
Example:
  
   
Rule: 

                               
                      
After figuring out what the product rule was we proved it.


The section outlined in yellow in step 3 is a funny form of zero. We used a funny form of zero instead of one here because we wanted to add something into the equation without changing the value, instead of multiply.

Quotient Rule:
 

In terms of u and v: 
We did not prove this in class but the following link goes through the proof step by step.
http://people.hofstra.edu/stefan_waner/proofs/quotientruleproof.html
The following video gives some examples of the quotient rule as well as a mnemonic device that might help you remember it.

IW #'s 1-4 due next class
Quiz on 1-4 next class
Our next scribe will be Alice!