Scribe post. 10/16.

Class on the day of October 16th, 2012 was unfortunately shortened to 35 minutes, due to a Pop Tech assembly. Due to this, we got right into class as soon as possible. Starting off with the hand back of the quiz that was taken on the 12th. We then (as a class) went over the quiz.

•UPDATE•
Cool video explaining derivatives of trigonometric functions, including graphs, examples, and algebra here



Quiz questions

1)  , find dy/dx.

With this question you can simply use the power rule to find the derivative, which includes taking the power (4) multiplying by the number in-front of x (3) and subtracting the power (4) by 1, which gives you...




2) Given  , find dy/dx.

It may look different, but this question is very similar to question know. If you know that 
is the same as , you can apply this knowledge to this question. If  = , then  = . Do the same thing for , and you will get , which would be WRONG! That is what I did on my quiz and got a point off because of a careless mistake. For , you can take out the constant and get , which gives you 1/5 multiplied by the derivative of x, which is simply 1. That leaves you with ...

  NOTE: The - 1/5 is already derived, so it is going to stay there in the last step, thats just the way I did it :) )

Now, do the same thing as question 1 and use the power rule to derive the function, getting you...

.
3) Find the points on the curve  where the tangent line is parallel to the x-axis. With this problem, having a calculator sure made it much more simple. If you want to find a tangent line that is parallel to the x axis, you can simply find the maximum and the minimum points of the function, and take the tangent of those two points on the function.

This shows tangent lines of the maximum and the minimum and how they are parallel to the x-axis. On the calculator, it does not give you exactly an x value of 3 and -1, but rounding gives you 3 and -1. To get the y value (because it specifically says points on the curve, which means x and y value) you can either simply look on the calculator, or plug in your x values of 3 and -1 into the given function.

4) Find the equation of the line perpendicular to the tangent to the curve   at the point (2,1).
        When you see the word tangent in this equation, you know you are going to be working with a derivative, so to get it out of the way you can take the derivative of the given function which gives you
. Now that you have the derivative of the function, you can plug in the given x value into the derivative, because you are finding something perpendicular to a tangent line at a specific point (2,3). When you plug this in and get an answer of 8, this gives you your slope of the tangent line. Now that we have the slope of the tangent line, we can use the perpendicular rule by flipping the number, and changing the sign. When you do this, you get .  Now we have our slope to our perpendicular line, and with a slope and points, we can use point slope to get a final equation which would be....



5) Suppose u and v are functions of x that are differentiable at x=3 and that u(3)=2, u'(3)=-5, v(3)=4, and v'(3)=6, Find the value (not values) of d/dx(uv) at x=3.
So this is saying, find the derivative of u * v. When we see a product of u and v, we can use the product rule which is UV'+VU'. We are given all of the values we need, so it is just a matter of plugging them in and solving, and when you do this using the product rule you get. (2)(6) + (4)(-5)
= 12 - 20
= -8

6) Use the x --> c form of the derivative to find the derivative of   at x = 5. Show appropriate working.

This was the longest question of all, and we were limited to using the x approaches c form of the derivative, the form is.................        MALFUNCTION!

                                        Malfunction fixed

7) Where are the four places that a function's derivative does not exist?

1. CUSP
2. CORNER
3. VERTICAL TANGENT
4. DISCONTINUITIES

There are no equations to this, just memorize it! (Shower time).

8) Given , find f'(x). (Hint: Use technology, and be sure you are in radian mode...)

This is pretty much stating, find the value of x = 2 on the derivative of 
On a calculator, you can find the derivative by using nDerive. First, you plug in  for Y1, and for Y2 you want the derivative, so plug in exactly "nDerive(Y1, X, X) and it should give you the derivative. There is really no need to even graph it, you can just go to the table and plug in x=2, and the number underneath Y2 will give you your answer.

9) Consider the graph of the function "h" below. Identify any values of x on the closed interval  where "h" is not differentiable.

This connects to question 7, where functions are non-differentiable. At zero, there are different derivatives of the left and right lines connecting to each other at zero, this means that it is considered a corner. At x = 2, this is discontinuity, which is also non-differentiable. -2 may not look like its even in the function (well it did for me) but the given information of   means you need to include it, and if you do, it is most obviously another discontinuity. So there are your three values where h is non-differentiable. X=0, X=2, and X=-2.

10) Find . No need to simplify.

When you see you need to find a derivative of a quotient, you can use the quotient rule, which is (VU' - UV') / U^2.



= 

= 

= 

=  

After we wen't over those problems on the quiz, we moved onto a couple questions from IW #5 including #23 and #27.

23) The position of a body at time "t" sec is . Find the body's acceleration each time the velocity is zero.

Velocity is ds/dt, so if we find the derivative of this equation, that gives us our velocity which is...



It is set equal to zero because the questions asks the acceleration when the velocity is zero, and we get...



So t = 3 and t = 1

We will now plug these into the acceleration equation which is the second derivative of s/t or the derivative of v which will give us..

Acceleration = 
(Plug in 1 and 3 for t)





This gives us our answers of -6 m/sec^2 and 6m/sec^2

27) Suppose that the dollar cost of producing x washing machines is 
a) Find average cost of producing 100 washing machines
b) Find marginal cost when 100 machines are produced
c) Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly.

a) To find the average of the cost when 100 washing machines are made you need to find the rise over run which is slope, which is the derivative. So, to do this you will do...

. 

Now to do this you can put the equation c(x) into Y1 on your calculator, go to the table and plug in 100 and then plug in 0, subtract the value you get for zero from the value you got for 100 and divide that by 100.

OR, you could do it algebraically like this...

 = 11000

 = 2000

11000 - 2000 = 9000

9000/100 = $90. Which is your answer

*Answer in back is wrong, it says $110*

b) Marginal cost is just the derivative of the original cost function. So to find the marginal cost when 100 machines are being produced, we can take the derivative of the cost and plug in 100 to get an exact value.

(plug in 100)

$80.00 is your answer.

c) To calculate how much one washing machine will cost, we will use instantaneous rate of change using 101 washing machines and 100 washing machines.

(It's supposed to be divided by 1 but the equation editor would not let me so pretend there is a 1 there, even though it doesn't matter)

= 11079.9 - 11000

= $79.90


Up to this point we didn't start anything new in class really, then we came into derivatives of trig functions. With these, it brought us back to Pre AP-Calc where O'B told us about who hangs out with who (referring to trig functions), where it can really help you while finding the derivatives of these equations.

HINT: Whenever you are finding the derivative of a "co" function, it will be negative!

(sinx)' = cos(x)

This is the graphic way of finding the derivative of (sinx). The thinner line, as you should be able to see is plain old sin(x). Below is a list of steps to go through to successfully come up with its derivative
*Make sure to set x-axis in terms of pie*
1. Make a point anywhere on the function sin(x)
2. Make a tangent line at that point to the function sin(x).
3. Once you have your point and tangent line on that point, click the slope tool, and then simply click on the tangent lines equation. This will give you that red triangle that appears in the image above, it simply gives you an image of the slope of that tangent line.
4. When you have this slope, you will go down to where you enter in functions and equations and put in "(x(A),m) which means you are taking the x value of that A point for your x value, and the slope "m" for the y value, which gives you a new point of "B".
5. Option click on that B point, and click trace, this will make it so when you move your A point around, the B point will be traced, which is your derivative for sin(x)!
 This can be done for all of the derivatives of the trig functions, but make sure you Command F it ;) before you start again, this will clear the trace that you did before on the previous function to get the derivative.

There is most definitely another way to solve for the derivative of sin(x), and that is algebraically and that is done as showed below...

*Always work with a sin(x)/x if you see one, as it is always equal to 1*

Up above in the picture you can see that it instantly goes from sin(x) multiplied by a limit, then right to sin(x) multiplied by zero. We used a "Lemma theorem" to prove that that limit of

 was equal to 0.

Theorem below

*Always work with a sin(x)/x if you see one, as it is always equal to 1*

cos(x)' = -sin(x)

At first thought, it may be that the derivative of cos(x) is just sin(x) as sin(x)' = cos(x). With a closer look, we saw that the traced function was most definitely sin(x), but had a bit of a twist to it, and that was a negated sin(x). The reason its -sin(x) is because instead of having a positive slope as it approaches zero from the left side, it has a negative slope as it is approaching 0 from the left side.

tan(x)' = sec(x)^2

Looking at this, we see the "U's" of a secant function. The only different thing is that all of the "U's" are going upwards and positive. A simple change can be made of squaring the sec(x) to ensure that all of those "U's" will for sure go upwards and positive!

Algebra:

(Focused to the left side equation)

Simplifying tan(x) into sin(x)/cos(x) gives us opportunities to work with, and we were given the quotient rule to work with. (When you see a quotient in a derivative, we know to use the quotient rule (stated above).

cot(x)' = -csc(x)^2

Taking a look at this one, we now see the "U's" are turned upside down going negative. We didn't talk about this one specifically too much. If you think about it, if you want to get the secant "U's" all going in the same direction, that is where you square it. To get them all facing down and going negative, you negate it. And because this is cotangent instead of tangent, it will be cosecant instead of secant.

sec(x)' = sec(x)tan(x)

This makes much more sense done algebraically.

Algebra:
 *Focusing on the right side equation*

Just like what was done with tan(x), we want to simplify it into cos(x) and sin(x)'s if we can. Once again, there is a quotient, so the quotient rule comes into play.

csc(x)' = -csc(x)cot(x)

Algebraically: Given the derivatives of sin(x), cos(x), and tan(x), it will make this a lot easier to simplify and solve, so lets put these here for reference
sin(x)' = cos(x)
cos(x)' = -sin(x)
tan(x)' = -sec(x)^2



We here have a quotient, USE QUOTIENT RULE!











A -cos(x)/sin(x) is the opposite of tangent, so it is cotangent (negated of-course), and 1/sin(x) is equal to co-secant. It does not matter which order the csc(x) and cot(x) are put in, because the whole thing is negated and multiplied.



That concluded our short 35 minute class of AP Calculus.

REMEMBER

QUIZ NEXT CLASS ON THURSDAY, OCOBER 18, 2012

*Also, do not forget to ask questions about IW #5 before the quiz!*

IW#6 is on page 146 and includes questions 1, 5, 7, 21, 26, 31, 43, 47, 49, and also page 148 questions 1-4.

Scribe for next class: C HART!