October 29, 2010

To start class, we corrected Quiz #4. We collectively did poorly enough on it that it will not be counted towards our grade unless we so choose. Mr. O’Brien noticed that some of us had trouble using the space provided, but he said that the lack of space was intentional. We will get plenty of room to convey our calculations on the AP Test, but if we can work in minimal space, our organization will benefit in plenty of space. Remember to stay as organized and as concise as you can.

During our pretty extensive correcting session, we also decided that besides knowing that and .

To check on our progress with Part 2 of the exploratory, we did some work with logarithm functions and their inverses in Geogebra.

So, our function was which looks like this when graphed:

Note the two points on the graph... (1,5) and (3,6). The first point can be found by taking . Remember that is the same as saying "What power do you raise 3 to to get 1?" Any number raised to zero is 1 so the log (at any base) of one is always zero. 0+5 is 5, so we have the first point. Doing the same thing, (what power do you raise 3 to to get 3?), you find the point (3,6).

Now, to graph the inverse function, we'll use our set of steps...

1. Replace f(x) with y

2. Replace the Ys with Xs and vice versa.

3. Solve for y.

If you work through it, When we graph it, we already know two points on the graph. When graphed, the inverse of a function looks like the graph of the original reflected across the line y=x, so when we reverse the x and y values of the points on f(x) we get points: (6,3) and (5,1) on g(x).

To make sure we knew how to find the derivative of a logarithm, we next found the derivate of f(x) at point (3,6).

so,

and, at the value of x=3.

Note: On the calculator portion of an exam, you can hit "2nd-->trace" to get to the calc function, then hit "dy/dx" for a quick deriv. value.

So, now we know a point on the line, and the slope at that particular point, so we know the equation for the line. Surely enough when graphed, the line is tangent to the function.

For the line tangent to g(x) at (6,3), we have two options. One is to use the derivative rule for exponents, but the other way is far simpler. We know that f(x) and g(x) are inverses of each other, so we know that the two inverse points have reciprocal slopes. Therefore the equation of line tangent to g(x) at (6,3), is .

So, with our various derivative rules, we had only one kind of function we could not take the derivative of: Inverse Trig Functions.

As some background information, we discussed that the inverse functions only exist in certain quadrants of the unit circle. Arcsin and arctan exist in quadrants four and one, while the arccos function, exists in the first and second quadrants. We also discussed that trig functions equal ratios (like ) while inverse trig functions equal angles (like 45 degrees). For a quick review of important inverse trigonometric info, here's some help from wolframalpha:

http://www.wolframalpha.com/input/?i=arcsin

http://www.wolframalpha.com/input/?i=arccos

http://www.wolframalpha.com/input/?i=arctan

Using the example problem cos(arcsin(1/4)), we remembered how to compose trigonometric functions. We know that arcsin is an angle, so we can say that: and therefore . This is where our trigonometric properties from last year come in handy. Remember that ? Well, it does, so we can use it to find cos(arcsin(1\4)). so, . Solve for cosine of the angle, and we get: .

Using this very tactic, we finally have everything we need to solve for the derivatives of inverse trigonometric functions...

In class, we started with y=arcsin(x), which we know is the same as sin(y)=x. Take the derivative with respect to x, and we get: or . Now, we can solve for cos(y) since we know that . We've already stated that sin(y)=x so, , and . Plug that back in for cos(y) in our derivative equation and we have the following rule:

Amazingly, it took implicit differentiation, the chain rule, and trigonometric rules to find the derivative of arcsin. For arccos, we went through a similar process:

so...

Two down...

And that's when we ran out of time... tune in next time for how to find the derivative of arctan... (or just look on http://www.wolframalpha.com/input/?i=derivative+of+arctan). Hint... it has to do with a certain other trigonometric rule...

The next scribe is Julia.

Unit 2 assignment solutions

Assignment #9 solutions
Assignment #10 solutions
Assignment #11 solutions
Assignment #12 solutions

Links for Monday-Wednesday

Exploration
Geogebra File
Solutions to Part 1 of Exploration (to be posted Tuesday)
Solutions to Part 2 of Exploration (to be posted Thursday)

Scribe Post Friday October 22nd, 2010

Class started out all normal

That is, it started half past nine

And we quizzed ourselves until 10:10

Using rules and signs and sines.

We then went over Kyle's post

A great job that he did

This lasted till 10:45

And then we worked derivs

We did Implicit chain rule

Until our brains did drool

And now I'm up at 3 o'clock

Because of this stupid school.

The first problem we faced was the infamous

This evil little guy sparked a flew of mathematical steps:

Using the chain rule (), we can find the derivative:

(NOTE that the derivative of y is not 1 in this case because it is the value we are trying to find)

Now, using the glorious product rule, we can make this little function more complicated:

Now, we know from the Power Rule that x' equals 1. So,

And using our awesome Algebra I skills which we all CLEARLY remember, if we move all of the y' to one side of the equation then we will be able to find out what y' equals:

Now if I pull out the y', I can isolate it:

LOOK AT THAT BEAUTIFUL ANSWER! It's a prettycool formula, that Implicit Differentiation. Yep.

So then we took a look at a few homework questions: Namely pg 174 #19 and #35b. If you want me to write these up, I'll do it over the weekend. But for now, you have the back of your book!

Finally, we took a look at inverses.

There are a few simple steps to find the inverse of a function:

1. Switch your x and y values. EX. y=3x would now be x=3y.

2. Isolate your "y" value. EX. in x=3y, to find y, I'd have to divide the x by 3. y=x/3.

3. Replace "y" with . EX. y=x/3 is now f^-1(x)=x/3.

The first inverse we took a look at was .

So, like in STEP 1, we switch the y and x values:

becomes .

Then solve:

becomes .

So, for the original function , we now have an inverse of . This is what this looks like graphically:

Now, I can take the same point on each function to see what relationship exists between a derivative and an inverse. I chose the points (1,2) for f(x) and (2,1) for g(x).

First I need to find the derivative (or the slope) to find the tangent line at point A. So, if I take and Power Rule all over it, then I get . At f(1), the slope is 3. If I then input this into Point-Slope formula, I get

which equals

.

The derivative (or slope, again) of g(x) must be found!

Now to nab the derivative using the Chain Rule:

Now to input the "x" of my point. In this case it would be "2".

.

Yay! So now that we know the slope of the new derivative, I can input it, once again, into point-slope form.

.

The derivatives are perfect reciprocals of each other!

Now, let's take a final look at them graphically to just get an eye to what this looks like:

NOTE: h(x) and k(x) are the tangent lines.

Look at those colors! Those tangent slopes are perfect reciprocals!

For an added bonus, here's some algebra:

PROVE:

The derivatives of inverse functions are reciprocals.

PROOF:

Let f(x) and g(x) be inverse functions to each other.

So,

Now I can take the derivs of both sides:

Q.E.D.

Another example:

Given that , find at x=3.

Let's freaking graph it!

NOTE: The function is cut off at x≥1 because it was a whole cubic function, then it wouldn't pass the horizontal line test. This type of function is called a 1 to 1 function.

Inverse: (3,??)

Original: (??,3)

So now because I know I will be using the value "3", I can set my function equal to it:

Now that it is set to 0, I can synthetically divide. The divisor is positive 2, which results in the function equaling . AAAAAAAAAAND that's as far as we got in class.

Anyways, if you need any help with finding the derivatives of inverse functions, click here. It does have some bits that we haven't covered yet, but ignore that.

This one is very similar to the stuff we've been doing.

And these guys are pretty awesome too :D

I thought I might add another link! *update*

Well, it's 3. I'll probably be grumpy tomorrow due to lack of sleep. Hello, Friday.

The next scribe post WILL NOT be anyone in the musical, because they would keel over more than I already am doing. So, I'm sorry, Marcel, but you're not involved in the musical and it's been a little while. Enjoy!