Friday, January 29, 2010

Super sick scribe post

Hey y'all. This is Wednesday's (27th) scribe post. Enjoy!


Firstly, we went over the quiz because a number of us did not do so hot. I was personally upset with myself for missing the first problem in which we were given a point of a function and asked what the derivative was. The derivative of a point is the slope of the tangent line so all I had to do was draw the tangent line and find its slope by counting squares. I do not know how I missed such a simple problem. Hurray to Bru for lowering the effect of quizzes on our grades.


Secondly, in class we went over what an Explicit function is. An explicit function or equation is the form that we normally write or see an equation. It is an equation that defines Y as a function of X.


 Ex. Y=2x+3


The expression on the right explains how Y is related to X. This is called an explicit equation because it comes from the same root as “explain”, which is precisely what it does.



An equation that is not set up in the “Y equals…” form, only implies that Y is a function of X and therefore, it is called an implicit function. Deriving an implicit function can be easier than deriving an explicit function depending on how the equation is set up. This is called implicit differentiation.


Ex.  Y-2x=3


To derive an implicit function: You do not have to reset up the equation. You simply derive the terms in the equation and replace Y with Y’. Then solve for Y’ and the result is your derivative. Lets try it out.


Y-2x=3     The equation


Y’-2=0       taking the derivative of the terms using the power rule.


Y’=2         Solving for the derivative



It is as simple as that, especially when you take the derivative of such an easy equation. And we can check if this is right by finding the derivative explicitly.


Y=2x+3       The explicit form of the equation


Y’=2           Finding the derivative using the power rule.



Now lets try the last problem on the exploration we were supposed to finish for homework. First let us solve the problem explicitly.


Y3=X7                         The equation


Y=X7/3                        We must isolate Y for the right side to explain how Y is related to X. Now we derive


Y’=7/3X4/3                                      Power Rule


So lets see if this matches up with the result we get from implicit differentiation.


Y3=X7                           The equation


3Y2 x Y’=7X6                                We’ve derived the equation and multiplied all the Y terms by Y’


Y’=7X6/3Y2                                     We isolate Y’ so as to figure out what it equals


Y’=7X6/3(X7/3)2             We know from the explicit equation that Y=X7/3


Y’=7X6/3X14/3                              Simplify and then simplify even more


Y’=7/3X4/3                                      Check


                                                                      We must’ve done it right because Y’ prime equals the same thing when we solved the equation both explicitly and implicitly. Yay




So another day in the world of calcultopia comes to an end. Another battle won but many more lay in our path to the ultimate answer of calculus. $@… Or something like that. Next scribe is Winn Plot


J-tron foteen
























  1. HI J-Tron Foteen,
    Haha, I'm the first who commends on this post. Lucky!!!
    I like this post. It is well organized. I like the color to highlight the critical parts. A little review of the quiz at the very beginning is good. A lot of details provide people a better sense of explicit form and implicit form. By the way, I think the first question on the quiz has another solution. It is pretty obvious that the function appeared in the group is f(x) = x^ (-1/2). So we can use the power rule to figure out the derivative of this function and find the exact value of the derivative of the function at x = 1. Algebra is actually a very neat thing to master. I like doing algebra a lot. And algebra can always give you a better sense of math. That’s what I think, maybe not that relevant.
    One thing I think you probably want to add into your post is that when you are trying to find the derivative of a very complex function, don’t forget to use the other rules such as product rule. Another thing is that y can be in the y’. So people don’t need to find the explicit form of a very complex function to try to put y’ in terms of x. you can just leave y in the y’.
    By the way, if the function is ((sin(y)cos(xy))^6)/xy = 10x, can you find the derivative? Really complex……..

  2. Hey J-Tron Foteen! I really enjoyed reading your scribe post. The colors and the spaces left in between paragraphs made it easy to understand.
    Everything was is detailed, like the steps used to solve the given problems. It was also helpful to include the solution of that homework problem in your post.

    aand, YDplus: that's a very complex one! Wow! where did you get that from? anyways, here is how I did it:

    [(sin(y)cos(xy))^6] xy = 10x

    {sin(y)}^6 {cos(xy)}^6 = 10(x^2)y

    6{sin(y)}^5 . cos(y) . y'. -6{cos(xy)}^5 . sin(xy) . (y+xy') = 10(2xy + x^2 .y')

    {-36[sin(y) . cos(xy)]^5 . cos(y) . sin(xy)} - 20xy = (x^2 . y') / [(y+xy') y']

    {-36[sin(y) . cos(xy)]^5 . cos(y) . sin(xy)} - 20xy = x^2 / (y+xy')

    1/[{-36[sin(y) . cos(xy)]^5 . cos(y) . sin(xy)} - 20xy] = (y+xy') / x^2

    y + xy' = x^2 /[{-36[sin(y) . cos(xy)]^5 . cos(y) . sin(xy)} - 20xy]

    xy' = [x^2 /[{-36[sin(y) . cos(xy)]^5 . cos(y) . sin(xy)} - 20xy]] - y

    y' = {[x^2 /[{-36[sin(y) . cos(xy)]^5 . cos(y) . sin(xy)} - 20xy]] - y}/x

    OMG! is there any way to do solve this problem in a short period of time? I feel like there should be, but I can't figure it out...
    By the way, thanks YDplus for reminding me all of these rules.! :)

  3. I thought that this post was quite good, I liked the formatting. At my first glance I was going to say that there should be more graphics, but once I read through the whole thing I don't think that it needed any graphics. Sometimes pictures can distract from the meaning of the actual post. I got a little confused when I started reading the procedure to find the derivative of the equation. A little confusion is inevitable though, and I think that you did a very good job describing what was going on. The large gap at the bottom of the post threw me off a little, but I got over that quickly. A job very well done.

  4. This is a really thorough scribe post! It covered everything we did in class, some subjects like explicit functions I had even forgotten already! But now I remember so thanks. Also it didn't seem like you rushed through it at all, the explanations are easy to understand and colorful! What resources did you use to make sure you covered all the material from class?

  5. Blitzen, I used the exploration from class... And my grey box. Both were helpful

  6. J-tron foteen,

    This is a really good scribe post. It is quite thorough and has a lot of great examples. I mean is there anybody from Calculoplis that doesn't love and appreciate a good calculus example? I think not. Also, as flying slug said earlier, the first thing I noticed for your post was no pictures. Even without pictures this post still explains everything that needs explaining. Without the pictures the post is a little shorter, which is always a plus. Despite not having pictures, it is still also entertaining. Good job.

    mc Casper

  7. Hey J-tron, good meaningful coloring and I thought that putting the examples of explicit and implicit equations alone and in bold was a good move. I think one thing that might help you as the equations get more complex is to use equation editor. I know it is a freakin pain but at least for the 2nd example, it could just make things more on the side of eye candy. (Same goes for you Marley). Also, I think that it could have been very helpful to include in the very beginning to use the chain rule and that y' is always alone (ie, no exponents or multiplication by an integer...) in the derivative (or is it?). In my opinion, this is one of the most difficult concepts we have covered so far, and you did a fantastic job of explaining it. However, I think that putting in other examples (not on the wksht) could really help to ingrain this concept into our minds (although, I found a whole bunch easily in the txt book). And personally, I don't feel that including an extraneous picture in hopes of getting an E is necessary to my learning calculus, so thank you for keepin it real.

  8. Dude thanks for making me the scribe post. I really like the fact that you used such a large number of equations in your scribe post. For me its super helpful to have a ton of examples to go off of in my learning. So i thought i'd share some more with everyone else.

    Theres a few practice problems, and expalnation, and some examples.

  9. JT, nice post, I am not part of this class so as an outsider and calc lover I followed the example very easily and made sense to me, even reminded me of a couple of things that I had forgotten (oh nooooo).

    Marley, I like your try to tackle the hard problem, try using this formula when doing it. Let a,b be two functions. If you want to take the derivative of a(b) = (a')b + a(b') and in the first line, I don't follow how you got 10(x^2(y)) on one side, you should look at that again.