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Homework Help: Log problems

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Find dy/dx:

    2. Relevant equations

    Answer: 2(2xyex2y)/(x2ex2y-2)

    3. The attempt at a solution

    I'm not interested in the steps towards the answer, but more into being able to graph it and confirm it on a ti-84.

    The way I've been checking if my answer was right with just regular derivatives was by graphing the original equation, then asking my calculator for dy/dx at some random point x and see if it matches up with what I get from solving it by myself and inputting that same random point x.

    I can't do that with this because ti-84s only take it terms of one variable x. I tried seeing if I could make it so that I can get y equal something, but I can't. I tried it on a ti-89, but it told me it wasn't possible.

    LN(ex2y)= ln(2x+2y)


    from there, I'm not quite sure where to go, or if it's even possible. I posted this calculus related problem here because what I want isn't really calculus it's just algebra type pre-cal stuff.

    Thank you for your time.
  2. jcsd
  3. Oct 2, 2009 #2
    Just to make sure, do you know how to get to the answer? From what I know, I don't think that the TI graphing calculators can graph such functions without parameterizing them (but not all can be parameterized probably, at least not easily), but I could be wrong. The trouble is that you can't explicitly solve for y in your equations, and that is why you have to use implicit differentiation. For example, the equation x2+y2=1 describes the circle of radius 1. You can't plug that into the TI-84's equation editor without splitting it up into two functions. If you don't see what I mean, try to solve for y.

    I also don't think it is a good habit to test if your derivative is right by just testing it at a few points. Use Wolfram Alpha to test your derivatives, or just make sure you know what you are doing. It is a very bad habit to get used to checking anything and everything with your calculator. I don't see how it will help when doing implicit differentiation, and I think it is probably counter-productive.
  4. Oct 2, 2009 #3
    yeah, I know how I got the answer, but all I do to be honest is just follow rules and formulas. I don't really KNOW what I'm doing.

    But...I do understand your point. I guess I won't be able to check my answer then? argh. Also, ti-89 platinums can be entered without parametrization, I tried it on a ti-89 and it told me it wasn't possible :/

  5. Oct 2, 2009 #4
    What you are doing is this: you are given an equation involving a variable x and the "variable" y which represents some function of x. There are curves in the plane, that are defined by such an equation, like x2+y2=1, in which you cannot explicitly solve for y as a single function of x (if you try this one you will get two functions after taking the square root after solving for y2). But we would still like to be able to do calculus with functions like this.

    For example, you will see this come up in related rates. You will have 2 or 3 functions that all depend on the variable t (representing time) and are related somehow (like volume and the radius that both change as time varies), but aren't explicitly defined using the variable t. We still need to be able to differentiate these expressions with respect to t to do calculus and analyze the model. This is why implicit differentiation is useful.

    Right now, you are just practicing the technique. Sometimes you just have to learn to do something, and then later you will know what you are actually doing once you have the technique down.

    By the way, I believe it is called the TI-89 Titanium. :)
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