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Homework Help: Derivative of an equation

  1. Aug 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx in terms of x and y

    2. Relevant equations

    x^2 + y^2 + ln(2) = xy

    3. The attempt at a solution

    I'm pretty sure I did it right, but I am wondering if I was supposed to take the derivative of everything:

    2x + 2y = 1

    Is that the correct derivative? If so, then I'm good to go...
  2. jcsd
  3. Aug 15, 2007 #2
    ln(2) and xy can be derived as well
  4. Aug 15, 2007 #3
    Didn't I already derive xy? It's 1 right?

    And isn't ln(2) a constant?
  5. Aug 15, 2007 #4
    xy = product rule ... are you doing implicit/explicit differentiation? y' if you are.

    ln(2) ... i forgot, been a while since i've taken Calculus
  6. Aug 15, 2007 #5
    It just says: "Find dy/dx"

    So, OK, let me correct the dy/dx xy:

    so it's y + x

    Now, can the answer be written thus:

    x + y = 1 ??

    And would I have to write it like this:

    dy/dx x+y = 1 dy/dx
  7. Aug 15, 2007 #6
    Where is this "1" coming from? And that still isn't the correct derivative of xy, you need to use the product rule. What is the derivative of y with respect to x?
  8. Aug 15, 2007 #7
    have you learned the product rule?

    f(x)g(x) = f'(x)g(x) + g'(x)f(x)
  9. Aug 15, 2007 #8
    You need to take partial derivatives, and in the process of doing this, you will get dy/dx 's in your equation. Then group like terms and solve for dy/dx. Have you learned about partial derivatives? The product rule mention above will be helpful in the process, as well.
  10. Aug 15, 2007 #9
    Alright hold up.
  11. Aug 15, 2007 #10
    I'd imagine this problm is showing up in the context of a calulus one class after a lesson on implicit differentiation, hence the original poster has probably never encountered partial derivatives before, and has no reason at all to find them for this problem, I don't understand why you think he needs to take partial derivatives.
  12. Aug 15, 2007 #11
    Crap, you're right, I made a typo. I meant to type implicit, not partial. Apologies.
  13. Aug 15, 2007 #12
    Not to be nitpicky, but I think what rocophysics meant was
    (fg)'(x)=f'(x)g(x) + g'(x)f(x)

    I don't go looking to correct peoples' mistakes, however I know how difficult learning new notations can be. If we all start out on the same page, it will be easier down the road.

  14. Aug 15, 2007 #13
    The thing that I'm not getting is how to get from

    2x + 2y + ln2 = xy


    d/dx (2x + 2y + ln2) = xy

    and what to do after that, and what to do with the other side of the equation...
  15. Aug 15, 2007 #14
    Can someone show me a small example of how to do these derivatives?
  16. Aug 15, 2007 #15
    If you take the derivative with respect to x of the left hand side of the equation, you have to take it of the right hand side, too.

    However, simply writing d/dx on the left side there isn't enough. You need to take the derivative with respect to x to each of the summands on the left side (and, of course, the right side of the equation too).

    For instance, the derivative w.r.t. x of [tex]5x^3[/tex] would be [tex]15x^2(\frac{dx}{dx})=15x^2.[/tex]
    The derivative w.r.t. x of [tex]4y^3[/tex], however would be [tex]12y^2(\frac{dx}{dy})[/tex]. The [tex]\frac{dx}{dy}[/tex] remains in this case.

    You have learned about implicit differentiation, correct?
    Last edited: Aug 15, 2007
  17. Aug 15, 2007 #16
    Let's start over. You are being asked to implicitly differentiate the function

    Now how about an easier one to start. Can you try to implicitly differentiate
    [tex]x^2+y^2=1[/tex] with respect to x?
  18. Aug 15, 2007 #17


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    Homework Helper

    You want d/dx(x^2 + y^2 + ln(2)) = d/dx(xy). Split this into four derivatives and I'm sure anyone here will jump on any one you get wrong.
  19. Aug 15, 2007 #18
    Yeah, I have. But I suck at life.

    So what about this for starters:

    d/dx (x^2 + y^2 + ln2) = (xy) d/dx

    would I then do:

    2 + 2y (d/dx) + 0 = xy d/dx

  20. Aug 15, 2007 #19


    2x + y^2 = 1

  21. Aug 15, 2007 #20
    And what about finding dy/dx in terms of x AND y?
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