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Homework Help: Dx ln problem

  1. Dec 5, 2007 #1
    1. The problem statement, all variables and given/known data

    e^(xy) + xy = 2, find y'

    the actual answer is -(y/x)

    3. The attempt at a solution

    xy + lnxy = ln2
    xy' + y + 1/x + y'/y = 1/2
    xy' + 1/y (y') = 1/2 - y -1/x
    y'(x + 1/y) = 1/2 - y - 1/x
    y' = (1/2 - y - 1/x)/(x+1/y)

    i simplified it further but could not get the right answer. help would be appreciated
     
  2. jcsd
  3. Dec 5, 2007 #2
    I see you are using implicit diferentiation, good. Now the derivative of
    ln(2) is NOT 1/2.

    Casey
     
  4. Dec 5, 2007 #3

    cristo

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    Yea, there's a few mistakes here. Firstly, you can't just take the logarithm of each term individually; you need to take the logarithm of each side. That's if you even want to take logs in the first place.

    To be honest, I can't see how to do this; the implicit derivative of e^xy is bothering me. I'm blaming it on the fact that it's late!
     
  5. Dec 5, 2007 #4
    Cristo he did take ln of both sides. It has just been simplified since the lne^(xy) is just xy. Have some tea!

    Then we have product rule+chain rule=0
     
  6. Dec 5, 2007 #5

    cristo

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    Is that aimed at my British-ness? :smile:

    However, he didn't. If I have an expression [itex]e^{xy}+a=b[/itex] then taking logarithm of both sides yields [itex]\ln(e^{xy}+a)=\ln(b)[/itex]. Now, [itex]\ln(v+w)\neq\ln(v)+\ln(w)[/itex] so the equation cannot be simplified as he has done in the OP.
     
  7. Dec 5, 2007 #6
    Oh i keep forgeting u cant distribute ln :rolleyes:

    ok i c what your saying so then i get:

    ln(e^xy(x)(y)) = ln2
    1/[e^xy(x)(y)] (e^xy) (xy' + y) + e^xy (xy' + y) = 0
    (xy' + y)(1/xy + e^xy) = 0
    xy' + y = 0
    y' = -(y/x)

    thank u all for your help!!
     
  8. Dec 5, 2007 #7

    cristo

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    How did you get your first line? Taking the logarithm of your expression gives [itex]\ln(e^{xy}+xy)=\ln 2[/itex]
     
  9. Dec 5, 2007 #8
    er oh.... i just realized i did that wrong, strangely it worked, ok i fixed it:

    1/(e^xy + xy) (e^xy) (xy' + y) + (xy' + y) = 0
    (xy' + y)[(e^xy)/(e^xy + xy) + 1] = 0
    xy' + y = 0
    y' = -(y/x)

    my brain gets big picture but misses the small details :cry:
     
  10. Dec 5, 2007 #9
    [tex]e^{xy}+xy=2[/tex]

    [tex]e^{xy}=2-xy[/tex]

    now take the ln of both sides and differentiate
     
  11. Dec 5, 2007 #10
    Yes. And apparently, I need the tea....oh wait, here is some over here:blushing:

    Casey
     
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