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Homework Help: Difficult differentiation question

  1. May 10, 2010 #1
    If xlny - y = 5e then find the value of dy/dxat the point(6e,e)



    2. Relevant equations



    So here is my attempt at the solution, please any advice on where im going wrong or if indeed it is correct greatly appreciated
    xlny - y - 5e
    Therefore dy/dx is:
    x/y + lny - 1 -5e
    Substituting in values for x and y:
    (6e/e) + lne -1 -5e
    Therefore
    5 -5e = -8.591
     
  2. jcsd
  3. May 10, 2010 #2

    Mark44

    Staff: Mentor

    The above is wrong. You started out with an equation, and each subsequent step must be an equation.

    Use implicit differentiation to get a new equation with all terms differented. That will give you an equation that you can solve algebraically for dy/dx.
     
  4. May 10, 2010 #3

    Cyosis

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    Homework Helper

    Don't forget the chain rule, also [itex]dy/dx \neq 1[/itex] in general.
     
  5. May 10, 2010 #4
    Could u expand on implicit differentiation please?
     
  6. May 10, 2010 #5

    Cyosis

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    Take the derivative with respect to x on both sides.

    [tex]
    \frac{d}{dx}(x \ln y-y)=\frac{d}{dx}5e
    [/tex]
     
  7. May 10, 2010 #6
    ok i tried that and got:

    1 x 1/y dy/dx = 5e

    therefore

    dy/dx = y/1 x 5e

    dy/dx = y5e
     
  8. May 10, 2010 #7

    Cyosis

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    You really have to brush up on your differentiation techniques. The derivative of a constant is 0, for x lny you need to use the product rule and the chain rule and for the entire expression you need to use the sum rule. Start out with using the sum rule, then differentiate the two left hand side terms separately.
     
  9. May 10, 2010 #8

    Mark44

    Staff: Mentor

    No, this isn't right. To differentiate xlny you need to use the product rule. It looks like you made up your own rule, one that isn't valid.

    Also, the derivative of any constant is zero. 5e is just a constant.
     
  10. May 10, 2010 #9
    Ah rite so using the product rule:
    u = x v = ln y
    du = 1 dv = 1/y . dy/dx

    Therefore:
    x/y ' dy/dx + lny -1 = 0

    |Correct?
     
  11. May 10, 2010 #10

    Cyosis

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    The answer you have is looking better, but it's still wrong. See post #3.

    Also I have no idea what you're doing here:

     
  12. May 10, 2010 #11
    That is my working for product rule:
    u = x
    du = 1

    v=ln y
    dv = 1/y .dy/dx

    I have no idea wot to do now?
     
  13. May 10, 2010 #12

    Mark44

    Staff: Mentor

    There are some technical problems with the above. If u = x, then du/dx = 1, not du = 1.
    It's close. Why do you have y ' and dy/dx? Are you using ' to indicate multiplication? That's not a good symbol to use for that.
     
  14. May 10, 2010 #13

    Mark44

    Staff: Mentor

    Don't show this stuff. Show what you're taking the derivative of and show what you get. What you have above is just clutter that gets in the way.

    Here's what I mean

    d/dx(x*lny) = x*d/dx(ln y) + 1 * lny = x*(1/y)*dy/dx + lny = (x/y)*dy/dx + lny
     
  15. May 10, 2010 #14
    Yes sorry I meant multiplication it should say:

    x/y.dy/dx +lny -1 = 0
     
  16. May 10, 2010 #15

    Mark44

    Staff: Mentor

    The "-1" is what's wrong.
     
  17. May 10, 2010 #16
    Ah ok would it be -1 dy/dx?
     
  18. May 10, 2010 #17

    Mark44

    Staff: Mentor

    Yes.

    Now solve your equation for dy/dx. Finally, evaluate dy/dx at the point (6e, e).
     
  19. May 10, 2010 #18
    Hint...

    Product rule (uv)' = u'v + uv'

    Now, consider the equation you are trying to differentiate implicitly

    [tex]\frac{d}{dx}(xlny \, - \, y) = \frac{d}{dx}5e[/tex]

    What do you get if you apply the product rule and implicit differentiation to the left hand side ?

    What do you get when you differentiate the constant term 5e on the right hand side ?
     
    Last edited: May 10, 2010
  20. May 10, 2010 #19

    Mark44

    Staff: Mentor

    Skins, you've come late to the party. The OP is well ahead of your hints.
     
  21. May 10, 2010 #20
    Ok solving for dy/dx I get:

    dy/dx = y(lny)/1-x

    subs in 6e and e:

    =e ln e/ 1-6e

    =e/-5e

    =-1/5

    Think that is correct.
     
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