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Implicit differentiation homework

  1. Nov 8, 2008 #1
    hi

    i couldn't solve this question
    i think there is a mistake in it
    anyone can check it please ?
    [​IMG]
     
  2. jcsd
  3. Nov 8, 2008 #2

    HallsofIvy

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    Re: differentiation

    Use implicit differentiation:
    The derivative of 5y, with respect to x, is 5 dy/dx and the derivative of 5xy, with respect to x, is 5y+ 5x dy/dx.

    Differentiate both sides of the equation with respect to x and solve for dy/dx.
     
  4. Nov 8, 2008 #3
    Re: differentiation

    [​IMG]
    then I got stuck
     
  5. Nov 8, 2008 #4

    gabbagabbahey

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    Re: differentiation

    Now solve your original equation for y and substitute it into your result.
     
  6. Nov 8, 2008 #5

    tiny-tim

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    And 5y = … ? :smile:
     
  7. Nov 12, 2008 #6
    Re: differentiation

    5y=2x/(x+1)
     
  8. Nov 12, 2008 #7

    tiny-tim

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    Yup! :smile:

    So dy/dx = (5y - 2)/-5(x+1) = … ? :wink:
     
  9. Nov 12, 2008 #8
    Re: differentiation

    Tiny Tim
    Ever heard of substitution: instead of 5y in the differential equation you may substitute
    2x/(x+1) while the first equation seems to be right.
     
  10. Nov 12, 2008 #9

    tiny-tim

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    Sorry, not following you. :confused:

    btw, are you the same person as UNknown 2010?
     
  11. Nov 12, 2008 #10
    Re: differentiation

    No I am knot!
     
  12. Nov 12, 2008 #11

    gabbagabbahey

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    Re: differentiation

    You're "a method for fastening or securing linear material such as rope by tying or interweaving."? :confused:\

    Disclaimer: (1) Comment said in jest. (2) Quote is the definition of the term 'knot', taken from wiki :smile:
     
  13. Nov 14, 2008 #12

    Mentallic

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    Re: differentiation

    I don't know if you might prefer this method, but this is how I would've done it since I don't have a clue about what everyone else has suggested:

    [tex]2x-5y=5xy[/tex]

    [tex]y(5x+5)=2x[/tex]

    [tex]y=\frac{2x}{5(x+1)}[/tex]

    Now just use the quotient rule. [tex]\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex] where u=numerator, v=denominator.

    EDIT: Felt like I was giving away the answer (literally)
     
    Last edited: Nov 14, 2008
  14. Nov 14, 2008 #13

    tiny-tim

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    Hi Mentallic! :smile:

    I think your way is better!! :biggrin:
     
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