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Chain,Product, Quotient Rule

  1. Jul 9, 2009 #1
    Hello everybody
    I would also like to solve the following problem using either the Chain,Product, or Quotient Rule but am unsure of the working stages to get to the given answers

    i) Find the equation of the tangent at the point with coordinates (1,1) to the curve with the equation y=(X2+3)/x+3

    ii) Given that y=xe-3x find dy/dx and hence find the coordinates of the stationary points on the curve y=xe-3x


    Again, help with the workings would be much appreciated.
    Answers:
    i) 4y=x+3
    ii) (1-3x)e-3x , (1/3,1/3e-1)
     
  2. jcsd
  3. Jul 9, 2009 #2

    rock.freak667

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    Homework Helper

    I'll give you the two formulas you will need to use but YOU need to do the working for them. When you post the working, then we can proceed to the rest of the problem.


    For [itex]y=\frac{u}{v}[/itex]

    [tex]\frac{dy}{dx} = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]


    for y=uv

    [tex]\frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}[/tex]


    And note that dy/dx is a gradient function, so at any x point in a curve,once you have dy/dx you can find the gradient at that point x.
     
  4. Jul 9, 2009 #3
    In the first equation, is just x in the denominator or should the equation be
    y=(x2+3)/(x+3) ?
     
  5. Jul 9, 2009 #4

    tiny-tim

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    Hi Timiop2008! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

    (you use the product rule for products, the quotient rule for quotients, and the chain rule for combinations, and you may have to use more than one)
     
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