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Derivative of Wait What?

  1. Mar 2, 2013 #1
    wtf.jpg

    I think the topic sums it up pretty well. I have no idea what I'm supposed to be doing here.
     
  2. jcsd
  3. Mar 2, 2013 #2
    Maybe I figured it out now.

    Is a)

    u'(1) = f(1) g'(1) + g(1) f'(1)

    = 0

    ?
     
    Last edited: Mar 2, 2013
  4. Mar 2, 2013 #3
    They are piecewise functions.
    So u(x) and v(x) will be piecewise functions aswell.
    The intervals are (-∞,0), (0,2) and (2,∞)
    (I assumed they don't change after the graph ends)
     
  5. Mar 2, 2013 #4
    That's not the question though, they want me to differentiate a combination of the two.
     
  6. Mar 2, 2013 #5
    I would do it in the following steps:
    1) Find the functions f(x) and g(x), they're piecewise and linear.
    2) Find the functions u(x) and v(x), they're also piece wise.
    3) Derivate the functions.
    4) Evaluate the u'(x) in 1 and v(x) in 5.
     
  7. Mar 2, 2013 #6
    The beginning is good, but why zero in the end?

    What is the geometric meaning of the derivative?
     
  8. Mar 2, 2013 #7

    CompuChip

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    That is correct. What are the values of f(1), f'(1), g(1) and g'(1)?
     
  9. Mar 2, 2013 #8
    The meaning is instantaneous rate of change.

    Lets see here

    (2)(-1)+(1)(2) = 0
    Still get it to zero.
    f(1) = 2
    g'(1) = -1

    g(1) = 1
    f'(1) = 2
     
  10. Mar 2, 2013 #9

    HallsofIvy

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    For x between 0 and 2, the graph of f is the straight line between (0, 0) and (2, 4): f(x)= 2x. For x between 0 and 2, the graph of y is the straight line between (0, 2) and (2, 0): g(x)= 2- x. Their product is fg(x)= 2x(2- x)= 4x- 2x^2. (fg)'= 4- 2x and at x= 1, that is, indeed, 0.
     
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