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Another, use definition as a derivative (as a limit) problem 2

  1. Oct 25, 2011 #1
    use definition as a derivative (as a limit) to show that

    Dx (x(x+1)) = 2x + 1, x [itex]\in[/itex] ℝ

    Would I need to find the derivative first?

    f'(x) = 2
     
  2. jcsd
  3. Oct 25, 2011 #2

    vela

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    No, you're supposed to calculate the derivative of x(x+1) using the definition of the derivative.
     
  4. Oct 25, 2011 #3
    So the derivative of x(x+1) is just 1?
     
  5. Oct 25, 2011 #4

    Mark44

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    No, it is not. Since you are supposed to show (by using the limit definition of the derivative) that the derivative of x(x + 1) = 2x + 1, why would you think that the derivative is 1?
     
  6. Oct 25, 2011 #5

    HallsofIvy

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    Do you know what the definition of derivative (as a limit) is?
     
  7. Oct 26, 2011 #6
    Is this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?
     
  8. Oct 26, 2011 #7
    Whoops I meant 2x+1 = 2x^1-1 + 0 = 2
     
  9. Oct 26, 2011 #8

    Mark44

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    This would give you f'(a). For f'(x), the definition is this limit:
    [tex]\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}[/tex]
     
  10. Oct 26, 2011 #9

    Mark44

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    2x + 1 ≠ 2x1 - 1.

    Note that what you wrote in the middle expression above is interpreted as 2x1 - 1 + 0 = 2x - 1, and that's probably not what you intended. If you have an exponent that is an arithmetic expression, USE PARENTHESES.

    Also, in what you wrote, you are apparently taking a derivative, but your notation gives no hint that you're doing this. What you should have written is
    d/dx(2x + 1) = 2 + 0 = 2.
     
  11. Oct 26, 2011 #10

    vela

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    The denominator should be h, not f(h). Mark provided you with the correct definition. So you want to evaluate
    [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{[(x+h)((x+h)+1)] - x(x+1)}{h}[/tex]Multiply out the top, cancel terms, simplify, and then take the limit. If you do it correctly, you'll find you end up with 2x+1.
     
  12. Oct 27, 2011 #11
    I dont know how but I ended getting (2x-x+1) using the derivative (as a limit) formula.

    [itex]\stackrel{Lim}{h→0}[/itex] [itex]\frac{(x^2+2xh+h^2+1)-x(x+1)}{h}[/itex]

    [itex]\stackrel{Lim}{h→0}[/itex] (2x-x+1)
     
    Last edited: Oct 27, 2011
  13. Oct 27, 2011 #12

    Mark44

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    What you're starting with is incorrect. It should be
    [tex]\lim_{h \to 0}\frac{(x + h)(x + h + 1) - x(x + 1)}{h}[/tex]

    It would make life easier if you rewrote x(x + 1) as x2 + x.
     
  14. Oct 27, 2011 #13
    Ahhh I ended getting 2x+1, thank you.
     
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