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Derivative Definition

  1. Jan 14, 2006 #1

    I am attempting to find the derivative of this function using the definition of derivative, however my solution is not the same as Mathematica's solution.

    Is my formulation incorrect on some protocol somewhere? :uhh:

    [tex]G(t) = \frac{4t}{t + 1}[/tex]
    [tex]G'(t) = \lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h}[/tex]

    [tex]\lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h} = \lim_{h \rightarrow 0} 4 [\frac{(t + h)(t + 1)}{(t + h + 1)(t + 1)} - \frac{t(t + h + 1)}{(t + 1)(t + h + 1)}] \frac{1}{h}[/tex]

    [tex]\lim_{h \rightarrow 0} 4[ \frac{(t + h)(t + 1) - t(t + h + 1)}{(t + h + 1)(t + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} 4[ \frac{(t^2 + ht + t + h - t^2 - ht - t)}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h}[/tex]

    [tex]\lim_{h \rightarrow 0} [ \frac{4h}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} ( \frac{4}{t^2 + ht + 2t + h + 1} ) = \frac{4}{t^2 + 2t + 1}[/tex]

    [tex]\boxed{G'(t) = \frac{4}{(t + 1)^2}}[/tex]

    Mathematica solution:
    [tex]- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}[/tex]
     
  2. jcsd
  3. Jan 14, 2006 #2

    Galileo

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    [tex]\frac{4}{(t+1)^2}=- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}[/tex]


    No mistakes :smile:

    Check it!
     
  4. Jan 14, 2006 #3
    It's the same result, so yours is also correct.
     
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