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Find the 1st derivative

  1. Mar 23, 2006 #1

    tony873004

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    Find the 1st derivative using "the long way" method. Show all the algebra.
    [tex]f(x)=x^2+\frac{4}{x}[/tex]

    [tex]
    \begin{array}{l}
    f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} = \\
    \\
    \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h} \right)^2 + 4\left( {x + h} \right)^{ - 1} - \left( {x^2 + 4x^{ - 1} } \right)}}{h} = \\
    \\
    \mathop {\lim }\limits_{h \to 0} \frac{{x^2 + 2xh + h^2 + 4\left( {x + h} \right)^{ - 1} - x^2 - 4x^{ - 1} }}{h} \\
    \end{array}

    [/tex]

    Here's where I get stuck. I don't know what to do with
    [tex]
    {\left( {x + h} \right)^{ - 1} }
    [/tex]

    I forget the algebra for this step. Am I even going in the right direction to bring this term up from the denominator?
     
  2. jcsd
  3. Mar 23, 2006 #2

    Galileo

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    Ack.. nasty. If you've done such limit problems on simpler functions before you find that you sometimes need to multiply top and bottom by some factor and sometimes you don't. The functions x^2 and 1/x require different treatments, so you really should split the limit in two:

    [tex]f'(x)=\lim_{h \to 0}\frac{(x+h)^2-x^2}{h}+4\lim_{h \to 0}\frac{1/(x+h)-1/x}{h}[/tex]
    And solve them seperately.

    EDIT: fixed typos: x-> x^2 and 1/x -> 4/x
     
    Last edited: Mar 23, 2006
  4. Mar 23, 2006 #3

    tony873004

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    I don't really follow you here. Don't post a follow up yet, because my brain is dead right now. I might look at this in the morning and understand it.

    I'm still concerned since this way is nothing like the lecture notes or the book introduces. And this teacher does take off points for the right answer if you didn't do it the way he wants.

    This question was from the test, and I did continue beyond what I posted, but using some very bad algebra. Then I put the correct answer (from the power rule which we couldn't use here). I was hoping he wouldn't catch it. No such luck.
     
  5. Mar 23, 2006 #4

    tony873004

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    I'm still lost. Sorry Galileo. There's some tricks you're doing that I don't understand.
     
  6. Mar 23, 2006 #5

    JasonRox

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    He separated the problem that's all.

    Simplifying the expression he gave will get you to the beginning of where you started. You might want to do this to show yourself that it is true.

    Basically, what he is saying is...

    If f(x) = h(x) + g(x), then f'(x) = h'(x) + g'(x).

    Note: The derivative must exist obviously.
     
  7. Mar 23, 2006 #6

    Curious3141

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    Since this was a prob in a test you've already done, I think it's OK to post a complete solution.

    For clarity, I'll give it in steps.

    [tex]f(x + h) = (x+h)^2 + \frac{4}{x+h} = x^2 + 2xh + h^2 + \frac{4}{x+h} [/tex]
    [tex]f(x+h) - f(x) = (x^2 + 2xh + h^2 + \frac{4}{x+h}) - (x^2 + \frac{4}{x}) = 2xh + h^2 + 4\frac{x - (x+h)}{x(x+h)} = 2xh + h^2 - 4\frac{h}{x(x+h)}[/tex]
    [tex]\frac{f(x+h) - f(x)}{h} = 2x + h - 4\frac{1}{x(x+h)}[/tex]
    [tex]f'(x) = lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} = 2x - \frac{4}{x^2}[/tex]

    Clear enough ? :smile:
     
  8. Mar 23, 2006 #7

    Hurkyl

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    Ethical issues with tests and homework aren't the only reasons to avoid posting complete solutions! In general, a student will learn more if they're walked through a problem than if they simply read a solution.

    Of course, it's even better to merely point them in the right direction -- but it's hard to give good hints, and sometimes people used to being hand-held aren't too receptive. :frown: (This is just a general comment on the philosophy of help! In particular, it's not directed at anyone in this thread)
     
    Last edited: Mar 23, 2006
  9. Mar 23, 2006 #8

    Curious3141

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    Sometimes, they can learn from worked solutions, so they can apply it to future problems. :smile:

    Tony's obviously not one of those who want to be spoon-fed. He's worked hard (I think too hard ! ;) ) and showed his working. I think it's OK to give a solution he can learn from for future reference.
     
  10. Mar 23, 2006 #9

    tony873004

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    Thanks, Curious3141. I finally figured it out in the other thread, but your way is shorter and helps clarify things. This really isn't that difficult of a problem now that I understand it. And now I understand what Galileo is saying too. Thanks Galileo.
     
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