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Another, what should be simple, derivative question

  1. Apr 10, 2006 #1
    This is not homework but it is more practise for me. We have to find the derivative of f(x) = sqrt(2x + 5) using the definition of a derivative, I used to remember how to do this but now I am not able to figure out what I am doing wrong.

    Here is what I have.

    y' = lim(h -> 0) [sqrt(2{x + h} + 5) - sqrt(2x + 5)] / h

    y' = (lim(h -> 0) [sqrt(2{x + h} + 5) - sqrt(2x + 5)] / h) * ([sqrt(2{x + h} + 5) + sqrt(2x + 5)] / [sqrt(2{x + h} + 5) + sqrt(2x + 5)])

    y' = lim(h -> 0) [2(x + h) + 5 - (2x + 5)] / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]

    y' = lim(h-> 0) [2x + 2h + 5 - 2x - 5] / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]

    y' = lim(h ->0) 2h / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]

    y' = lim(h->0) 2 / [sqrt(2{x + h} + 5) - sqrt(2x + 5)]

    I have no idea how to format the questions to make them look nice.
     
  2. jcsd
  3. Apr 10, 2006 #2

    AKG

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    On line 2, you have a + in the denominator which you turn into a - in line 3. It should stay as a +, not beomce a -. You'll then be able to evaluate the limit properly, and you'll get the answer you'd expect had you used the power and chain rules.
     
  4. Apr 10, 2006 #3
    Lol, that would just be an error here and not on my paper, as well we are learning the chain rule this week. But back to the question.

    y' = lim(h->0) 2 / [sqrt(2{x + h} + 5) + sqrt(2x + 5)]

    This would then be?

    y' = 2 / [sqrt(2{x + 0} + 5) + sqrt(2x + 5)]

    y' = 2 / [sqrt(2x + 5) + sqrt(2x + 5)]

    and that would be the final answer?
     
  5. Apr 10, 2006 #4

    Galileo

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    Yes, that's very much correct.
    The two terms in the denominator are the same, so you can simplify one bit more:

    y'=1/sqrt(2x+5)
     
  6. Apr 10, 2006 #5

    dextercioby

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    Another way to do it is to use this definition

    [tex] f'(x)=:\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} [/tex].

    Daniel.

    Edit. OOps, you used this one, i thought you used the other. It's simpler with the other one.
     
    Last edited: Apr 10, 2006
  7. Apr 10, 2006 #6
    Just out of curiousity and wanting to learn about calculus... what is the other way to do it?

    As well, how do you format the math to look like that?
     
  8. Apr 10, 2006 #7

    dav2008

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    To format the math, just click on the picture above and it will give you the source.

    You can also learn the code here: https://www.physicsforums.com/showthread.php?t=8997

    As far as calculus goes, you will learn that the derivative with respect to x of xa is a*xa-1. You can then use the rules of derivatives to find derivatives of more complicated functions.
     
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