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Homework Help: Differential help

  1. Dec 26, 2012 #1
    The problem I am struggling with is differentiating f(r)=r/sqrt(r^2 + 1)

    I rewrote it as: r(r^2 +1)^(-1/2)
    Split this up to get:
    g(r)=r
    h(r)=(r^2 + 1)^(-1/2)

    By product rule, f'(r)=gh'+hg'

    Obviously
    g' = 1

    Now by chain rule, I find h'= -r(r^2 + 1)^-(3/2)

    Finally, I get f'= [(r^2 + 1)^(-1/2)] - [(r^2)((r^2 + 1)^(-3/2))]

    However, Several sources say this is not correct. I followed what I believe to be the right procedures, so any feedback on where I went wrong would be great. I'm doing this from mobile, so I can't show quite as much of my work as I want. Thanks in advance.
     
    Last edited: Dec 26, 2012
  2. jcsd
  3. Dec 26, 2012 #2
    This isn't true.

    This is true.

    This isn't true (likely because you have the incorrect form of g.
     
  4. Dec 26, 2012 #3
    Also, did you mean to write [itex]f(r)=...[/itex] or [itex]f(x)=...[/itex].
    If you write

    [tex]f(x)=\frac{r}{\sqrt{r^2+1}}[/tex]

    then r is a constant, so f'(x)=0.
     
  5. Dec 26, 2012 #4
    I meant g' = 1, about to edit it. Everything else was in fact calculated assuming g'=1, it was simply a typing error
     
  6. Dec 26, 2012 #5
    fixed this too. it's f(r)
     
  7. Dec 26, 2012 #6
    OK. But please don't edit your posts have there have been replies. It makes the thread difficult to read for future readers. Just make a new post with the corrections.

    Anyway:

    OK. Looking back at this, I think that it was correct after all. What sources say that it is incorrect??
     
  8. Dec 26, 2012 #7
    Solutions in the back of the book, wolframalpha. Apparently the answer is (r^2 +1)^(-3/2)
     
  9. Dec 26, 2012 #8
    Your answer is the same as that answer. Try to add up the fractions in

    [tex]\frac{1}{\sqrt{r^2 + 1}} - \frac{r^2}{\sqrt{(r^2+1)^3}}[/tex]
     
  10. Dec 26, 2012 #9
    Thanks! I didn't think to do that, thank you so much!
     
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