Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Using definition of derivative to calculate, stuck, cant find a cancellation

  1. Jul 27, 2011 #1
    use the definition of the derivative to compute the derivative of the function at x=1

    f(x) = 1/(1 + x 2) for all x.

    3. The attempt at a solution
    i put it into lim[ (f(x)-f(xo))/x-xo ] form and got

    [(1/1+x2) - (1/2)] <---------numerator
    ___________________________ <----- division sign ("over")
    x-1 <----------- denomenator

    form and from there cancelled to
    2-1-x2 / 2(x2+1)

    and then again to -x2 + 1 / 2(x2+1)

    but now im stuck and cant find a cancellation to calculate the limit at x=1 without getting a 0 in there soemwhere

    if i stick this into the limit calculation formula itll just give me a 0 and thats no good. i already know the answer is -1/2 but i cant quite get there. any help is appreciated, thanks.
  2. jcsd
  3. Jul 27, 2011 #2
    Okay, if I've read this correctly, so far you have
    [tex]\lim_{x\to 1} \frac{\frac{-x^2 + 1}{2(x^2 + 1)}}{x-1}[/tex]

    What happens if you factor out a [itex]-1[/itex] from the top fraction's numerator? Is there any algebra you can do on that to divide out the [itex]x-1[/itex]?
  4. Jul 27, 2011 #3
    well, actually right now i have

    [tex]\lim_{x\to 1} \frac{-x^2 + 1}{2(x^2 + 1)}[/tex]
  5. Jul 27, 2011 #4
    Where did the [itex]x-1[/itex] in the denominator go?

    [tex]\lim_{x\to 1} \frac{\frac{1}{1+x^2} - \frac{1}{2}}{x-1} =
    \lim_{x\to 1} \frac{\frac{2}{2(1+x^2)} - \frac{1+x^2}{2(1+x^2)}}{x-1} =
    \lim_{x\to 1} \frac{\frac{2 - 1 - x^2}{2(1+x^2)}}{x-1} =
    \lim_{x\to 1} \frac{\frac{-x^2 + 1}{2(1+x^2)}}{x-1}[/tex]
  6. Jul 27, 2011 #5
    omg your right.

    careless mistake got me again. dang
    yeah, taking that -1 out brings out an x-1 in the numerator that cancels with the denominator that I completely forgot about

    and then i can take my limit to get -2/4, which is right.

    BIG THANKS, man, seriously.
  7. Jul 27, 2011 #6
    No problem, little mistakes like these happen to the best of us. :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook