Results 1 to 3 of 3

Thread: Need help with derivative

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    1

    Need help with derivative

    Hello, this is my first post. I need help finding the derivative of $\displaystyle 1/x^2$ at x=2. The answer is supposed to be $\displaystyle -2/x^3$ and i need to find the derivative by definition. I have worked this out a million times and i keep making mistakes in my algebra i suppose because no matter what i try, i cannot get $\displaystyle -2/x^3$. any help would be greatly appreciated.

    also, this is the first time ive ever heard of latex so hopefully i got it right, if i didnt i apologize
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    $\displaystyle \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{1}{(x+h)^2}-\frac{1}{x^2}\right)$

    $\displaystyle \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{1}{x^2+2xh+h^2}-\frac{1}{x^2}\right)$

    $\displaystyle \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{x^2-(x^2+2xh+h^2)}{x^2(x^2+2xh+h^2)}\right)$

    $\displaystyle \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{-2xh-h^2}{x^2(x^2+2xh+h^2)}\right)$

    $\displaystyle \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{h(-2x-h)}{x^2(x^2+2xh+h^2)}\right)$

    $\displaystyle \displaystyle \lim_{h\to 0}\frac{-2x-h}{x^2(x^2+2xh+h^2)}$

    $\displaystyle \displaystyle \frac{-2x+0}{x^2(x^2+2x(0)+(0)^2)} = \dots$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    I always use substitution when it comes to finding derivatives using the limit definition:

    We have $\displaystyle \displaystyle f'(x) = \lim_{h \to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}.$ Let $\displaystyle x+h = a$ and $\displaystyle x = b$, then $\displaystyle h = b-x = a-b$, thus:

    $\displaystyle \begin{aligned} f'(x) & = \lim_{a \to b}\frac{\frac{1}{a^2}-\frac{1}{b^2}}{a-b} = \lim_{a \to b}\frac{b^2-a^2}{a^2b^2(a-b)} = \lim_{a\to{b}}\frac{-(a-b)(a+b)}{a^2b^2(a-b)} \\& = -\lim_{a\to{b}} \frac{a+b}{a^2b^2} = -\frac{2b}{b^4} = -\frac{2}{b^3} = -\frac{2}{x^3} ~ \diamond \end{aligned}
    $



    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: Jan 24th 2011, 11:40 AM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 23rd 2010, 06:33 AM
  4. Replies: 2
    Last Post: Nov 6th 2009, 02:51 PM
  5. Replies: 1
    Last Post: Jan 7th 2009, 01:59 PM

Search Tags


/mathhelpforum @mathhelpforum