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Homework Help: Need help on derivatives

  1. Dec 1, 2006 #1
    1. The problem statement, all variables and given/known data
    [tex]D_x(\frac{1}{x^2}-x)[/tex]


    2. Relevant equations
    [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]


    3. The attempt at a solution
    [tex]\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}[/tex]
     
    Last edited by a moderator: Dec 1, 2006
  2. jcsd
  3. Dec 1, 2006 #2

    HallsofIvy

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    I don't see much of a "problem statement". If I were asked to find the derivative of [itex]\frac{1}{x^2}- x[/itex], I would write it as [itex]x^{-2}- x[/itex], and then use the power rule: the derivative is [itex]-2x^{-3}-1= \frac{-2}{x^3}- 1[/itex].

    If the problem itself requires that you use the definition of the derivative, say so!

    What is [tex]\frac{1}{(x+h)^2}- \frac{1}{x^2}[/tex]?
     
  4. Dec 1, 2006 #3
    Show more steps but...

    If you have to use the definition of a derivative to solve,

    [tex]f(x+h)[/tex] = what? Plug in x+h into the original equation and show us what you get. Then show us what f(x+h)-f(x) looks like.

    edit::Sorry HallsofIvy, didn't see your post before I posted :P
     
  5. Dec 1, 2006 #4
    hint [tex] (x+h)^{2}=x^2+2hx+h^{2}[/tex]
     
  6. Dec 1, 2006 #5
    thank you for all the hint
    i was really told to do it the long cut way

    where i was stuck was:
    [tex]\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - x - h - \frac{1}{x^2} + x}{h}[/tex]
    then i cancel the x.....

    then
    [tex]\lim_{h \to 0} \frac{\frac{1}{x^2+2xh+h^2} - h - \frac{1}{x^2}{h}[/tex]

    then i cancel the [tex]\frac{1}{x^2}[/tex]

    then this was the part where i got stuck:
    [tex]\lim_{h \to 0} \frac{\frac{1}{2xh}+\frac{1}{h^2}-h}{h}[/tex]
     
    Last edited: Dec 1, 2006
  7. Dec 1, 2006 #6
    Notice that

    [tex]f(x+h)=\frac{1}{(x+h)^{2}}-(x+h)[/tex]
    and
    [tex]f(x)=\frac{1}{x^{2}}-x[/tex]

    so

    [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
    [tex]=\lim_{h \to 0} \frac{(\frac{1}{(x+h)^{2}}-(x+h))-(\frac{1}{x^{2}}-x)}{h}[/tex]
    [tex]=\lim_{h \to 0} \frac{\frac{1}{x^{2}+2hx+h^{2}}-\frac{1}{x^{2}}-h}{h}[/tex]

    After this point you will need to rewrite both fractions on the numerator as one giant fraction and divide each term by h and then evaluate
     
    Last edited: Dec 1, 2006
  8. Dec 1, 2006 #7
    [tex](x+y)^{2} = x^{2}+2xy+y^{2}[/tex] it is NOT [tex]x^{2}+y^{2}[/tex]

    [tex] \frac{1}{(x+h)^{2}} = \frac{1}{x^{2}+2hx+h^{2}}[/tex] it is NOT [tex]\frac{1}{x^{2}}+\frac{1}{h^{2}}[/tex] or anything like that.
     
  9. Dec 1, 2006 #8
    I got it...
    [tex]\frac{-2-x^3}{x^3}[/tex]

    i am very thankful for all your help
     
  10. Dec 1, 2006 #9
    Correct.

    No problem at all, happy to help you. I would rewrite it though.

    [tex]\frac{x^{2}-1}{x^{2}} = 1-\frac{1}{x^{2}}[/tex] (Just demonstrating how, you should be able to apply it and simplify)

    But thats up to you and I would think that the format you have it in now is okay.
     
    Last edited: Dec 1, 2006
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