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Calculus n00b stuck

  1. May 12, 2005 #1
    One of my biggest griefs is the fact that I'm a complete n00b when it comes to maths.

    I'm supposed to find the second derivate of

    [tex]f(x) = \frac{2-3x}{x^2}[/tex]

    I started out with this:

    [tex]f'(x) = \frac{-3}{x^2} + \frac{2-3x}{2x}[/tex]

    [tex]f''(x) = \frac{3}{2x} + \frac{2-3x}{2}[/tex]

    But it didn't take long until I found I was just making gibberish... :frown:
    Any ideas?
  2. jcsd
  3. May 12, 2005 #2


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    Nope.Both are incorrect.U should apply the rules properly.It's easier,if u decompose.

    [tex] f(x)=2x^{-2}-3x^{-1} [/tex]

    Can u compute the derivatives now...?


    P.S.HINT:Just the power rule involved.
  4. May 12, 2005 #3
    Are you familiar with the quotient rule?
  5. May 12, 2005 #4
    This should be correct, then:

    [tex] f(x)=2x^{-2}-3x^{-1} [/tex]

    [tex] f'(x)=-4x^{-3}+3x^{-2} [/tex]

    [tex] f''(x)=12x^{-4}-6x^{-3} [/tex]

    The textbook says it's [itex] \frac{12-6X}{X^4}[/itex], which, according to my graphing calculator identical to what I got above, but unfortunately both me and my father are lacking in skill to figure that out.

    I have heard about the quotient rule, but I haven't been able to fully understand it. I'm going to ask my teacher for a tutor lesson tomorrow.
  6. May 12, 2005 #5


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    [tex] \frac{12-6x}{x^{4}}=\frac{12}{x^{4}}-\frac{6x}{x^{4}}=12x^{-4}-6x^{-3} [/tex]


  7. Jun 9, 2005 #6
    If you divide your final answer for f'' by x^4 you will find it gives the textbook answer ^_^
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