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Derivative problem

  1. Oct 8, 2006 #1

    first I did this:

    I used the quotient rule to get:

    1.x^2(2x^3-1) + 2x^3-1(x^2)

    2.x^2(6x^2) + 2x^3-1(x^2)

    3.6x^4 + 4x+4 -2x


    and I arrived with my final answer:


    I know I got it wrong. But I would love to know why
    and what I should have done instead. Thank you so
    much for your help!
    Last edited: Oct 8, 2006
  2. jcsd
  3. Oct 8, 2006 #2


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    Homework Helper

    First of all, is the function [tex]f(x)=\frac{2e^3}{x^2}[/tex] or [tex]f(x)=\frac{2x^3}{x^2}[/tex]? It seems to me that you applied the quotient derivative rule wrong. The rule is, for some function [tex]f(x)=\frac{g(x)}{h(x)}[/tex]: [tex]f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}[/tex].
  4. Oct 8, 2006 #3
    What do you use to put in the functions so it appears that way? that is neat!

    I'll try doing what you say and I'll see if it works. Thanks a bunch!

    BTW, i editted the question. I'm sorry about the typo!!!

    EDIT -

    I just noticed... I used addition instead of subtraction... DOH! This is what happens when you're up for 48 hours...
    Last edited: Oct 8, 2006
  5. Oct 8, 2006 #4
    so this is what I did...

    x^2(6x^2) - 2x^3-1(2x)/x^4

    then i got:


    I still think I did something wrong. Thanks
  6. Oct 8, 2006 #5


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    Just slow down. Identify what g(x) and h(x) are and use the rule I gave you.
  7. Oct 8, 2006 #6
    Radou - with the equation of you gave me, I noticed you said first take g'(x)h(x) - g(x)h'(x)

    should it not be g(x)h'(x) - h(x)g'(x)????
    edit- ahhhhhhhhhhhhh my fault read it wrong. thx radou.
    Last edited: Oct 8, 2006
  8. Oct 8, 2006 #7


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    I wouldn't give you wrong equations, don't worry. :smile:
  9. Oct 8, 2006 #8
    Thanks Radou. I solved the problem successfully. I just had many nub mistakes hahaha. Much appreciated!
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