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How to differentiate (1/y^2 - 3/y^4)(y + 5y^3) for y

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data[/b]

    F(y)= (1/y[tex]^{}2[/tex] - 3/y[tex]^{}4[/tex])(y+5y[tex]^{}3[/tex])

    2. Relevant equations
    the answer is: F[tex]\acute{}(y)[/tex]=5+14/y[tex]^{}2[/tex]+9/y[tex]^{}4[/tex]
    3. The attempt at a solution

    In order to use the product formula I brougth the first parenthesis to a common denominator y[tex]^{}4[/tex] and then tried to use the product formula to find the derivative (it didnt work), then I tried subracting 1/y[tex]^{}2[/tex] from 3/y[tex]^{}4[/tex]and then tried to use the product rule. I can't figure out how to use the product/quotient formula to find the derivative, although I have the answer I want to know how to work the problem out.
  2. jcsd
  3. Mar 8, 2008 #2


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    (Yes, that seems to be the right answer.)

    Hi jimen! :smile:

    No way would I ever use the product rule for this … it's much simpler just to multiply it out, and then differentiate!

    But if you want to do it with the product rule as an exercise, that's fine …

    I expect you just got a minus in the wrong place - it's easy to do that with inverse powers. :frown:

    Can you show us your working (the line f´g + fg´, before expanding), so that we can see what might have gone wrong? :smile:
  4. Mar 8, 2008 #3
    thanks, more questions

    Thanks for your immediate reply.
    I'm new to the site, so I'm not familiar with typing the math problems:redface:. Basically, I'm having problems typing in the 1st parenthesis in fraction format. At any rate, I found the derivitative of the 1st parenthesis (did it using the power rule)= (-2y[tex]^{}-3[/tex]+12y[tex]^{}-5[/tex]. Then I found the derivative for 2 parenthesis, which is (1+5y[tex]^{}2[/tex]), Now Im stuck, don't know if I should add derivatives together or to multiply them. Suggestions?:confused:
  5. Mar 8, 2008 #4
    where did I go wrong?

    RE: F(y)= (1/y - 3/y)(y+5y)
    the answer is: F=5+14/y+9/y

    F(y)= (f*g)[tex]\acute{}[/tex]

    so, I get (14/y[tex]\hat{}2[/tex] +9/y[tex]\hat{}4[/tex])+5y,
    which is very close to the answer in the book, except that instead of 5y they have the answer as just (14/y[tex]\hat{}2[/tex] +9/y[tex]\hat{}4[/tex])+5
  6. Mar 8, 2008 #5
    Why do you post twice the same problem????
  7. Mar 9, 2008 #6


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    Hi jimen! :smile:
    I suggest you "bookmark" http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000
    and http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000
    for all the symbols and techniques.
    Yes - I thought so - you've just put [tex]y^{-3}[/tex] in the second part instead of [tex]y^{-2}[/tex]

    Try it again! :smile:
    [size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
  8. Mar 9, 2008 #7
  9. Mar 9, 2008 #8


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    ooh! what's mathtype? :smile:
  10. Mar 9, 2008 #9
  11. Mar 9, 2008 #10
    Thanks for your help, it worked!
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