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Differentiate part 2

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


    RE: F(y)= (1/y[tex]\hat{}2[/tex] - 3/y[tex]\hat{}4[/tex])(y+5y)
    the answer is: F=5+14/y[tex]\hat{}2[/tex]+9/y[tex]\check{}4[/tex]





    3. The attempt at a solution


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

    (-2y[tex]\hat{}-3[/tex]+12y[tex]\hat{}-5[/tex])*(y+5y[tex]\hat{}3[/tex])+(y[tex]\hat{}-3[/tex]-3y[tex]\hat{}-4[/tex])*(1+15y[tex]\hat{}2[/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
    ??
     
  2. jcsd
  3. Mar 8, 2008 #2
    I cannot clearly read what you did, but you first need to apply the product rule tha is
    let [tex] f(y)=y^{-2}-3y^{-4}, \ \ and \ \ \ g(y)=6y[/tex] so

    [tex]F(y)=f(y)*g(y)=>F'(y)=f'(y)*g(y)+g'(y)*f(y)[/tex]

    or you could merely foil everyghing out and you would end up with:

    [tex]F(y)=6y^{-1}-18y^{-3}[/tex] and then take the derivative of this one.


    However i think that you originally gave us the wrong function, for there is no way you can get the answer you provided us. check it again, for i won't do it for you. Because by just integrating the result you gave us, one cannot get the function you provided.
     
    Last edited: Mar 8, 2008
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