Differentiate Part 2 Homework: F(y)=5+14/y\hat{}2+9/y\check{}4

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SUMMARY

The discussion focuses on differentiating the function F(y) = (1/y² - 3/y⁴)(y + 5y) and verifying the solution against a textbook answer. The correct derivative, as derived through the product rule, is F'(y) = f'(y)g(y) + g'(y)f(y), where f(y) = y⁻² - 3y⁻⁴ and g(y) = 6y. The final simplified form of the function is confirmed to be F(y) = 6y⁻¹ - 18y⁻³, indicating a discrepancy in the original function provided by the user.

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Homework Statement




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]





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
??
 
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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.
 
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