Question

help: differentiate F(y)=(1/y^2-3/y^4)(y+5y^3), i got 13/y^2+12/y^4, and the book gives a different answer

Answer 1

\(F(y)=(\frac{1}{y^2}-\frac{3}{y^4})(y+5y^3)\)
\(F(y)=(y^{-2}-3y^{-4})(y+5y^3)\)
Use product rule
\(F'(y)=(y^{-2}-3y^{-4})'(y+5y^3)+(y^{-2}-3y^{-4})(y+5y^3)'\)
Did start off like this?

Answer 2

yes and after taking the derivatives i got -45y^-2-2y^-2+12y^-4+60y^-2

Answer 3

\(F'(y)=(-2y^{-3}+12y^{-5})(y+5y^{3})+(y^{-2}-3y^{-4})(1+15y^{2})\)
Foiling and then combine like terms
\((-2y^{-3}+12y^{-5})(y+5y^{3})= (-2y^{-2}-10+12y^{-4}+60y^{2})+(y^{-2}+15-3y^{-4}-45y^2)\)
\(=(9y^{-4}+15y^2-y^{-2}+5)\) my answer <--

Answer 4

What does the answer in the book has?

Answer 5

@petrina313

Answer 6

Do you see the part when i foiled it ? It doesn't show completely to me.

Answer 7

\[F'(y)=5+14/y^2+9/y^4\]

Answer 8

Another way to solve this is by foiling the equation first before taking the derivative. \(F(y)=(1/y^2-3/y^4)(y+5y^3)\)
i got \(5+9y^{-4}\) correct, i'll foil it again.

Answer 9

So here's what i did, but my answer is different from what i got and what the book has:
\(F'(y)=(-2y^{-2}-10+12y^{-4}+60y^{2})+(y^{-2}+15-3y^{-4}-45y^{-2})\)
\(F'(y)=-2y^{-2}+y^{-2}-45y^{-2}-3y^{-4}+60y^{-2}-10+15\)
\(F'(y)=5+14y^{-2}-3y^{-4}\)

Answer 10

thank you so much i just need to look it over to see what i got wrong

Answer 11

No problem. I think the best wau to solve this is by foiling this first F(y)=(1/y^2-3/y^4)(y+5y^3) and then take the derivative. Way simpler than taking the derivative and then foiling, too much foiling.

Answer 12

\[(y ^{-2}-3y ^{-4})(1+15y ^{3})\] when you multiply \[(y ^{-2})(15y ^{3})\] isnt the y supposed to stay?

Answer 13

@Zale101

Answer 14

you got 15y^3 hmmm, i got 15 y^2
when you take the derivative of (\(y+5y^3)\) that how i got \((1+15y^2)\)

Answer 15

wow i completely forgot to subtract it, thanks again

Answer 16

Np