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

In summary, the conversation discusses solving for the function F(y) given the function F(y)= (1/y^2 - 3/y^4)(y+5y) and provides a solution of F=5+14/y^2+9/y^4. The conversation also mentions the use of the product rule and the possibility of a mistake in the original function provided.
  • #1
jimen113
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0

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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  • #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:

1. What is the formula for F(y)?

The formula for F(y) is 5 + 14/y2 + 9/y4.

2. How do you differentiate F(y)?

To differentiate F(y), we use the power rule and the chain rule. First, we bring the exponent down and subtract 1 from the original exponent. Then, we multiply by the derivative of the inner function.

3. What is the derivative of F(y)?

The derivative of F(y) is -28/y3 - 36/y5.

4. What is the purpose of differentiating F(y)?

Differentiating F(y) allows us to find the rate of change of the function at any given point. It also helps us find the slope of the tangent line at a specific point on the graph of the function.

5. Can this formula be simplified?

Yes, this formula can be simplified by factoring out a common factor of 1/y2, giving us F(y) = 5/y2 + 14 + 9y2.

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