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

In summary, the conversation discusses a problem with finding the derivative of a function using the product rule. The function in question is F(y)= (1/y^{}2 - 3/y^{}4)(y+5y^{}3) and the resulting derivative should be 5+14/y^{}2+9/y^{}4. The conversation also includes a suggestion to use the website http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000 and http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 for help with typing mathematical symbols and
  • #1
jimen113
67
0
1. Homework Statement [/b]

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


Homework Equations


the answer is: F[tex]\acute{}(y)[/tex]=5+14/y[tex]^{}2[/tex]+9/y[tex]^{}4[/tex]

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.
 
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  • #2
jimen113 said:
F(y)= (1/y[tex]^{}2[/tex] - 3/y[tex]^{}4[/tex])(y+5y[tex]^{}3[/tex])
the answer is: F[tex]\acute{}(y)[/tex]=5+14/y[tex]^{}2[/tex]+9/y[tex]^{}4[/tex]

(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:
 
  • #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 I am stuck, don't know if I should add derivatives together or to multiply them. Suggestions?:confused:
 
  • #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]
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
??
 
  • #5
Why do you post twice the same problem?
 
  • #6
Hi jimen! :smile:
jimen113 said:
Basically, I'm having problems typing in the 1st parenthesis in fraction format.
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.
jimen113 said:
(-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])
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]​
 
Last edited by a moderator:
  • #7
tiny-tim said:
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.

either that or just use mathtype and its translation tool ;)
 
Last edited by a moderator:
  • #8
steven10137 said:
either that or just use mathtype and its translation tool ;)

ooh! what's mathtype? :smile:
 
  • #9
steven10137 said:
either that or just use mathtype and its translation tool ;)
Thanks!
 
  • #10
Thanks for your help, it worked!
 

1. What is the first step in differentiating (1/y^2 - 3/y^4)(y + 5y^3) for y?

The first step is to use the product rule, which states that the derivative of a product is the first function times the derivative of the second function plus the second function times the derivative of the first function.

2. How do you simplify the expression (1/y^2 - 3/y^4)(y + 5y^3) before differentiating?

You can simplify the expression by factoring out a 1/y^2 term from the first set of parentheses, and a y term from the second set of parentheses, to get (1 - 3/y^2)(1 + 5y^2).

3. What is the derivative of (1/y^2 - 3/y^4)(y + 5y^3) for y?

The derivative is (1 - 3/y^2)(5y^2) + (1/y^2 - 3/y^4)(3/y^3 + 15y^2).

4. How can you simplify the derivative of (1/y^2 - 3/y^4)(y + 5y^3) for y?

You can simplify the derivative by distributing the 5y^2 and 3/y^3 terms, and combining like terms to get (5y^2 - 15/y^2 + 3/y^3 - 9/y^5).

5. What is the final simplified expression when differentiating (1/y^2 - 3/y^4)(y + 5y^3) for y?

The final simplified expression is (5y^2 - 15/y^2 + 3/y^3 - 9/y^5).

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