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

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1. Homework Statement [/b]

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


Homework Equations


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

The Attempt at a Solution



In order to use the product formula I brougth the first parenthesis to a common denominator y^{}4 and then tried to use the product formula to find the derivative (it didnt work), then I tried subracting 1/y^{}2 from 3/y^{}4and 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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jimen113 said:
F(y)= (1/y^{}2 - 3/y^{}4)(y+5y^{}3)
the answer is: F\acute{}(y)=5+14/y^{}2+9/y^{}4

(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:
 
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^{}-3+12y^{}-5. Then I found the derivative for 2 parenthesis, which is (1+5y^{}2), Now I am stuck, don't know if I should add derivatives together or to multiply them. Suggestions?:confused:
 
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)\acute{}
f\acute{}*g)+(f*g\acute{})

(-2y\hat{}-3+12y\hat{}-5)*(y+5y\hat{}3)+(y\hat{}-3-3y\hat{}-4)*(1+15y\hat{}2)
so, I get (14/y\hat{}2 +9/y\hat{}4)+5y,
which is very close to the answer in the book, except that instead of 5y they have the answer as just (14/y\hat{}2 +9/y\hat{}4)+5
??
 
Why do you post twice the same problem?
 
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\hat{}-3+12y\hat{}-5)*(y+5y\hat{}3)+(y\hat{}-3-3y\hat{}-4)*(1+15y\hat{}2)
Yes - I thought so - you've just put y^{-3} in the second part instead of y^{-2}

Try it again! :smile:
[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​
 
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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:
steven10137 said:
either that or just use mathtype and its translation tool ;)

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