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

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$$^{}4$$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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tiny-tim
Homework Helper
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! 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. Can you show us your working (the line f´g + fg´, before expanding), so that we can see what might have gone wrong? thanks, more questions

I'm new to the site, so I'm not familiar with typing the math problems . 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 Im stuck, don't know if I should add derivatives together or to multiply them. Suggestions? where did I go wrong?

RE: F(y)= (1/y - 3/y)(y+5y)

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????

tiny-tim
Homework Helper
Hi jimen! 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 [Broken]
and http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 [Broken]
for all the symbols and techniques.
(-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! [size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​

Last edited by a moderator:
Hi jimen! I suggest you "bookmark" http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000 [Broken]
and http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 [Broken]
for all the symbols and techniques.
either that or just use mathtype and its translation tool ;)

Last edited by a moderator:
tiny-tim
ooh! what's mathtype? 