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

1. Mar 8, 2008

### jimen113

1. The problem statement, all variables and given/known data[/b]

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

2. Relevant equations
the answer is: F$$\acute{}(y)$$=5+14/y$$^{}2$$+9/y$$^{}4$$
3. 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.

2. Mar 8, 2008

### tiny-tim

(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?

3. Mar 8, 2008

### jimen113

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?

4. Mar 8, 2008

### jimen113

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
??

5. Mar 8, 2008

### sutupidmath

Why do you post twice the same problem????

6. Mar 9, 2008

### tiny-tim

Hi jimen!
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.
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]​

7. Mar 9, 2008

8. Mar 9, 2008

### tiny-tim

ooh! what's mathtype?

9. Mar 9, 2008

### jimen113

Thanks!

10. Mar 9, 2008

### jimen113

Thanks for your help, it worked!