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

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Homework Help Overview

The problem involves differentiating the function F(y) = (1/y^2 - 3/y^4)(y + 5y^3) with respect to y. The context is calculus, specifically focusing on the application of the product rule for differentiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the product rule versus expanding the expression before differentiation. Some express confusion about the correct application of differentiation rules and the handling of negative signs in inverse powers.

Discussion Status

Participants are actively sharing their attempts and reasoning. Some have provided guidance on potential errors in the differentiation process, while others are exploring different methods to approach the problem. There is no explicit consensus on the best method yet.

Contextual Notes

There are indications of difficulties with notation and formatting of mathematical expressions, which may be affecting the clarity of the discussion. Participants are also navigating the challenge of ensuring accurate differentiation of terms involving negative exponents.

jimen113
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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]​
 
Last edited by a moderator:
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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