# problem using the product rule in calculus

Discussion in 'Community Discussion' started by macman2790, Feb 22, 2007.

1. ### macman2790 macrumors 6502a

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Sep 4, 2006
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#1
im having trouble getting the same answer as it says in the back of the book for this one:

f(x)=(1/(x^2)-3/(x^4))(x + 5x^3)
im pretty sure that you have to use the product rule on this one.
the answer in the back of the book is:
f`(x)= 5 + 14/(x^4)+ 9/(x^2)

i've also tried multiplying it out and then differentiating it and it gets nowhere near that answer.

thanks

2. ### pianoman181 macrumors regular

Joined:
Jan 6, 2004
#2
Don't forget the chain rule!

If I weren't in bed right now, I'd do it out for you.

3. ### macman2790 thread starter macrumors 6502a

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Sep 4, 2006
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#3
too bad i cant use it, since it requires the product rule or the other one i stated. i dont know the chain rule, we aren't there yet. sorry

4. ### macman2790 thread starter macrumors 6502a

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#4
by the way the product rule is:
d/dx [f(x)g(x)] = f(x) d/dx[g(x)] + g(x) d/dx[f(x)]

5. ### Abstract macrumors Penryn

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#5
My advice: While I haven't done this sort of maths for years and years, I think it's best to convert it to this form:

[(x^-2) - (3x^-4)]*[x + 5x^3]

This way, you don't need to deal with 1/x type values, as the form above is much easier to deal with when taking derivatives. Can you solve it now?

6. ### macman2790 thread starter macrumors 6502a

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#6
thank you very much

7. ### siurpeeman macrumors 603

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Dec 2, 2006
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#7
are you sure it isn't f'(x) = 5 + 14/(x^2) + 9/(x^4) ?

8. ### ®îçhå®? macrumors 68000

Joined:
Mar 7, 2006
#8
Unless i am readig it incorrectly, the Chain rule is completely pointless in this case

Have you ever thought that the book is wrong?? Happens a lot in my book

9. ### Abstract macrumors Penryn

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#9

OK....just did the question myself, and got the same answer as you.

It actually brought a smile to my face that I could still do it. I had to cheat and use the Product Rule definition given by the OP, but still.