# Derivative by using product rule

• Oct 13th 2011, 12:41 PM
jordanchamp07
Derivative by using product rule
I have been doing a promplem but i am currently lost.I want to know how to do these type of problems since they will be on a test.The probles is as follows

x^2(2/x-1/(x+1))
plz help and show and explain each step.I have looked everywhere for help but nobody wants to or knows how to.Please help
• Oct 13th 2011, 12:46 PM
e^(i*pi)
Re: Derivative by using product rule
Quote:

Originally Posted by jordanchamp07
I have been doing a promplem but i am currently lost.I want to know how to do these type of problems since they will be on a test.The probles is as follows

x^2(2/x-1/(x+1))
plz help and show and explain each step.I have looked everywhere for help but nobody wants to or knows how to.Please help

No wonder since that's an awful way to write it

$x^2 \left(\dfrac{\frac{2}{x-1}}{x+1}}\right) = x^2 \left(\dfrac{2}{(x-1)(x+1)}\right)$

OR

$x^2 \left(\dfrac{2}{\frac{x-1}{x+1}}\right) = x^2(x+1) \left(\dfrac{2}{x-1}\right)$
• Oct 13th 2011, 12:51 PM
jordanchamp07
Re: Derivative by using product rule
well its x^2 ( (2/X) - (1/x+1))
• Oct 13th 2011, 01:05 PM
e^(i*pi)
Re: Derivative by using product rule
Quote:

Originally Posted by jordanchamp07
well its x^2 ( (2/X) - (1/x+1))

$x^2 \left(\dfrac{2}{x} - \dfrac{1}{x+1}\right) = \dfrac{x^2}{2x} - \dfrac{x^2}{x+1}$

The first term is simple enough to do. For the second write it as $u = x^2$ and $v = (x+1)^{-1}$ and use the product rule
• Oct 13th 2011, 01:09 PM
jordanchamp07
Re: Derivative by using product rule
man im still lost sorry i am not very bright :(
• Oct 13th 2011, 01:12 PM
mr fantastic
Re: Derivative by using product rule
Quote:

Originally Posted by jordanchamp07
man im still lost sorry i am not very bright :(

Then you need to say exactly what part or parts of post #4 you don't understand. It may well be that you need to go back and review easier questions before attempting this question. You might also need to go back and review your algebra - most difficulties in calculus are caused by a lack of algebraic competency (competency in algebra will be assumed by your instructor).

Sine the name of your post is use the product rule, I'd suggest you start by noting that if you have $x^2 \left( \frac{2}{x} - \frac{1}{x+1}\right)$ then $u = x^2$ and $v = \frac{2}{x} - \frac{1}{x+1}$.

Now go and look up the product rule to see what to do. Now go to your notes and review how to differentiate standard forms.
• Oct 13th 2011, 01:16 PM
jordanchamp07
Re: Derivative by using product rule
Well i really dont have much time to go back in review(if i had time i would go back).They teach me to do fs'+sf' but i do not know how to find the derivitive of second term.Is their a rewrite for it? my notes doesnt have this type of problems
• Oct 13th 2011, 02:54 PM
HallsofIvy
Re: Derivative by using product rule
What "second term" are you talking about? If you mean the denominator in $(x+1)^{-1}$ just use the power rule: $(x^n)'= nx^{n-1}$ together with the chain rule: $[(x+1)^{-1}]'= (-1)(x+1)^{-2}$ times the derivative of x+ 1. Do you know how to differentiate x+ 1?