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gerald
 5 years ago
find the derivatives of x4+x32x2/x+2
gerald
 5 years ago
find the derivatives of x4+x32x2/x+2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first do the long divison and then use partial fraction method to integrate

gerald
 5 years ago
Best ResponseYou've already chosen the best response.0pliz can you do that for me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where are you avnis chat box is broken

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that length question i think i solved it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so working on this question i know you can find common terms in this question i am just trying to find that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean i am refering to gerald's question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets do long division

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm okay, but i was thinking if we just rearrange and ge trid of the common terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry here http://www.skrbl.com/166859043

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.scriblink.com/index.jsp?act=phome&roomid=36&KEY=F71D8E165FCC9B3AA0E17782EDB164F0

gerald
 5 years ago
Best ResponseYou've already chosen the best response.0can those link lead me to answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes im typing solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.scriblink.com/index.jsp?act=phome&roomid=36&KEY=F71D8E165FCC9B3AA0E17782EDB164F0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here im over here http://www.scriblink.com/index.jsp?act=phome&roomid=36&KEY=F71D8E165FCC9B3AA0E17782EDB164F0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i solve dit on the the other one :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you get integral x^3  x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so basically gotta integrate x^3 x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that will give you ((x^4)/4)(x^3/3) + C

gerald
 5 years ago
Best ResponseYou've already chosen the best response.0those site are not active

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u just want to find the first derivative right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are we talking about derivatives or integrals? because you just integrated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uhm cantorset...do you know what you are talking about?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0according to the questions, it says "derivatives" , right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can use the quotient rule to solve this if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do ther derivative with quotient rule. ((x+2)(4x^3+3x^22x^2)(x^4+x+32x^2))/(x+2)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is the correct approach, ignore whatever cantorset mentioned its nonsense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0d/dx of f(x)/g(x)=( f'(x)g(x)f(x)g'(x))/(g(x))^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, so when u want to find the derivative, here , its better if u use the 'u/v' rule.. which is \[(v u \primeu v \prime )/v ^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wher u =numerator & v=denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0exactly, thats correct @avalanche :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great minds think alike :)
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