Mathematics 24 Online
OpenStudy (anonymous):

d 2x^2 + x + 1 _ ----------- dx 1 - 2x =?

OpenStudy (aravindg):

:)

OpenStudy (aravindg):

OpenStudy (aravindg):

u see many dont care to giv medal after i tell answer

OpenStudy (aravindg):

just use quotient rule....

OpenStudy (anonymous):

can show me the steps using differentiate first time keep second term keep first term differentiate second term =D

OpenStudy (anonymous):

coz that method can replace quotient rule and product rule

OpenStudy (anonymous):

what do you need? the derivative of $\frac{2x^2+x+1}{1-x}$?

OpenStudy (anonymous):

1 - 2x yes

OpenStudy (anonymous):

ok use $(\frac{f}{g})'=\frac{gf'-fg'}{g^2}$ with $f(x)=2x^2+x+1,f'(x)=4x+1$ $g(x)=1-2x$ $g'(x)=-2$

OpenStudy (anonymous):

can you solve it for me coz i'm not sure whether i get the correct answer

OpenStudy (anonymous):

i can write it out if you like, but it is a matter of removing the parentheses in the numerator and collecting like terms

OpenStudy (anonymous):

yea i understand

OpenStudy (anonymous):

so how do i leave it in my final answer

OpenStudy (anonymous):

hold on i think i messed up. let me go slow

OpenStudy (anonymous):

yeah i copied wrong. it should be $\frac{(1-2x)(4x+1)-(2x^2+2x+1)(-2)}{(1-2x)^2}$

OpenStudy (anonymous):

i forgot the 2 in front of the x^2 in the second part.

OpenStudy (anonymous):

so now the numerator is $-4x^2+4x+3$ my mistake. so "final answer" is $\frac{-4x^2+4x+3}{(1-2x)^2}$

OpenStudy (anonymous):

Oh yes you i got the answer woots :)

OpenStudy (anonymous):

let me delete the wrong stuff so as not to confuse more

OpenStudy (anonymous):

congrats!