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

:)

first giv medal thn answer

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

just use quotient rule....

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

coz that method can replace quotient rule and product rule

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

1 - 2x yes

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

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

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

yea i understand

so how do i leave it in my final answer

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

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

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

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

Oh yes you i got the answer woots :)

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

congrats!

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