Question

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

Answer 1

:)

Answer 2

first giv medal thn answer

Answer 3

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

Answer 4

just use quotient rule....

Answer 5

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

Answer 6

coz that method can replace quotient rule and product rule

Answer 7

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

Answer 8

1 - 2x yes

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

yea i understand

Answer 13

so how do i leave it in my final answer

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Oh yes you i got the answer woots :)

Answer 19

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

Answer 20

congrats!