Question

Given that y = 2x+1/1-x, find dy/dx. Hence, or otherwise find ∫1/(1-x)^2 dx.

Answer 1

quotient rule for this one

Answer 2

let me try that part

Answer 3

k

Answer 4

ooops!! i made a mistake , so don't try it

Answer 5

should be \[y'=y'=\frac{2(1-x)+(2x+1)}{(1-x)^2}\]

Answer 6

\[(\frac{f}{g})'=\frac{f'g-g'f}{g^2}\] with
\[f(x)=2x+1,f'(x)=2,g(x)=1-x,g'(x)=-1\] i forgot the \(2\)

Answer 7

oh, now I understand tht part, so what's the next step

Answer 8

what next step?

Answer 9

you mean the algebra, or the integral?

Answer 10

is it just end like that? i mean, no expansion or something? and what should I do with this one : ∫1/(1-x)^2 dx.

Answer 11

\[y'=\frac{2(1-x)+(2x+1)}{(1-x)^2}\]needs cleaning up with algebra up top

Answer 12

you get
\[y'=\frac{2-2x+2x+1}{(1-x)^2}\] or
\[y'=\frac{3}{(1-x)^2}\]

Answer 13

now it is easy to find
\[\int \frac{1}{(1-x)^2}dx\] since you know
\[\int \frac{3}{(1-x)^2}dx=\frac{2x+1}{1-x}\]

Answer 14

divide by 3 and you have it

Answer 15

i get this instead of yours. how come?

Answer 16

the denominator should be squared, not cubed

Answer 17

opps. i wrong. but how about this one

Answer 18

the numerator has \(2x+1\) so that should be \(2x+1\) as well
they are the same

Answer 19

u mention earlier that, g′(x)=−1 so it suppose (1-x)(2)+(2x+1)(-1)

Answer 20

in the formula for the quotient rule there is a minus sign in the numerator
it is \(f'g-g'f\) the two minus signs makes is a plus

Answer 21

oh, alright. let me try now.

Answer 22

@satellite73 I get the answer. thank you for your help... :D