Question

take derivative:

Answer 1

\[(x+1\1-x )^{2}\]

Answer 2

will i use quotient rule first then chain rule

Answer 3

Chain rule then quotient rule

Answer 4

move the 2 down first then quotient rule the inside.

Answer 5

will i move 2 down multiply(inside) then multiply by quotient rule
did that make sense?

Answer 6

yes.
So it ends up being 2(inside)(quotient-ruled-fraction)

Answer 7

I am working on this and I know it is not correct, but too difficult to type the problem out.
2(x+1/1-x) (2/(1-x)^2) ????????

Answer 8

Quotient rule: for simplification: T=top numerator, B=bottom denominator.

(T/B)'= ((T')*B-(B')*T)/(B^2)

Answer 9

2(x+1/1-x) is correct so far...
just follow the quotient rule for the inside, multiply and you're done.

Answer 10

i did that and got [this is after the 2(x+1/1-x) ]]
(1-x)(1)-(x+1)(-1)/(1-x)^2

Answer 11

then after that i get
2(x+1/1-x) (1-x+x+1/(1-x)^2
is that ok so far

Answer 12

Yeah that's correct.

Answer 13

ok then i get
2(x+1/1-x) [2/(1-x)^2]

what about that

Answer 14

Yeah. that's good too. Then you can combine fractions again.

Answer 15

ok then I get

2(2x+2/x^2-2x+1

Answer 16

No. Im not sure what you did there.

Answer 17

2(x+1/1-x) [2/(1-x)^2]
top: 2* (x+1)* 2
bottom: (1-x) * (1-x)^2

Answer 18

so should i have 4x+4 on top

Answer 19

Yup.

Answer 20

then bottom should be x^2-2x+1

Answer 21

never mind one 1-x is ^2

Answer 22

yeah. all the bottom components are the same, so just make it (1-x)^3

Answer 23

thanks for your help, i knew more than i thought. i just needed a little guidance