## anonymous 5 years ago take derivative:

1. anonymous

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

2. anonymous

will i use quotient rule first then chain rule

3. anonymous

Chain rule then quotient rule

4. anonymous

move the 2 down first then quotient rule the inside.

5. anonymous

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

6. anonymous

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

7. anonymous

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) ????????

8. anonymous

Quotient rule: for simplification: T=top numerator, B=bottom denominator. (T/B)'= ((T')*B-(B')*T)/(B^2)

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

Yeah that's correct.

13. anonymous

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

14. anonymous

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

15. anonymous

ok then I get 2(2x+2/x^2-2x+1

16. anonymous

No. Im not sure what you did there.

17. anonymous

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

18. anonymous

so should i have 4x+4 on top

19. anonymous

Yup.

20. anonymous

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

21. anonymous

never mind one 1-x is ^2

22. anonymous

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

23. anonymous

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