take derivative:

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take derivative:

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[(x+1\1-x )^{2}\]
will i use quotient rule first then chain rule
Chain rule then quotient rule

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move the 2 down first then quotient rule the inside.
will i move 2 down multiply(inside) then multiply by quotient rule did that make sense?
yes. So it ends up being 2(inside)(quotient-ruled-fraction)
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) ????????
Quotient rule: for simplification: T=top numerator, B=bottom denominator. (T/B)'= ((T')*B-(B')*T)/(B^2)
2(x+1/1-x) is correct so far... just follow the quotient rule for the inside, multiply and you're done.
i did that and got [this is after the 2(x+1/1-x) ]] (1-x)(1)-(x+1)(-1)/(1-x)^2
then after that i get 2(x+1/1-x) (1-x+x+1/(1-x)^2 is that ok so far
Yeah that's correct.
ok then i get 2(x+1/1-x) [2/(1-x)^2] what about that
Yeah. that's good too. Then you can combine fractions again.
ok then I get 2(2x+2/x^2-2x+1
No. Im not sure what you did there.
2(x+1/1-x) [2/(1-x)^2] top: 2* (x+1)* 2 bottom: (1-x) * (1-x)^2
so should i have 4x+4 on top
Yup.
then bottom should be x^2-2x+1
never mind one 1-x is ^2
yeah. all the bottom components are the same, so just make it (1-x)^3
thanks for your help, i knew more than i thought. i just needed a little guidance

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