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find 3rd derivate x/x+1
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Okay now jjanelle, see what I meant, i doubt this problem meant x/x + 1, or it would just be 2... That's why I like to keep parenthesis all the time..
I would solve it like this: \[y=x*(x+1)^{-1}\]
and use the product and chain rules..
\[y' = x(-1)(x+1)^{-2} + (x+1)^{-1}*1\]
\[6/(x+1)^{4}\]
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\[y' = (x+1)^{-1} - x(x+1)^{-2}\]
\[y'' = - (x+1) - [x(-2)(x+1)^{-3} + (x+1)^{-2}]\]
\[y" = -x - 1 +2x(x+1)^{-3} - (x+1)^{-2}\]
\[y''' = -1 + 2x(-3)(x+1)^{-4} + (x+1)^{-3}*2 + 2(x+1)^{-3}\]
\[y''' = -1 -6x/(x+1) + 2/(x+1)^3 + 2/(x+1)^3\]
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\[y''' = -1 -6x/(x+1) + 4/(x+1)^3\]
thank you
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