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I think I am doing it the correct way but I am not sure.

is it $$(-1)^{n+1}n!(1+x)^{-n} ?$$

thank you guys.

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It should be $ (-1)^{n-1}(n-1)!(1+x)^{-n}\, \forall n\geq 1 $. – Mhenni Benghorbal Mar 18 '13 at 14:54
may be, you should show us how you found that so we can help you. When the derivative of your expression for n it doesn't gives the expression for n+1. So it must be wrong ... – wece Mar 18 '13 at 14:58
problem solved . thanks for the help guys – nicolas Mar 18 '13 at 15:07
up vote 3 down vote accepted

This is how I would do it

$$f(x) = \ln(1+x)$$ $$f'(x) = \frac{1}{x+1}$$ $$f''(x) = -\frac{1}{(x+1)^2}$$ $$f'''(x) = (-1)(-2)\frac{1}{(x+1)^3}$$ $$f'^v(x) = (-1)(-2)(-3)\frac{1}{(x+1)^4}$$ $$f'^v(x) = (-1)^3\cdot \frac{3!}{(x+1)^4}$$ $$.....................$$ $$f'^n(x) = (-1)^{n-1}\cdot \frac{(n-1)!}{(x+1)^{n}}\forall n\ge 1$$

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