Calculating the nth derivative of $\frac{x}{x+1}$ I was asked to calculate the nth derivative of $f(x) =\frac{x}{x+1}$.
My solution:
$$ f'(x) = (x+1)^{-2}$$
$$f''(x) = (-2)(x+1)^{-3}$$
$$f'''(x) = (-2)(-3)(x+1)^{-4}$$
$$f^{n}(x) = n!(x+1)^{-(n+1)} . (-1)^{n+1}$$
I was confident that I got the right answer, but when I check the memo, it said the answer was the same as mine except for the $(-1)$ part. It said $(-1)^{n-1}$ instead of what I said. My question is is it still right anyways?

Of course it's right, you only have to realize that
$$(-1)^{n+1}=(-1)^{n-1+2}=(-1)^{n-1} (-1)^2 = (-1)^{n-1}$$
As a suggestion, you could try to prove your expression for $f^{(n)}(x)$ in a more rigorous way using induction, if you haven't done that so far! ;)
