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?

Note that
$$\frac x{x+1}=1-\frac1{x+1}=1-(x+1)^{-1}$$
So the $n$-th derivative is immediately seen to be
$$-(-1)(-2)\ldots(-n)(x+1)^{-n-1}=\frac{(-1)^{n+1}n!}{(x+1)^{n+1}}$$
So you see that both your answer and the solution is right, but $(-1)^{n+1}$ seems to fit nicer with the exponent of $(x+1)$.
