Calculating the nth derivative of $\frac{x}{x+1}$

Question

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?

Answer 1

Hint: $\dfrac{x}{x+1}=1-\dfrac1{x+1}$

In other words, your solution looks correct. Note that $(-1)^{n-1}=(-1)^{n+1}$

Answer 2

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 $(x+1)$.

Answer 3

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! ;)