Prove derivative $1/x$ by induction I'm trying to solve prove
$g(x)=\frac{1}{x}$.
$g^{(n)}(x)=\frac{(-1)^n(n!)}{x^{n+1}}$.
So for proving it I tried
Base case:
$g'(x)=\frac{-1}{x^2}=\frac{(-1)^1(1!)}{x^{1+1}}=\frac{-1}{x^2}$.
Inductive step:
$$ g^{(n)}(x)=\frac{(-1)^n(n!)}{x^{n+1}}$$
$$ g^{(n+1)}(x)=\frac{(-1)^{(n+1)} (n+1)\color{blue}{!}}{x^{n+2}}$$
So the way I was thinking about it is,
$g^n(x)'=g^{n+1}(x)$
but I can't continue from here:
$$g^{(n)}(x)'=\frac{n(-1)^{n-1}(n!)}{x^{n+1}}+\frac{(-1)^n-n!}{x^{2n+2}}.$$
 A: You are taking a derivative with respect to $x$.  You seem to have use the product rule, but there are not two expressions containing $x$ multiplied here.
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} g^{(n)}(x) 
    &= \frac{\mathrm{d}}{\mathrm{d}x} \left(  (-1)^n \frac{n!}{x^{n+1}} \right)  \\
    &= \frac{\mathrm{d}}{\mathrm{d}x} \left(  (-1)^n n! x^{-(n+1)} \right)  \\
    &= (-1)^n n! (-(n+1)) x^{-(n+1) -1}  \\
    &= (-1)^{n+1} n! (n+1) x^{-(n+1) -1}  \\
    &= (-1)^{n+1} (n+1)! x^{-(n+1) -1}  \\
    &= (-1)^{n+1} \frac{(n+1)!}{x^{(n+1)+1}}  \\
    &= (-1)^{n+1} \frac{(n+1)!}{x^{n+2}}
\end{align*}
A: $$(g^{(n)})'=(-1)^{n}*n!*(x^{-(n+1)})'=(-1)^n*n!*(-(n+1)))(x^{-(n+2)}=(-1)^{(n+1)}*(n+1)!*x^{-(n+2)}$$
your derivative $$(g^{(n)})'$$ is wrong.
A: Hint:
It can be done in a much simpler way, if you know this elementary formula, parallel to $\;(x^n)'=nx^{n-1}$:
$$\biggl(\frac1{x^n}\biggr)'=-\frac n{x^{n+1}}.$$
Therefore
$$\Bigl(g^{(n)}(x)\Bigr)'=\biggl(\frac{(-1)^n n!}{x^{n+1}}\biggr)'=(-1)^n n!\biggl(\frac{1}{x^{n+1}}\biggr)'. $$
Can you proceed?
A: For the inductive step we have
$$g^{n+1}(x)=(-1)^{n+1} \frac{(n+1)!}{x^{n+2}}=-\frac{n+1}{x}g^n(x)$$
which is true indeed
$$(g^n(x))'=\left(-\frac{n}{x}g^{n-1}(x)\right)'=\frac{n}{x^2}g^{n-1}(x)-\frac{n}{x}g^{n}(x)=-\frac{1}{x}g^{n}(x)-\frac{n}{x}g^{n}(x)=-\frac{n+1}{x}g^n(x)$$
