n-th derivative of $\frac{\ln x}{x}$. 
Let $f(x)=\frac{\ln x}{x},x>0$. Show that $$f^{(n)}(1)=(-1)^{n+1}(n!)(1+\frac{1}{2}+\cdot+\frac{1}{n})$$

Trial: n-th derivative of $\ln x$ is $$(-1)^{n-1}(n-1)! x^{-n}$$ and n-th derivative of $\frac{1}{x}$ is $$(-1)^n n! x^{-(n+1)}$$Then If I use Leibnitz's Theorem I need to face a big calculation.Please help.
 A: Using the product rule with $g(x)=1/x$ and $f(x)=\ln(x)$, we have
$$ f^{(n)}(x)=\sum_{i=0}^{n} {n\choose i} g^{(i)}(x)h^{(n-i)}(x)$$
$$= (-1)^n n!\, x^{-(n+1)}\ln(x) +\sum_{i=0}^{n-1} {n\choose i}(-1)^i i!\, x^{-i-1}(-1)^{n-i-1}(n-i-1)!\,x^{-n+i} $$
$$=(-1)^n n!\, x^{-(n+1)}\ln(x) + (-1)^{n-1}x^{-n-1}\sum_{i=0}^{n-1} {n\choose i}i!(n-i-1)!  $$
$$ =(-1)^n n!\, x^{-(n+1)}\ln(x) + (-1)^{n-1}x^{-n-1}\sum_{i=0}^{n-1} \frac{n!}{i!(n-i)!}i!(n-i-1)! $$
$$\implies f^{(n)}(x) = (-1)^n n!\, x^{-(n+1)}\ln(x) +(-1)^{n-1}n!\,x^{-n-1}\sum_{i=0}^{n-1} \frac{1}{(n-i)} .$$
Now, you can substitute $x=1$ to get the desired answer.
A: Let $g(x)=f(x+1)$ so that $g^{(n)}(0)=f^{(n)}(1)$.
Then since $\log(1+x) = \int dx/(1+x) =\sum_{n=0}^\infty {(-1)^n x^{n+1}/ (n+1)}$
$$g(x)={\log(1+x)\over1+x}=\sum_{n=0}^\infty(-1)^nx^n \sum_{n=0}^\infty {(-1)^n x^{n+1} \over n+1}=\sum_{n=0}^\infty(-1)^nx^n \sum_{m=1}^\infty {(-1)^{m-1} x^{m} \over m} =
\sum_{n=1}^\infty\left(\sum_{k=1}^n{(-1)^{n-1} \over k}\right) x^n$$
and apply Taylor's formula (as well as the identity $(-1)^{n-1}=(-1)^{n+1}$).
Alternate I
Let $I(x) = \int f(1+x) \; dx = {1\over2}(\log(1+x))^2$, so that $I^{(n+1)}(0)=f^{(n)}(1)$.
Then
$$I(x) = {1\over2}\left(\sum_{m=1}^\infty {(-1)^{m-1} x^{m} \over m}\right)^2
=\sum_{m=2}^\infty\left(\sum_{k=1}^m{(-1)^{m-2}\over2k(m-k)}\right)x^n
=\sum_{m=2}^\infty\left(\sum_{k=1}^m{(-1)^{m-2}\over2m}\left({1\over k}+{1\over m-k}\right)\right)x^m\,,$$
which, since $k$ and $m-k$ each run over the integers $1,2,\dots,m$ in the inside sum, is equal to
$$=\sum_{m=2}^\infty\left(\sum_{k=1}^m{(-1)^{m-2}\over mk}\right)x^m
=\sum_{n=1}^\infty\left(\sum_{k=1}^{n+1}{(-1)^{n-1}\over(n+1)k}\right)x^{n+1}$$
and again apply Taylor's formula.
Alternate II
It looks so much like an impossible induction problem, I could not resist trying...
Let $y=f(x)=(\log(x)/x)$, so that the derivative $D(xy) = x\,y'+y = D(\log x) = 1/x$.  By induction then, we have that the $n^{th}$ derivative is given by
$D^n(xy) = x\,y^{(n)}+n\,y^{(n-1)} = (-1)^{n-1} (n-1)!/x^n$, so that at $x=1$, we have
the equation
$$y^{(n)}+n\,y^{(n-1)} = (-1)^{n-1} (n-1)!\,.$$
Now we're ready to prove the formula for $f^{(n)}(1)$. That $f'(1)=1$ is easily verified.  Assume that at $x=1$, the $n^{th}$ derivative $y^{(n)}=f^{(n)}(1)$ satisfies
$$y^{(n)}=(-1)^{n-1}(n!)(1+\frac{1}{2}+\cdots+\frac{1}{n})\,.$$
Then applying the equation above for the $n+1^{st}$ derivative, we have
$$y^{(n+1)}=(-1)^{n} n! -(n+1)y^{(n)}
={(-1)^{n} (n+1)!\over n+1}-(n+1){(-1)^{n-1}(n!)(1+\frac{1}{2}+\cdots+\frac{1}{n})}
={(-1)^{n}(n+1)!(1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1})}$$
So by induction the formula for the derivative at $x=1$ is established.
A: A generalized product rule (Leibniz rule) for $n$th derivatives states that the $n$th derivative of a product of two $n$-times differentiable functions $f$ and $g$ is given by  $$(fg)^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x).$$
Using this, and the calculations you've already made for $\ln x$ and $1/x$, you should be able to get the desired result.
