# Finding the $n$-th derivatives of $x^n \ln x$ and $\frac{\ln x}{x}$.

How can I prove the following identities:

$$\left( x^n \ln x \right)^{(n)}= n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}$$

$$\left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1} \left( \ln x - 1 - \frac{1}{2} - \cdots -\frac{1}{n} \right), \quad x>0,\quad n\ge 1, \tag{b}$$

Here $(\cdot )^{(n)}$ indicates the $n$-th derivative with respect to $x$.

I don't know where to start!

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Is induction on $n$ not an option? –  Jyrki Lahtonen Dec 25 '11 at 11:41
For example you could use induction. Call $I_n (x)$ the RHside of the first formula; you can write the following recursion formula for $I_{n+1}(x)$: $$\tag{1} I_{n+1}(x)=(n+1)\ I_n (x) +n!\; .$$ Now, call $J(n)$ the LHside of the same formula: in order to prove $J_n (x)=I_n (x)$ for each $n\in \mathbb{N}$, it suffices to prove that $J_1 (x)=I_1 (x)$ and that also $J_n(x)$ satisfies recursion (1). I didn't make the computations, but I think this technique should work... If it doesn't, you could try with different recursion formulae. –  Pacciu Dec 25 '11 at 12:01

I prefer using induction to prove the statement. For (a), when $n=1$, by product rule $$\left( x \ln x \right)'=\ln x+x\cdot\frac{1}{x}=1!(\ln x+1),$$ which shows that (a) is true for $n=1$. Now suppose that (a) is true for $n=k$, i.e. $$\left( x^k \ln x \right)^{(k)}= k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right).$$ Now, for $n=k+1$, we have $$\left( x^{k+1} \ln x \right)^{(k+1)}=\left[ \big(x^{k+1} \ln x\big)' \right]^{(k)}= \left[ (k+1)x^k \ln x+x^{k+1}\cdot\frac{1}{x}\right]^{(k)}$$ $$=(k+1)\left( x^k \ln x \right)^{(k)}+(x^k)^{(k)}.$$ Note that $(x^k)^{(k)}=k!$, by induction assumption, we have $$\left( x^{k+1} \ln x \right)^{(k+1)}=(k+1)k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right)+k!$$ $$=(k+1)! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k}+\frac{1}{k+1} \right),$$ which shows that (a) is true for $n=k+1$.

I will let you try (b) using induction.

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To finish off Yuval's argument, we try an inductive proof on

$$\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}=-\sum_{k=1}^n \frac1{k}=-H_n$$

The equality is easily verified for $n=1$. We now try to simplify

$$\sum_{k=1}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}$$

which goes a bit like this:

\begin{align*} &\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \left(\binom{n+1}{k}-\binom{n}{k}\right)\frac{(-1)^k}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\sum_{k=1}^n (-1)^k\binom{n+1}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+\sum_{k=0}^{n+1} (1)^{n-k+1}(-1)^k\binom{n+1}{k}\right)\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+(1-1)^{n+1}\right)=-\frac1{n+1}\\ \end{align*}

and since we also have $-H_{n+1}-(-H_n)=-\dfrac1{n+1}$, the equality is established.

Here's an explicit application of Leibniz's product rule for the second problem, which uses the identity $\dfrac{\mathrm d^n}{\mathrm dx^n}\dfrac1{x}=\dfrac{(-1)^n n!}{x^{n+1}}$:

\begin{align*} \frac{\mathrm d^n}{\mathrm dx^n}\frac{\ln\,x}{x}&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\binom{n}{k}\left(\frac{\mathrm d^k}{\mathrm dx^k}\ln\,x\right)\left(\frac{\mathrm d^{n-k}}{\mathrm dx^{n-k}}\frac1{x}\right)\\ &=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\frac{n!}{\color{blue}{k!}\color{magenta}{(n-k)!}}\left(\frac{\color{red}{(-1)^{k-1}}\color{blue}{(k-1)!}}{\color{orange}{x^k}}\right)\left(\frac{\color{red}{(-1)^{n-k}}\color{magenta}{(n-k)!}}{\color{orange}{x^{n-k+1}}}\right)\\ &=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\frac{\color{red}{(-1)^{n-1}} n!}{\color{orange}{x^{n+1}}}\sum_{k=1}^n\frac1{\color{blue} k}=\frac{(-1)^n n!}{x^{n+1}}\left(\ln\,x-\sum_{k=1}^n\frac1{k}\right) \end{align*}

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The familiar formula $(fg)' = f'g + fg'$ generalizes to the $n$th derivative in the following way: given a product of $m$ functions $f_1,\ldots,f_m$, the $n$th derivative is a sum of $m^n$ terms, each corresponding to a sequence of length $n$ listing which function is differentiated at every "step".

Write $x^n\ln x$ as a product of $n+1$ factors, $n$ of them equal to $x$ and the other one equal to $\ln x$. The $n$th derivative is a sum of terms, each corresponding to a sequence of length $n$. If any of the first $n$ functions appears twice, then the corresponding term is zero, since $x''=0$. So the general term lists $k$ different functions out of the first $n$, and the other one ($\ln x$) appears in the remaining $n-k$ positions. Call that a term of type $T_k$.

The term $T_n$ appears $n!$ times, and contributes $n!\ln x$ to the total sum. For $k < n$, the term $T_k$ appears $n(n-1)\cdots(n-k+1) \binom{n}{k}$ times. It is easy to prove by induction that for $m > 0$, $(\ln x)^{(m)} = (-1)^{m+1} (m-1)! x^{-m}$. Substituting $m = n-k$, we see that each appearance of $T_k$ contributes $(-1)^{n-k+1} (n-k-1)!$ to the sum. So all terms $T_k$ contribute $(-1)^{n-k+1} n!/(n-k) \binom{n}{k}$. Therefore the $n$th derivative is \begin{align*} &n!\ln x + n!\sum_{k=0}^{n-1} (-1)^{n-k+1} \frac{1}{n-k} \binom{n}{k} \\ =&n!\ln x + n!\sum_{k=1}^n (-1)^{k+1} \frac{1}{k} \binom{n}{k}. \end{align*} Now all that remains is to prove the identity $$\sum_{k=1}^n (-1)^{k+1} \frac{1}{k} \binom{n}{k} = \sum_{k=1}^n \frac{1}{k}.$$

That should prove you with a good start on the first one. I trust that the second one is very similar (perhaps after substituting $y = 1/x$).

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