# 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.

• The idea implicit in the answers is to use Taylor series for your function to find the derivatives of the function. Don't make the mistake of thinking you can only go the other way! – user14972 Jan 2 '13 at 7:00
• Nice observation @Hurkyl. I hadn't realized that the generalized product rule for $n$th derivatives could be thought of that way. – Ted Jan 3 '13 at 4:20

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.

• No doubt it is a brilliant answer.Thank you. – Argha Jan 2 '13 at 10:15
• Loved the Alternate II part, well done! – Ellie Aug 4 '15 at 13:11

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.

• Thanks for the calculation. It is helpful. – Argha Jan 2 '13 at 10:15

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.