Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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! – Hurkyl 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
up vote 15 down vote accepted

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.

share|cite|improve this answer
No doubt it is a brilliant answer.Thank you. – Argha Jan 2 '13 at 10:15
Loved the Alternate II part, well done! – Phonon 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.

share|cite|improve this 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.