Find general formula for $n$-th derivative of $y = f(\ln x)$.
To start with I found couple of derrivatives:
\begin{align} y' &={1 \over x}f'(\ln x) \\ y'' &={1 \over x^2}(f''(\ln x)-f'(\ln x)) \\ y''' &={1 \over x^3}(f'''(\ln x)-3f''(\ln x)+2f'(\ln x)) \\ y'''' &={1 \over x^4}(f''''(\ln x)-6f'''(\ln x)+11f''(\ln x)-6f'(\ln x)) \end{align} Thought it would be easy to notice a certain pattern and prove it by induction. But I can not observe anything useful.
Maybe there is something like $n \choose k$ in general formula, if it exsists.
I need a wise hint.

  • $\begingroup$ the coefficients of $f^{(k)}$ of $y^{(n)}$ are called the stirling numbers of the first kind, $s(n,k)$ (see riordan's combinatory analysis page 48 table 1.) $\endgroup$
    – abel
    Nov 28, 2015 at 14:27

2 Answers 2


You might find Faà di Bruno's formula useful with $g(x)= \ln (x)$. Since

$$g^{(j)}(x)=(-1)^{j+1}\frac{(j-1)!}{x^j} $$ we get

$$ {d^n \over dx^n} f(\ln x)=\sum \frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}\cdot f^{(m_1+\cdots+m_n)}(\ln x)\cdot \prod_{j=1}^n\left((-1)^{j+1}\frac{(j-1)!}{x^j}\right)^{m_j}$$

where the sum is taken over the constraint

$$1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n. $$

  • $\begingroup$ Is there any way to compute this by generating functions, combinatoric argument, etc? $\endgroup$
    – Gyumin Roh
    Nov 28, 2015 at 13:46
  • $\begingroup$ Very elegant view. +1 $\endgroup$
    – Mikasa
    Nov 30, 2015 at 3:40
  • $\begingroup$ @GyuminRoh I don't know unfortunately. $\endgroup$
    – user1337
    Dec 5, 2015 at 11:50


We have $$y^{(n)} = \frac{1}{x^n} \sum_{i=1}^n a_{(n,i)} f^{(i)}(\ln x)$$ Where $a_{(n,i)}$ is the coefficient of $x^{i-1}$ in $$\prod_{k=1}^{n-1} (x-k)$$

I will skip the base since it is given in the problem.

Assume that this is true for $n=k \ge 4$. I will prove it for $n=k+1$.

From the induction hypothesis, we have $$y^{(k)} = \frac{1}{x^k} \sum_{i=1}^k a_{(k,i)} f^{(i)}(\ln x)$$

Differentiating using $(uv)'=u'v+uv'$, we have $$y^{(k+1)} = -\frac{k}{x^{k+1}} \sum_{i=1}^k a_{(k,i)} f^{(i)}(\ln x) + \frac{1}{x^k} \sum_{i=1}^k a_{(k,i)} f^{(i+1)}(\ln x) \cdot \frac{1}{x}$$

Now rearranging, we have $$y^{(k+1)} = \frac{1}{x^{k+1}} \sum_{i=1}^{k+1} (-ka_{(k,i)}+a_{(k,i-1)})f^{(i)}(\ln x)$$

Therefore, it suffices to prove that $$-ka_{(k,i)}+a_{(k,i-1)} = a_{(k+1,i)}$$

However, this is trivial upon closer inspection, since $$(\cdots + a_{(k+1,i)}x^{i-1} + \cdots) = \prod_{j=1}^k (x-j) = \prod_{j=1}^{k-1} (x-j) \cdot (x-k)= (\cdots + a_{(k,i)}x^{i-1} + a_{(k,i-1)}x^{i-2} + \cdots )(x-k) = (\cdots -ka_{(k,i)}x^{i-1}+a_{(k,i-1)}x^{i-1} + \cdots)$$

We are done. $\blacksquare$

The motivation? I looked at the coefficients and the polynomial just came to me.

  • $\begingroup$ Fantastic. How did you come up with that? $\endgroup$
    – Arty
    Nov 28, 2015 at 13:21
  • 1
    $\begingroup$ Added a full proof via Induction. Darn this is definitely the coolest problem I solved today. $\endgroup$
    – Gyumin Roh
    Nov 28, 2015 at 13:39

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