# Higher-order derivatives of $(e^x + e^{-x})^{-1}$

I am currently trying to build the derivatives of $$f(x) = \frac{1}{e^x+e^{-x}}.$$ It is fairly straightforward to obtain $$\frac{d^n f}{dx^n} = \frac{P_n(e^x)}{e^{(n-1)\cdot x} (e^x+e^{-x})^{n+1}},$$ where $$P_n(x)$$ is given by the recursive relationship $$P_0(x) = 1$$ and $$P_{n+1}(x) = P_n'(x) \cdot x \cdot (x^2+1) - P_n(x)((2\cdot n +1)\cdot x^2-1).$$ If we represent $$P_n(x)$$ by $$\sum_{i = 0}^n a^{(n)}_{2i} x^{2i}$$, then we have $$a_0^{(n)} = 1$$ for all $$n$$, $$a_{2k}^{(n)} = 0$$ for all $$k > n$$, and $$a_{2i}^{(n+1)} = (2i+1) \cdot a_{2i}^{(n)} - (2(n-i)+3) \cdot a_{2 \cdot (i-1)}^{(n)}$$ for all $$n \geq 0$$ and $$0 < i \leq n+1$$. So we can obtain with this relationship, that $$a_{2n}^{(n)} = (-1)^{n}$$ and that $$a_2^{(n)} = -3^n +k +1$$. However, I am wondering whether we can say something about the maximum of $$|P_n(e^x)/e^{2n}|$$ for each $$n$$ or the maximum of $$\max_{0 \leq i \leq n} |a_{2i}^{(n)}|$$ over each $$n$$.

Edit: It seems that $$\sum_{i=0}^n |a_{2i}^{(n)}| = n! \cdot 2^n$$.

• I think it has direct relation with this Oct 4, 2020 at 9:21
• Notice that $$f(Ix) = \frac{\sec(x)}2.$$ Oct 5, 2020 at 3:38
• Other formulas for derivatives are given at functions.wolfram.com/ElementaryFunctions/Sec/20/02 Oct 6, 2020 at 4:24

Using the tried-and-true method of calculating small examples and plugging them into the OEIS, one finds that the $$P_n(x)$$ are, up to sign, known as MacMahon polynomials, and their coefficients are given by Eulerian numbers of type B. The OEIS also has a separate entry for the maximal coefficients that you are asking about, although it doesn't list a formula for the asymptotic growth. But there is a long list of references that will hopefully be helpful.