# Finding $\frac{d^{n-1}}{dx^{n-1}}(1+\sqrt x)^{2n}|_{x=1}$

How can we find $$\frac{d^{n-1}}{dx^{n-1}}(1+\sqrt x)^{2n}\Big|_{x=1},$$where $$n\in\mathbb{N}$$?

Attempt of complex integration
\begin{aligned}&\frac{d^{n-1}}{dx^{n-1}}(1+\sqrt x)^{2n}\Big|_{x=1}\\ &=\frac{(n-1)!}{2\pi i}\int_C\frac{(1+\sqrt z)^{2n}}{(z-1)^n}dz\text{ (Cauchy's formula)}\\ &=\frac{(n-1)!}{\pi i}\int_{C'}\frac{z(1+z)^{2n}}{(z^2-1)^n}dz\text{ (Substitute z\to z^2)}\\ &=\frac{(n-1)!}{\pi i}\int_{C'}\frac{z(1+z)^{n}}{(z-1)^n}dz\\ &=\frac{(n-1)!}{\pi i}\int_{C''}-\frac{2 (w+1) w^n}{(w-1)^3}dw\text{ (Substitute w=\frac{1+z}{z-1})}\\ &=\frac{(n-1)!}{\pi i}2\pi i(-2n^2)=-4n\cdot n!\end{aligned} But the correct answer is $$4n\cdot n!$$, I don't know where the mistake is.
Also, I wonder if there is a solution without using complex analysis.

• The substitution $z\to z^2$ could be dangerous... – Szeto Dec 10 '18 at 9:07

The transformation from $$z$$ to $$w$$ reverses the orientation of the closed curve. Therefore, you need an extra minus sign, leading to the correct result.
Alternatively, let $$u:=z-1$$, so you will not need to worry about the orientation of the curve, and the required result is $$\frac{(n-1)!}{\pi \text{i}}\,\oint_{C'''}\,\frac{(u+1)(u+2)^n}{u^n}\,\text{d}u$$ for some positively oriented closed curve $$C'''$$ about the point $$u=0$$. Since $$(u+1)(u+2)^n=u^{n+1}+2n\,u^{n}+2n^2\,u^{n-1}+\mathcal{O}(u^{n-2})\,,$$ we get the required result.