Differentiating a function with a factorial

Question

How can I differentiate the following function given that it has a factorial in the denominator.

$\frac{x^{2x-1}}{(2x-1)!}$

Answer 1

I would use logarithmic differentiation. Let $y=\dfrac{x^{2x-1}}{(2x-1)!}.$ Then \begin{align*} \ln(y)&=\ln\!\big(x^{2x-1}\big)-\ln((2x-1)!)\\ &=(2x-1)\ln(x)-\sum_{j=1}^{2x-1}\ln(j)\\ \frac{y'}{y}&=2\ln(x)+\frac{(2x-1)}{x}-2 \psi(2 x). \end{align*} Here $\psi$ is the digamma function. Then we simply multiply back to obtain the answer $$y'=\dfrac{x^{2x-1}}{(2x-1)!}\left[2\ln(x)+\frac{(2x-1)}{x}-2 \psi(2 x)\right].$$

Answer 2

In the same spirit as in Adrian Keister's answer, write $$y=\dfrac{x^{2x-1}}{(2x-1)!}=\frac{x^{2 x-1}}{\Gamma (2 x)}\implies \log(y)=(2x-1)\log(x)-\log(\Gamma (2 x))$$ $$\frac{y'}y=2\log(x)+\frac{2x-1}x-\frac{\Gamma' (2 x) }{\Gamma (2 x)}=2\log(x)+\frac{2x-1}x-\frac{2 \Gamma (2 x) \psi (2 x)}{\Gamma (2 x)}$$ Simplify and use $y'=y \times \frac{y'}y$