User:Henry kong/Sandbox

From Weierstrass Form:
 * $\ds \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left[{\left({1 + \frac z n}\right] e^{-z / n} }\right)$

We can take the reciprocal of the both side and obtain:
 * $\ds \map \Gamma z = \frac {e^{-\gamma z}} z \prod_{n \mathop = 1}^\infty \frac{e^{\frac z n}}{1 + \frac z n}$

Take the derivative of both side:

Divide both side by $\map \Gamma z$: