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> Find the Var$(X)$ given that $$m\_X(u)=\Big(\frac{p}{1-(1-p)e^u)}\Big)^r \ \ \ \ \ \ \ u<\text{ln}((1-p)^{-1})$$
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I have found $\mathbb{E}(X)$ to be
$$\mathbb{E}(X)=m'\_X(u)$$
$$\mathbb{E}(X)=r\Bigg(\frac{p}{(1-(1-p)e^u)}\Bigg)^{r-1}\times\frac{p(e^u-pe^u)}{(1-(1-p)e^u)^2}$$
substituting $u=0$
$$\mathbb{E}(X)=r\Bigg(\frac{p}{p}\Bigg)^{r-1}\times\frac{p(1-p)}{p^2}=\frac{r(1-p)}{p}$$

My question is, how do I find $\mathbb{E}(X^2)$? Differentiating $\mathbb{E}(X)$ again seems like it's impossible.