Why is the regularized beta function differentiable?

Question

We have the definition of Regularized Incomplete Beta Function: $$I_x (a,b) = \frac{B(x;a,b)}{B(a,b)},$$ where $B(x;a,b)$ is the incomplete Beta function, $B(a,b)$ is the Beta function and $x \in [0,1], a,b > 0$.

Is the Regularized Incomplete Beta Function differentiable with respect to x ? If yes why ?

I think Regularized Incomplete Beta Function is differentiable since I found differentiation of Regularized Incomplete Beta Function on functions.wolfram.com website but I don't know why. Thanks in advanced.

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$$B(x;a,b) := \int_{0}^{x}{t^{a-1} (1-t)^{b-1} dt}$$ $$B(a,b) := \int_{0}^{1}{t^{a-1} (1-t)^{b-1} dt}.$$

Answer 1

By the fundamental theorem of calculus,$$I_x^\prime=\frac{B_x^\prime}{B(a,\,b)}=\frac{x^{a-1}(1-x)^{b-1}}{B(a,\,b)}.$$This at least ensures differentiability on $(0,\,1)$. As $x\to0^+$, $B_x\sim\int_0^xt^{a-1}dt=x^a/a$, so $B_0=0=\lim_{x\to0^+}B_x$. We can treat $x\to1^-$ similarly.