= Derivative of an Abel integral
{title2=$\frac{d}{dx}\int_0^x\frac{h(t)}{\sqrt{x-t}}dt=\frac{h(0)}{\sqrt x}+\int_0^x\frac{h'(t)}{\sqrt{x-t}}dt$}
Suppose $h$ is continuously <differentiable> on $[0,L]$. For $0<x\leq L$, the <improper integral> $g(x)=\int_0^xh(t)(x-t)^{-1/2}dt$ exists, but substituting directly into the ordinary <Leibniz integral rule> produces an undefined endpoint term. First use <integration by parts> to obtain
$$
g(x)=2h(0)\sqrt x+2\int_0^xh'(t)\sqrt{x-t}\,dt.
$$
The new endpoint term at $t=x$ vanishes, and the differentiated kernel is locally <integrable>. Differentiating gives the displayed formula. One rigorous justification rewrites the second term as $\int_0^x[\int_0^s h'(t)(s-t)^{-1/2}dt]ds$ by <Fubini's theorem>; the inner <integral> is continuous for $s>0$, so the <fundamental theorem of calculus> applies. Moreover $g(x)=2h(0)\sqrt x+O(x^{3/2})$: if $h(0)\ne0$, the right <difference quotient> at zero diverges rather than defining a finite <derivative>.
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