Derivative of singular integrand

Question

I am trying to differentiate this integral with respect to $x$: $$T(x,t) = {1\over\sqrt\pi} \int_0^t {g(s) \over \sqrt{t-s} } e^{-{x^2\over 4(t-s)}} ds$$ According to this paper the derivative with respect to $x$ at $x=0$ is $T_x(0,t) = g(t)$. (The equation in the paper has extra terms, which I've simplified here.) However, if I pull the derivative through the integral $$ \partial_x T(x,t) = {1\over\sqrt\pi} \int_0^t {g(s) \over \sqrt{t-s} } e^{-{x^2\over 4(t-s)}} {-x\over 2(t-s)} ds $$ which would be zero at $x=0$. I suspect one cannot exchange the derivative with the integral here because of the singular integrand. So how do I verify that $T_x(0,t) = g(t)$? I should add that all variables are real and $g$ is well-behaved (continuous and bounded).