I have a hypergeometric function with two time dependent arguments
${}_2F_1(a(t),b(t),\frac{1}{2},x)$
and I want to take the derivative with respect to $t$. To be specific, I have
$a(t) = \frac{-\sqrt{1-\frac{4it}{v_0}}-1}{4}$
$b(t) = \frac{\sqrt{1-\frac{4it}{v_0}}-1}{4}$
($i$ is the imaginary unit, $v_0 = constant$, and after taking the derivative I will set $t=0$. Is it possible to do this and find a sensible, analytic answer? Thanks!
We have $a(0) = -1/2$ and $b(0) = 0$. There is a simple closed form because we need to take the derivative of ${_2 F_1}$ wrt a parameter only at zero, where termwise differentiation gives $$\lim_{b \to 0} \frac {(b)_k - (0)_k} b = \cases {\Gamma(k) & $k > 0$ \\ 0 & $k = 0$},\\ \frac d {db} \, {_2 F_1}(a, b; c; x) \bigg\rvert_{b = 0} = \frac {a x} c \, {_3 F_2}(1, 1, a + 1; 2, c + 1; x).$$ The other partial derivative disappears because ${_2 F_1}(a, 0; c; x) = 1$. Therefore $$\frac d {dt} \, {_2 F_1}(a(t), b(t); 1/2; x) \bigg\rvert_{t = 0} = -b'(0) x \, {_3 F_2}(1, 1, 1/2; 2, 3/2; x) = \\ \frac i {2 v_0} (\ln(1 - x) + \sqrt x \,(\ln(1 + \sqrt x) - \ln(1 - \sqrt x))).$$