How would one go about approximating the value of
$$\sum^{\infty}_{n=0}\frac{d^n\left(J_0(2i\sqrt x)\right)}{dx^n},$$
where $J_0$ is a Bessel function of the first kind, order zero. I expect some numerical approximation would give a result, but maybe there is a more efficient approach to it?
I shall derive an asymptotic formula for large positive $x$. Using the hint in the comments, we obtain $$ \sum\limits_{n = 0}^\infty {\frac{{{\rm d}^n J_0 (2{\rm i}\sqrt x )}}{{{\rm d}x^n }}} = \sum\limits_{n = 0}^\infty {\frac{{{\rm d}^n I_0 (2\sqrt x )}}{{{\rm d}x^n }}} = \int_0^{ + \infty } {I_0 (2\sqrt {x + t} ){\rm e}^{ - t} {\rm d}t} $$ for $x>0$. The substitution $t=xs$ yields $$ \sum\limits_{n = 0}^\infty {\frac{{{\rm d}^n J_0 (2{\rm i}\sqrt x )}}{{{\rm d}x^n }}} = x\int_0^{ + \infty } {I_0 (2\sqrt{x}\sqrt {1 + s} ){\rm e}^{ - xs} {\rm d}s} . $$ Now we apply the asymptotics $$ I_0 (z) = \frac{{{\rm e}^z }}{{\sqrt {2\pi z} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right), \quad z\to+\infty, $$ to derive \begin{align*} \int_0^{ + \infty } {I_0 (2\sqrt{x}\sqrt {1 + s} ){\rm e}^{ - xs} {\rm d}s} =\;& \frac{{x^{3/4} {\rm e}^{2\sqrt x } }}{{2\sqrt \pi }}\left( \int_0^{ + \infty } {{\rm e}^{ - xs + 2\sqrt{x}(\sqrt {1 + s} - 1)} \frac{{{\rm d}s}}{{(1 + s)^{1/4} }}}\right. \\ &+\left. \mathcal{O}\!\left( {\frac{1}{{x^{1/2} }}} \right)\int_0^{ + \infty } {{\rm e}^{ - xs + 2\sqrt{x}(\sqrt {1 + s} - 1)} \frac{{{\rm d}s}}{{(1 + s)^{3/4} }}} \right) \end{align*} as $x\to+\infty$. By Theorem $2.1$ of Chapter $9$ in F. W. J. Olver's book Asymptotics and Special Functions, we find $$ \int_0^{ + \infty } {{\rm e}^{ - xs + 2\sqrt{x}(\sqrt {1 + s} - 1)} \frac{{{\rm d}s}}{{(1 + s)^{k/4} }}} = \frac{1}{x} + \mathcal{O}\!\left( {\frac{1}{{x^{3/2} }}} \right),\quad k = 1,3 $$ as $x\to+\infty$. Therefore $$ \sum\limits_{n = 0}^\infty {\frac{{{\rm d}^n J_0 (2{\rm i}\sqrt x )}}{{{\rm d}x^n }}} = \frac{{{\rm e}^{2\sqrt x } }}{{2\sqrt \pi x^{1/4} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^{1/2} }}} \right)} \right) $$ as $x\to+\infty$. Using more terms from the asymptotic expansion of the modified Bessel function and computing a more precise error term in Olver's theorem, the asymptotic result can be improved. Alternatively, we can repeatedly differentiate the asymptotic expansion of the modified Bessel function and formally sum the result. With either of these methods, it is found that $$ \sum\limits_{n = 0}^\infty {\frac{{{\rm d}^n J_0 (2{\rm i}\sqrt x )}}{{{\rm d}x^n }}} \sim \frac{{{\rm e}^{2\sqrt x } }}{{2\sqrt \pi x^{1/4} }}\left( 1 + \frac{{17}}{{16x^{1/2} }} + \frac{{425}}{{512x}} + \frac{{347}}{{8192x^{3/2} }}+\ldots\right) $$ as $x\to+\infty$.
Addendum. If we employ the power series $$ I_0 (z) = \sum\limits_{k = 0}^\infty {\frac{{\left( {\frac{1}{4}z^2 } \right)^k }}{{k!^2 }}} $$ in the integral representation, we obtain, after term-wise integration, $$ \sum\limits_{n = 0}^\infty {\frac{{{\rm d}^n J_0 (2{\rm i}\sqrt x )}}{{{\rm d}x^n }}} = {\rm e}^{x}\sum\limits_{k = 0}^\infty {\frac{{Q(k + 1,x)}}{{k!}}} = \sum\limits_{k = 0}^\infty {\frac{{e_k (x)}}{{k!}}}= \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{j = k}^\infty {\frac{1}{{j!}}} } \right)\!\frac{{x^k }}{{k!}}}. $$ Here $Q$ is the normalised incomplete gamma function and $e_k(x)$ is the $k$th partial sum of the exponential power series. This expansion is valid for all complex $x$.