How is the derivative of differential equation containing error function?

Question

Consider, $$c(x,t)= \exp\big(\frac{-x^2}{4Dt}\big)-\frac{x\pi^{1/2}}{2\sqrt{Dt}}\operatorname{erfc}\frac{x}{2\sqrt{Dt}}$$ For calculating $$\left|\frac{\partial c}{\partial t}\right|_{x=0}$$ how is the derivative of the above equation involving $\operatorname{erfc}$?

Answer 1

Since $$\frac d {du}\text{erfc}(u)=-\frac{2 e^{-u^2}}{\sqrt{\pi }}$$ just apply the chain rule to get $$\frac d {dt}c(x,t)=\frac{\sqrt{\pi } D x }{4 (D t)^{3/2}}\text{erfc}\left(-\frac{x}{2 \sqrt{D t}}\right)+\frac{x^2 }{2 D t^2}e^{-\frac{x^2}{4 D t}}$$ Now, what happens when $x\to0$ ?

Edit

Just added for your curiosity : for small values of $x$, using Taylor you would get $$\frac{\partial c(x,t)}{\partial t}=\frac{\sqrt{\pi } D x}{4 (D t)^{3/2}}+\frac{3 x^2}{4 D t^2}+O\left(x^4\right)$$