differentiation of $\operatorname{erfc}(\sqrt{ax})$

Question

I need your help to figure out the derivative of $\operatorname{erfc}(\sqrt{ax})$ with respect to $x$. Based on my knowledge on Wolfram references, they cite that: $$\frac{d \operatorname{erfc}(z)}{dz}=-\frac{2\exp(-z^2)}{\sqrt{\pi}}$$ I've done an analogy so I made change of variable and I assigned: $z=\sqrt{ax}$ then $dx=\frac{2z\,dz}{a}$ then I progressed on the calculus. Is my reasoning right? Please give me your proposals.

Answer 1

$$ \mathrm{erfc}(x) = 1-\mathrm{erf}(x)= \frac{2}{\sqrt\pi}\int_x^{\infty}\mathrm e^{-t^2}dt. $$ Using the Leibniz's rule for differentiation under the integral sign we have $$ \Big(\mathrm{erfc}(\sqrt{ax})\Big)'=-\frac{2}{\sqrt\pi}\mathrm e^{-(\sqrt{ax})^2}\Big(\sqrt{ax}\Big)'=-\frac{2}{\sqrt\pi}\mathrm e^{-ax}\frac{\sqrt{a}}{2\sqrt{x}}=-\Big(\frac{a}{\pi x}\Big)^{1/2}\mathrm e^{-ax} $$

Answer 2

Just use the chain rule: $$ \frac{d}{dx}f(g(x))=f'(g(x))g'(x) $$ In your case $f(z)=\operatorname{erfc}(z)$ and $g(x)=\sqrt{ax}$, so $f'(z)=-\frac{2}{\sqrt{\pi}}\exp(-z^2)$ and $g'(x)=\frac{a}{2\sqrt{ax}}$. Hence $$ \operatorname{erfc}'(\sqrt{ax})=-\frac{2}{\sqrt{\pi}}\exp(-ax)\frac{a}{2\sqrt{ax}} $$

Answer 3

We do not know your results, so we cannot say whether your reasoning is right or wrong, but: In the context of derivatives, your wording is unusual. Normal lingo is to say: We apply the https://en.wikipedia.org/wiki/Chain_rule, i.e. with $$f(x)=\mathrm{erfc}(x), \; g(x)=\sqrt{ax}, \, F(x) = f(g(x))$$ you compute $$F'(x)= f'(g(x))g'(x)=\left(-\frac{2}{\sqrt{\pi}}e^{-(\sqrt{ax})^2}\right)\left(\frac{1}{2 \sqrt{ax}}\right)a$$ $$\frac{d}{dx} \mathrm{erfc}(\sqrt{ax})=-\sqrt{\frac{a}{\pi x}}\;e^{-ax}$$