I would like to ask if it's possible to solve for the derivative of the error function of $x^3$.
$$\frac{d}{dx} \operatorname{erf}(x^3)$$
I only have little understanding about this topic but I know that $$ \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt$$
So, if I solve for $\operatorname{erf}(x^3)$ first, is it correct to just change the bounds from 0 to $x^3$ and then solve it from there?
On
With a simple change of integration variables $$ \operatorname{erf}(x^3 ) = \frac{2}{{\sqrt \pi }}\int_0^{x^3 } {e^{ - t^2 } dt}\; \mathop = \limits^{\; t = s^3 } \; \frac{6}{{\sqrt \pi }}\int_0^x {s^2 e^{ - s^6 } ds} . $$ Thus $$ \frac{d}{{dx}}\operatorname{erf}(x^3 ) = \frac{6}{{\sqrt \pi }}x^2 e^{ - x^6 }. $$
Error function is actually defined as the integral of the Gaussian function as $$ \text{erf}(x)\triangleq\frac{2}{\sqrt \pi}\int_0^x \exp(-t^2)dt, $$ hence you can find the derivative by using $$ \frac{d}{dx}f(g(x))=g'(x)f'(g(x)) $$