Take the following function:
$$\frac{dn}{d[\log(x)]} = a\exp{\frac{-(\log(x) - b)^2}{c}}$$
I'm interested in obtaining the form for $dn/dx$, so I take:
$$\frac{d[\log(x)]}{dx} = \frac{1}{x} \rightarrow d[\log(x)] = \frac{dx}{x}$$
I use this in the first Eq. to obtain:
$$\frac{dn}{d[\log(x)]} = \frac{dn}{dx} x$$
which means:
$$\frac{dn}{dx} = \frac{1}{x} a\exp{\frac{-(\log(x) - b)^2}{c}}$$
Is this correct? I feel like I'm missing something.
$\frac{dn}{dx} = \frac{1}{x} a\exp{\frac{-(\log(x) - b)^2}{c}}\ \ $ is correct.
But it doesn't give the solution of the equation $\frac{dn}{d[\log(x)]} = a\exp{\frac{-(\log(x) - b)^2}{c}}$.
Better, change of variable $X=\log(x)$ which leads to $dn = a\exp{\frac{-(X - b)^2}{c}}dX$.
The antiderivative is : $n=\frac{a\sqrt{\pi c}}{2}\text{erf}(\frac{X-b}{\sqrt c})+constant$ $$n=\frac{a\sqrt{\pi c}}{2}\text{erf}(\frac{e^x-b}{\sqrt c})+constant$$