So, I have the following Brownian distribution that models the number of molecules arriving at some distance $d$. If we take the derivative of the given equation w.r.t to the time $t$ we obtain the time taken to propagate the molecules to the distance $d$. Is my differentiation correct?
$$N(t)= \frac{{VQ}}{(4\pi{Dt})^{3/2}} {\exp\left(-\frac{d^2}{4{Dt}}\right)}$$
if I take the derivative of the above equation and set it to $0$, then
$$\frac d{dt}N(t)=\frac d{dt}\left(\frac{{VQ}}{(4\pi{Dt})^{3/2}} {\exp\left(-\frac{d^2}{4{Dt}}\right)}\right) = 0$$
$$=\frac{{VQ}}{(4\pi{D})^{3/2}} \frac d{dt}\left(\frac{1}{{t}^{3/2}} {\exp\left(-\frac{d^2}{4{Dt}}\right)}\right) = 0$$ Next using product rule and transfering $VQ/(4\pi{D})^{3/2}$ on the RHS, $$= \frac d{dt}\left(\frac{1}{{t}^{3/2}} \right) {\exp\left(-\frac{d^2}{4{Dt}}\right)} + \left(\frac{1}{{t}^{3/2}} \right)\frac d{dt}{\exp\left(-\frac{d^2}{4{Dt}}\right)} = 0$$
$$= \left(\frac{-3}{{2t}^{1/2}} \right) {\exp\left(-\frac{d^2}{4{Dt}}\right)} + \left(\frac{1}{{t}^{3/2}} \right){\exp\left(-\frac{d^2}{4{Dt}}\right)}\left(\frac{-d^2}{4D}\right)\left(\frac{-1}{t^{1/2}} \right) = 0$$
After removing the common terms and transfering them to the RHS we are left with, $$ \frac{3}{2}- \frac{d^2}{4Dt^{3/2}} = 0$$ $$ \frac{3}{2} = \frac{d^2}{4Dt^{3/2}}$$ $$ t^{3/2} = \frac{d^2}{6D}$$
Is this correct? What bothers me is the power of ${3/2}$. I think that should not be there. Correct me if I am wrong.