I need to find the velocity at which the fraction is maximum in Maxwell–Boltzmann distribution
$$P(v) = \left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}4\pi v^2 \exp{\left(-\frac{mv^2}{2kT}\right)}$$
I know that occurs when $\mathrm dP(v)/\mathrm dv = 0$, but can anyone show the steps which lead to the answer?
I don't think it's on topic, but it's no skin off my back to write a short answer. You need to use the chain rule $(fg)' = fg' + gf'$:
$$\begin{align} \frac{\mathrm dP}{\mathrm dv} &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\frac{\mathrm d}{\mathrm dv}(\mathrm e^{-mv^2/2kT}) + \mathrm e^{-mv^2/2kT}\frac{\mathrm d}{\mathrm dv}(v^2)\right] \\ &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\left(\frac{-2mv}{2kT}\mathrm e^{-mv^2/2kT}\right) + \mathrm e^{-mv^2/2kT}(2v)\right] \\ &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi\mathrm e^{-mv^2/2kT}\left(\frac{-mv^3}{kT} + 2v\right) = 0 \end{align}$$
Since an exponential cannot be zero, the polynomial factor must be equal to zero:
$$\begin{align} \frac{-mv^3}{kT} + 2v &= 0 \\ v\left(2 - \frac{mv^2}{kT}\right) &= 0 \\ v &= 0, \pm \sqrt{\frac{2kT}{m}} \end{align}$$
You're not interested in the $v = 0$ stationary point, and the negative square root is unphysical (we're talking about a distribution of speeds, which must be non-negative). So the maximum occurs at $v = \sqrt{2kT/m}$.
From a chemist's point of view this is the physically important quantity, as it represents the most probable speed of a gas molecule. You could substitute it back into $P(v)$ and find out the value of $P$ for this value of $v$, but I don't know what circumstances this would be of interest in.
