Finding the derivative Maxwell Boltmann distribution

Question

I need to basically show that the slope is $0$ when $\epsilon = kT$

$f(\epsilon)=\left(\frac{8 \pi}{m}\right)\left(\frac{m}{2 \pi kT}\right)^{\frac{3}{2}}\epsilon e^{\frac{-\epsilon}{kT}}$

So I broke the function up into two parts and used the product rule:

$f(\epsilon)=\left(\frac{8 \pi}{m}\right)\left(\frac{m}{2 \pi kT}\right)^{\frac{3}{2}}\epsilon$

$f'(\epsilon)=\left(\frac{8 \pi}{m}\right)\left(\frac{m}{2 \pi kT}\right)^{\frac{3}{2}}$

$g(\epsilon)=e^{\frac{-\epsilon}{kT}}$

$g'(\epsilon)=e^{\frac{-\epsilon}{kT}}\left(\frac{-kT+\epsilon}{\left(kT\right)^2}\right)$

Putting it together and solving:

$fg'+gf'$

I got:

$f'(\epsilon)=e^\frac{-\epsilon}{kT}\left(\frac{8 \pi}{m}\right)\left(\frac{m}{2 \pi kT}\right)^\frac{3}{2}\left[\epsilon\left(\frac{-kT+\epsilon}{\left(kT\right)^2}\right)+1\right]$

But plugging in $f'(kT)$ does not give me $0$.

Do you guys see where I did anything wrong?

Answer 1

$$g'(\epsilon)=-\dfrac{1}{kT}g(\epsilon)$$

Answer 2

$$\ \frac{d}{d\epsilon}\bigg[\bigg(\frac{8\pi}{m}\bigg)\bigg(\frac{m}{2\pi kT}\bigg)^{\frac{3}{2}}\epsilon e^{\frac{-\epsilon}{kT}}\bigg]=\bigg(\frac{8\pi}{m}\bigg)\bigg(\frac{m}{2\pi kT}\bigg)^{\frac{3}{2}}\frac{d}{d\epsilon}\bigg[\epsilon e^{\frac{-\epsilon}{kT}}\bigg]=$$ $$\ =\bigg(\frac{8\pi}{m}\bigg)\bigg(\frac{m}{2\pi kT}\bigg)^{\frac{3}{2}}e^{\frac{-\epsilon}{kT}}\bigg[1-\frac{\epsilon}{kT}\bigg]$$