Answer
$$kb^{\frac{1-m}{m}} (b-a)$$
Work Step by Step
Given $$M(t)=\left(a+(b-a)\left(1+k m t+\frac{1}{2}(k m t)^{2}\right)\right)^{1 / m}$$
Since
\begin{align*}
M'(t)&= \frac{1}{m} \left(a+(b-a)\left(1+k m t+\frac{1}{2}(k m t)^{2}\right)\right)^{(1 / m)-1} \left[km(b-a) +k^2m^2(b-a) t \right]\\
&= \left(a+(b-a)\left(1+k m t+\frac{1}{2}(k m t)^{2}\right)\right)^{(1 / m)-1} \left[k(b-a) +k^2m(b-a) t \right]
\end{align*}
Then
\begin{align*}
M'(0)&= \left(a+(b-a)\left(1+k m (0)+\frac{1}{2}(k m (0))^{2}\right)\right)^{(1 / m)-1} \left[k(b-a) +k^2m(b-a)(0) \right]\\
&=kb^{\frac{1-m}{m}} (b-a)
\end{align*}
