Answer
$$k(b-a) \left( b \right)^{\frac{1}{m}-1}$$
Work Step by Step
Given $$M(t)=\left(a+(b-a) e^{k m t}\right)^{1 / m}$$
Since
\begin{align*}
M^{\prime}(t)&=\frac{1}{m}\left(a+(b-a) e^{k m t}\right)^{\frac{1}{m}-1}\left(k m(b-a) e^{k m t}\right)\\
&=k(b-a) e^{k m t}\left(a+(b-a) e^{k m t}\right)^{\frac{1}{m}-1}
\end{align*}
Then
\begin{align*}
M^{\prime}(0) &=k(b-a) e^{k m (0)}\left(a+(b-a) e^{k m (0)}\right)^{\frac{1}{m}-1}\\
&= k(b-a) \left( b \right)^{\frac{1}{m}-1}
\end{align*}
