Solution
= Solution
Put $s=b^at^ax$ and write $y'$ for <differentiation> with respect to $s$. Work on a real branch where the powers are defined and <differentiable>; $b>0$, $t>0$ suffices for arbitrary real $a$, and the parameter choice below has $b>0$. At fixed $x$, $\partial s/\partial t=as/t$. The <product rule> and <chain rule> therefore give
$$
\boxed{\frac{\partial u}{\partial t}=at^{a-1}[y(s)+s y'(s)].}
$$