Fill in the details of the derivation of the formula for the acceleration. There are three ways this is typically approached: the transport theorem, the chain rule, and a direct calculation using Taylor expansions to two terms (i.e., linear approximations). We shall use the last.
Consider the path of a fluid particle. Suppose that at time $t$ it is at the point $x$ in space. Its velocity is then $u(x, t)$. At later time $t+\Delta t$ it has moved along the particle path. If the path is smooth and $\Delta t$ is small, the point it has moved to is approximately $x+u(x, t) \Delta t$ $\left(+O\left(\triangle t^2\right)\right)$.
(a) Draw a schematic of a particle path. Indicate on it the two points and the velocity vector. The velocity at this second point is $u(x+u(x, t) \Delta t, t+\Delta t)$. The acceleration is the change in velocity per unit change in time. Thus,
$$
a(x, t)=\lim _{\Delta t \rightarrow 0} \frac{u(x+u(x, t) \Delta t, t+\Delta t)-u(x, t)}{\Delta t} .
$$
(b) Recall the Taylor expansion: $u(x+h)=u(x)+h \cdot \nabla u+O\left(|h|^2\right)$. Use this expansion in the above limit to show that the fluid's acceleration is
$$
a=\frac{\partial u}{\partial t}+(u \cdot \nabla) u .
$$
Another approach is via the chain rule.
(c) Reconsider the above difference quotient. Add and subtract terms and write the difference quotient as
$$
\frac{u(x+u(x, t) \Delta t, t+\Delta t)-u(x, t)}{\Delta t}=A+B,
$$
where as $\Delta t \rightarrow 0, A \rightarrow \frac{\partial u}{\partial t}, B \rightarrow(u \cdot \nabla) u$.
This proof is equivalent to using the chain rule!