If $$\mathbf{q_1} = w_1 + x_1\mathbf{i} + y_1\mathbf{j} + z_1\mathbf{k}$$ is an unit quaternion, $$w_1^2 + x_1^2 + y_1^2 + z_1^2 = 1$$representing the first (earlier) rotation, and $$\mathbf{q_2} = w_2 + x_2\mathbf{i} + y_2\mathbf{j} + z_2\mathbf{k}$$ is a second (later) rotation, then their Hamilton product represents the combined rotation,
$$\begin{align}\mathbf{q} = \mathbf{q_2} \mathbf{q_1} = \; & w_1 w_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\\
+ \; & (w_1 x_2 + w_2 x_1 - y_1 z_2 + y_2 z_1) \; \mathbf{i}\\
+ \; & (w_1 y_2 + w_2 y_1 + x_1 z_2 - x_2 z_1) \; \mathbf{j}\\
+ \; & (w_1 z_2 + w_2 z_1 - x_1 y_2 + x_2 y_1) \; \mathbf{k}
\end{align}$$
Note that the Hamilton product is not commutative; the order of the quaternions matters. The result is also an unit quaternion, except for any numerical errors that might creep in. Fortunately, you can always normalize the rotation quaternion,
$$\mathbf{q'} = \frac{w + x \;\mathbf{i} + y \;\mathbf{j} + z \;\mathbf{k}}{\sqrt{w^2 + x^2 + y^2 + z^2}}$$
to avoid compounding errors. (It is basically safe to do after each operation, but usually necessary only after a few products.)
The inverse of a rotation is
$$\mathbf{q}^{-1} = w - x \;\mathbf{i} - y \;\mathbf{j} - z \;\mathbf{k}$$
ie. negating all components of an unit quaternion, except for the scalar component $w$, inverts the rotation. (Negating all components does not change the rotation it represents.)
You can also interpolate between two unit quaternions, $0 \le p \le 1$,
$$\begin{align}\mathbf{q'} = (1-p)\mathbf{q_1} + p\mathbf{q_2} & = w_1 + p (w_2 - w_1)\\
& + \left( x_1 + p (x_2 - x_1) \right ) \; \mathbf{i} \\
& + \left( y_1 + p (y_2 - y_1) \right ) \; \mathbf{j} \\
& + \left( z_1 + p (z_2 - z_1) \right ) \; \mathbf{k}\end{align}$$
if you normalize the result to unit length,
$$\mathbf{q} = \frac{w + x' \;\mathbf{i} + y' \;\mathbf{j} + z' \;\mathbf{k}}{\sqrt{w'^2 + x'^2 + y'^2 + z'^2}}$$
This is very useful in camera movement between two orientations. To get a really smooth change, use e.g.
$$p = 3 p'^2 - 2 p'^3$$
or
$$p = 6 p'^5 - 15 p'^4 + 10 p'^3$$
with $0 \le p' \le 1$. Both start and stop with zero velocity, but the former has a fixed rate of change of acceleration ("jerk"), and the latter starts and stops with zero acceleration.
