A quaternion is just a point in $4$-space. You've got a function from, say, $[0, 2\pi]$ to $4$-space defined by
$$
\gamma: \theta \mapsto (\cos(\theta/2), 0, 0, \sin(\theta/2)).
$$
The derivative of that function is
$$
\gamma': \theta \mapsto \frac{1}{2}(-\sin(\theta/2), 0, 0, \cos(\theta/2)).
$$
Since $\gamma(\theta)$ is a unit quaternion for all $\theta$, its derivative should be a tangent vector to the unit sphere at $\gamma(\theta)$, i.e., the dot product (not quaternion product!) of $\gamma(\theta)$ and $\gamma'(\theta)$ should be zero...and in fact it is.
It's possible that what you want is the body-centered derivative, which is just
$\gamma(\theta)^{-1}\gamma'(\theta)$, where the "-1" denotes quaternion inverse, and the multiplication here is quaternion multiplication; that'll get you a unit quaternion in the tangent space to $(1,0,0,0)$, which consists of all pure-vector quaternions...but maybe that's not what you want. If it is, it happens to be easy to compute in this case.
I'm going to just write $\gamma$ for $\gamma(\theta)$ from now on, and "c" and "s" for the cosine and sine of $\theta/2$, OK?
Because $\gamma$ is a unit quaternion, its inverse is its conjugate, so
$$
\gamma^{-1} = \bar{\gamma} = (c, 0, 0, -s).
$$
That makes the body-centered derivative be
$$
\gamma^{-1} \gamma' = (c, 0, 0, -s) \star (-s, 0, 0, c),
$$
where I've used $\star$ to denote quaternion multiplication. The result is
\begin{align}
\gamma^{-1} \gamma' &= (c, 0, 0, -s) \star \frac{1}{2}(-s, 0, 0, c)\\
&= \frac{1}{2}(-sc + sc; c(0,0,c) -s (0,0,-s) - (0,0,c) \times (0,0,-s)) \\
&= \frac{1}{2}(0; (0,0,c^2 + s^2) \\
&= \frac{1}{2}(0, 0,0,1).
\end{align}
In other words, your body-centered derivative is, at all times, $\frac{1}{2} {\mathbf k}$.
The geometric interpretation of quaternion multiplication is fundamentally 4-dimensional (unlike quaternion conjugation, which can be considered as an action on $\Bbb{R}^3$).
Let's start with an easy case. Say $q=a+bi$ with $b \neq 0$, $a^2+b^2=1$. That is, $q$ is a non-real unit quaternion in the subalgebra of $\Bbb{H}$ generated by $i$. What affect does multiplying by $q$ have on an arbitrary quaternion $r$?
First of all, if $r$ also lies in the subalgebra generated by $i$, then we can consider multiplication of $q$ and $r$ to be ordinary complex multiplication; that is, multiplication by $q$ rotates $r$ by $\theta$ in the $\{1, i\}$ plane.
Secondly, if $r$ lies in the orthogonal complement of that subalgebra, $r=cj+dk$, we can write $r=(c+di)j=j(c-di)$. The first of these representations can be used to left-multiply by $q$ via ordinary complex multiplication; the second one can be used to right-multiply. In either case, multiplying by $q$ rotates $r$ by $\theta$ in the $\{j, k\}$-plane; however, the sign difference means that the two multiplications rotate in opposite directions from each other.
We can then find the effect of multiplying $q$ by an arbitrary quaternion by projecting that quaternion into these two planes. That is, an arbitrary quaternion will have its $\{1, i\}$-projection and $\{j, k\}$-projection both rotated by $\theta$ when it is multiplied by $q$; the direction of the $\{j, k\}$-rotation, but not of the $\{1, i\}$-rotation, will be affected by whether we're left-multiplying or right-multiplying by $q$.
The general case works similarly. For any unreal unit quaternion $q$ that makes an angle $\theta$ with the real axis, multiplication by $q$ rotates by $\theta$ in the $\{1, q\}$-plane, and also rotates by $\theta$ in its orthogonal complement. The direction of the first rotation is fixed, but the direction of the second rotation depends on whether we're multiplying by $q$ on the left or on the right. You can see this just by noticing that any unreal quaternion generates a 2-dimensional subalgebra isomorphic to $\Bbb{C}$, making the previous few paragraphs work in general after some relabeling.
This also gives you a way to see why quaternion conjugation works the way it does on $\Bbb{R}^3$. If $q$ makes an angle $\theta$ with the real axis, then the map $r \mapsto qrq^{-1}$:
- is the identity map in the $\{1, q\}$-plane, since that plane forms a commutative subalgebra of $\Bbb{H}$
- involves a rotation by $2\theta$ in its orthogonal complement. Both $q$ and $q^{-1}$ rotate quaternions in $\{1, q\}^{\perp}$ by $\theta$. If they were multiplied on the same side, those rotations would have to cancel out; since left-multiplication behaves oppositely to right-multiplication, this means they must reinforce each other. Since the orthogonal complement of $\{1, q\}$ is orthogonal to $1$, it is pure imaginary, so we've reproduced the fact that quaternion conjugation corresponds to a double-angle rotation in $\Bbb{R}^3$ (when identified with $\Im(\Bbb{H})$).
Note also that multiplying by a general quaternion involves scaling by the norm of that quaternion, but conjugation conveniently causes the norms of $q$ and $q^{-1}$ to cancel.
Best Answer
The same way de Moivre's is proven for $i$. The square roots of $-1$ in the quaternions $\mathbb{H}$ are precisely the unit imaginary quaternions $u$. (Exercise: prove this.) Therefore,
$$\begin{array}{ll} \exp(\theta u) & \displaystyle =1+\theta u+\frac{\theta^2}{2!}u^2+\frac{\theta^3}{3!}u^3+\cdots \\ & \displaystyle =\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right)+\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\right)u \\ & =\cos(\theta)+\sin(\theta)u. \end{array} $$