For a measured variable $x$ with error $\delta x$, the error on the function $y=f(x)$ is defined to be:
$$ \delta y = \Biggl| \frac{\mathrm{d}y}{\mathrm{d}x} \Biggr| \space \delta x $$
My problem is how to apply this to the following function:
$$ \alpha (t) = \frac{\mathrm{d}}{\mathrm{d}t}\Bigl(\mathrm{ln}\bigl(x(t)\bigr)\Bigr) $$
i.e. what is the error $\delta \alpha$ arising from the measurement error $\delta x$? I'm assuming it is:
$$ \delta \alpha = \Biggl| \frac{\mathrm{d}}{\mathrm{d}x} \Biggl( \frac{\mathrm{d}}{\mathrm{d}t} \mathrm{ln}(x) \Biggr) \Biggr| \space \delta x$$
but I am struggling with differentiating a derivative with respect to a different variable. I imagine it involves applying the chain rule but my maths is not strong enough to work this out.
In your case as the input the function $\alpha$ is $t$, I guess you want to calculate error in $\alpha$ for an error in $t$. So it has to be,
$\delta \alpha = \Biggl| \frac{\mathrm{d}}{\mathrm{d}t} \Biggl( \frac{\mathrm{d}}{\mathrm{d}t} \mathrm{ln}(x(t)) \Biggr) \Biggr| \space \delta t$
The way you simplify $\frac{\mathrm{d}}{\mathrm{d}t} \mathrm{ln}(x(t))$ is by chain rule of differentiation see, https://en.wikipedia.org/wiki/Chain_rule
and you get $= \frac{x'(t)}{x(t)}$ which you need find derivative again with respect to $t$ again by chain rule.
EDIT after your comment:
It is better to start with the simplified form of the given function,
$\alpha (t) = \frac{x_{t}(t)}{x(t)}$ now as you suggested there is measurement error with respect to $x$ so this implies,
$\delta \alpha = \frac{\mathrm{d}}{\mathrm{d}x} \Biggl( \frac{x_{t}(t)}{x(t)} \Biggr) \space \delta x$
Evaluating $\frac{\mathrm{d}}{\mathrm{d}x} \Biggl( \frac{x_{t}(t)}{x(t)} \Biggr) $ gives
$= \Biggl(\frac{d(x_t)}{dx}x - x_{t} \Biggr)/x^2 = \frac{-x_t}{x^2}$
by $x_t$, I mean derivative of $x$ with respect to $t$