$$β=arctan [ (1+tan^2 \theta ) K\sin \alpha + tan \theta \sqrt {1+(1+tan^2 \theta )K^2 \sin^2 \alpha} ]$$
This is the general straight line (circle) equation on the sphere. $\alpha$ is longitude, $\beta$ is latitudes, $K$ is the slope of a straight line (circle) and $\theta $ is the distance between the point of the straight line and the equator when alpha equals zero.
Diagram, the horizontal line (circle) is the equator, and its $K=0$. The $K$ values of other lines (circles) are $-0.3$, and their $\theta $ values are $0.45pi, \ 0.25pi, \ 0, \ 0.$
The calculation formula of the slope of the spherical curve is:
$$K={\sinβ_2-\sinβ_1\over\sinα_2\cosβ_2-\sinα_1\cosβ_1}$$
In the sphere very small area, this formula is approximate to the formula of the slope of the plane curve. So the slope of a plane curve is only a special case of the slope of a spherical curve.
According to this formula, we can find the derivative of the spherical curve and the calculus operation for the sphere.
You could correlate with the following result that has a slightly different notation ($\varphi$ latitude, $\theta $ longitude, $ \psi $ angle small circle makes to longitude, $\alpha $ inclination of small circle to plane of equator).
$$ \tan \alpha = \dfrac{\sin \theta \cos \varphi + \cos \theta \tan \psi}{ \sin \varphi}{} $$