What is the difference between partial and normal derivatives?
The (calculus-of-variations) tag seems to be not the most popular one, so maybe it needs some more advertising (-:
Serious. The Euler-Lagrange equations associated with calculus of variations provide an example,
where both partial and common differentiation are involved. That's why it might help here.
Let there be given a curve $\vec{q}(t)$ and a real valued function $L$ with the following arguments:
this curve, the time derivative $\dot{\vec{q}}(t)$ of the curve and the time $t$ itself.
Minimize the following integral as a function/functional of the curve $\vec{q}(t)$:
$$
W\left(\vec{q},\dot{\vec{q}}\right) = \int_{t_1}^{t_2} L\left(\vec{q},\dot{\vec{q}},t\right) dt
= \mbox{minimum}
$$
It is proved in the reference that the curve minimizing the integral $W$ is given by the following system
of mixed partial-common differential equations, one for each of the coordinates $q_k(t)$ of the curve $\vec{q}(t)$:
$$
\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}_k}\right) = 0
$$
These are the well known Euler-Lagrange equations. They are specified for the following problem: find all curves
in the Euclidean plane for which the length $W$ between two given end-points is minimal.
This makes $\vec{q} = (x,y)$ and $\dot{\vec{q}} = (\dot{x},\dot{y})$ in:
$$
W = \int_{t_1}^{t_2} L(\dot{x},\dot{y}) dt = \mbox{minimal} \qquad \mbox{with} \quad
L(\dot{x},\dot{y}) = \sqrt{\dot{x}^2 + \dot{y}^2}
$$
Giving for the Euler-Lagrange equations:
$$
\frac{\partial L}{\partial x} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) = 0 \\
\frac{\partial L}{\partial y} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{y}}\right) = 0
$$
Partial derivatives. Obviously:
$$
\frac{\partial L}{\partial x} = \frac{\partial L}{\partial y} = 0
$$
Somewhat less obviously:
$$
\frac{\partial \sqrt{\dot{x}^2 + \dot{y}^2}}{\partial \dot{x}} = \frac{\dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}} \\
\frac{\partial \sqrt{\dot{x}^2 + \dot{y}^2}}{\partial \dot{y}} = \frac{\dot{y}}{\sqrt{\dot{x}^2 + \dot{y}^2}}
$$
Common derivatives:
$$
\frac{d}{dt} \frac{\dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}} =
\frac{ \ddot{x} \sqrt{\dot{x}^2 + \dot{y}^2}
- \dot{x} \left( \dot{x} \ddot{x} + \dot{y} \ddot{y} \right) / \sqrt{\dot{x}^2 + \dot{y}^2}}
{\left(\sqrt{\dot{x}^2 + \dot{y}^2}\right)^2} =
\dot{y} \frac{\dot{y}\ddot{x} - \dot{x}\ddot{y}}{\left(\dot{x}^2 + \dot{y}^2\right)^{3/2}}
= - \kappa \, \dot{y} \\
\frac{d}{dt} \frac{\dot{y}}{\sqrt{\dot{x}^2 + \dot{y}^2}} =
\frac{ \ddot{y} \sqrt{\dot{x}^2 + \dot{y}^2}
- \dot{y} \left( \dot{x} \ddot{x} + \dot{y} \ddot{y} \right) / \sqrt{\dot{x}^2 + \dot{y}^2}}
{\left(\sqrt{\dot{x}^2 + \dot{y}^2}\right)^2} =
\dot{x} \frac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{\left(\dot{x}^2 + \dot{y}^2\right)^{3/2}}
= + \kappa \, \dot{x}
$$
Where $\kappa$ is recognized as the
curvature.
The Euler-Lagrange equations thus say that $- \kappa\, \dot{x} = +\kappa \, \dot{y} = 0$ ,
with can only be fulfilled if $\kappa = 0$ : the curvature is zero.
Indeed, the shortest path between two points in the Euclidean plane is a straight line.