The Coriolis and centrifugal forces appear when looking at Newton's 2nd Law in a rotating frame of reference. To derive it, it's most convenient to use Hamilton's principle. Let \vec{x}=(x_1,x_2,x_3) be the coordinates of a particle with respect to Cartesian coordinates of an inertial reference frame. The Lagrangian with a force that can be derived from a scalar potential field V(\vec{x}), is given by
L=\frac{m}{2} \dot{\vec{x}}^2-V(\vec{x}),
and the equations of motion by the Euler-Lagrange equations
\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{\vec{x}}}=\frac{\partial L}{\partial \vec{x}},
i.e., with the above given Lagrangian
m \ddot{\vec{x}}=-\vec{\nabla} V(\vec{x})=\vec{F}(\vec{x}),
which is Newton's 2nd law written in an inertial frame.
Now we consider a frame of reference, which rotates with respect to the inertial frame around the z axis. The Cartesian coordinates with respect to the rotating frame are (x',y',z'). The inertial coordinates read in terms of the rotating ones
x=x' \cos(\omega t) - y' \sin (\omega t), \quad y=x' \sin(\omega t) +y' \cos(\omega t), \quad z=z'.
Now we have to express the Lagrangian in terms of the rotating coordinates. For that we need
\dot{x}=\dot{x}' \cos(\omega t) - x' \omega \sin(\omega t) - \dot{y}' \sin \omega t-y' \omega \cos(\omega t),
\dot{y}=\dot{x}' \sin(\omega t) + x' \omega \cos(\omega t) +\dot{y}' \cos(\omega t) -y' \omega \sin(\omega t).
Now you have to evaluate the kinetic energy. It turns out that after the dust has settled you can write
T=\frac{m}{2} \dot{x}^2=\frac{m}{2} (\dot{\vec{x}}'{}+\vec{\omega} \times \vec{x}')^2,
where
\vec{\omega}=(0,0,\omega).
Evaluating the Euler-Lagrange equations in the new coordinates finally leads to
m [\ddot{\vec{x}}'+2 \vec{\omega} \times \dot{\vec{x}'} + \vec{\omega} \times (\vec{\omega} \times \vec{x}')]=\vec{F}'(\vec{x}').[/tex]
Since \vec{F}(\vec{x}) is a vector field, you have
\vec{F}'(\vec{x})'=\hat{D} \vec{F}(\vec{x})=\hat{D} \vec{F}(\hat{D}^{-1} \vec{x}'),
where \hat{D} denotes the rotation matrix
\hat{D}=\begin{pmatrix}<br />
\cos(\omega t) & \sin(\omega t) &0 \\<br />
-\sin(\omega t) & \cos \omega(t) &0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}.
Now you rewrite this equation as
m \ddot{\vec{x}}'=\vec{F}'(\vec{x}')-2 m \vec{\omega} \times \dot{\vec{x}}' - m \vec{\omega} \times (\vec{\omega} \times \vec{x}'.
You get an equation of motion which looks like Newtons 2nd Law in an inertial frame but with two additional forces, the Coriolis force and the centrifugal force. They are "inertial forces", only appearing because we are looking at the motion from the point of view of an observer at rest wrt. a non-inertial frame of reference.
