Deriving EOM of cantilever beam using specific Lagrangian

[JasonMech]

Homework Statement

The specific Lagrangian for a cantilever beam is given by:

[tex]
\overline{L}=\frac{1}{2}m[\dot{u}^2(s,t)+\dot{v}^2(s,t)]-\frac{1}{2}EI[\psi ^{\prime}(s,t)]^2
[/tex]

where [itex] m,EI [/itex] are mass and bending stifness, respectively. [itex] \dot{u},\dot{v} [/itex] are velocities in [itex] u,v [/itex] directions. [itex] \psi^{\prime} [/itex] is the angular velocity, found via:
[tex]
\tan \psi = \frac{v^{\prime}}{1+u^{\prime}}
[/tex]

and where [itex] s,t [/itex] are position and time. Prime and dot refers to differentiation wrt. position and time, respectively.

Homework Equations

The question is: how do you differentiate wrt. [itex] \psi [/itex]? That is, what is [itex] \frac{\partial \overline{L}}{\partial \psi} [/itex]?

The Attempt at a Solution

To me, it should be trivial (=0) but it seems not to be the case (since an article postulates it differently - but I cannot see where I go wrong!? It's probably basic math? Help would be appreciated a bunch!

 

[Valkarie]

My attempt:\frac{\partial \overline{L}}{\partial \psi}=\frac{\partial}{\partial \psi}\left(\frac{1}{2}m[\dot{u}^2(s,t)+\dot{v}^2(s,t)]-\frac{1}{2}EI[\psi ^{\prime}(s,t)]^2\right)\\=\frac{\partial}{\partial \psi}\left(\frac{1}{2}m[\dot{u}^2(s,t)+\dot{v}^2(s,t)]\right)-\frac{\partial}{\partial \psi}\left(\frac{1}{2}EI[\psi ^{\prime}(s,t)]^2\right)\\=0-EI\cdot2\psi^{\prime}(s,t)\cdot\frac{\partial \psi^{\prime}(s,t)}{\partial \psi}\\=0-EI\cdot2\psi^{\prime}(s,t)\cdot\frac{\partial}{\partial \psi}\left(\frac{v^{\prime}}{1+u^{\prime}}\right)\\=0-EI\cdot2\psi^{\prime}(s,t)\cdot\left(-\frac{v^{\prime\prime}}{(1+u^{\prime})^2}\right)=0