How to get a time derivative in symbol form

Hello

Im currently working with the Lagrange Equation to create an equation of motion which i will solve with an odesolver.

In order to use the Lagrange Equation i need the time derivative of the kinetic energy that has been derived to the dq's

q(t) = the generalized coordinates in matrix form (they are a function of time)

q(t) = [q1; q2]

dq(t) = the first time derivative of the q's, also in matrix form (normally indicated with a dot above the q)

d/dt(q(t)) = dq(t) = [dq1; dq2]

ddq(t) = the second time derivative of the q's, also in matrix form (normally indicated with two dots above the q)

d/dt(dq(t)) = ddq(t) = [ddq1; ddq2]

T = kinetic energy

T(q(t),dq(t)) T is a function of q(t) and dq(t)

then it will be derived to the dq(t) and transposed

(T,dq)^T (^T = transpose, in matlab .')

This is how far I get with my limited knowledge of Matlab, now it gets tricky for me

I need the time derivative of (T,dq)^T

I call (T,dq)^T = Tdq

Tdq(q,dq) is a function of q and dq

d/dt(Tdq) Time derivative of Tdq where q becomes dq, dq becomes ddq.

Example:

            (9/4)*(m1+4*m2)*L^2*dq1+3*m2*L*dq2*cos(q1+q2)
Tdq       = 
            3*m2*L^2*dq1*cos(q1+q2)+(4/3)*m2*L^2*dq2

And the time derivative should be:

            (9/4)*(m1+4*m2)*L^2*ddq1+3*m2*L^2*dqq2*cos(q1+q2)-3*m2*L^2*dq2*(dq1+dq2)*sin(q1+q2)
d/dt(Tdq) =
            3*m2*L^2*ddq1*cos(q1+q2)-3*m2*L^2*dq1*(dq1+dq2)*sin(q1+q2)+(4/3)*m2*L^2*ddq2

This Time derivative is pretty large, but still can be solved by hand, the ones i have now are nearly impossible to solve with the hand. Thats why i need matlab to solve it.

So basically i need matlab to recognise that d/dt(q) = dq and that d/dt(cos(q)) = -dq*sin(q) (chain rule) ect

Their has to be a simple solution ( I can't be the first one who tries to use the lagrange equation of motion in matlab), But i just can't seem to find it. I am new to matlab and still a student, so a simple solution with a simple example would be much appreciated.

Please help me solve this

Thanx

Maarten