Thanks Alec Mihailovs It worked. Roozbeh

But there is a problem if I use this method; Assume after I obtain Ypr:=diff(Y,t) which is equal to: X' + L1 cos(θ1) θ1' + L2 cos(θ1 - θ2) (θ1' - θ2') now, I need to calculate the partial differentiation of answer in respect to theta[1], If I write: diff(eval(Ypr,theta[1](t)=theta[1]),theta[1]); the Maple assume θ1' as zero and will give me: 0 So, Whay should I do? Thanks, Roozbeh

It can be done in a few different ways. One way is diff(eval(Y,theta[1](t)=theta[1]),theta[1]); L1 cos(theta[1]) + L2 cos(theta[1] - theta[2])It's an interesting question. Why didn't you post it in the forum? Alec > Hi again, > > with your solution: > > PDEtools[declare](X(t),theta(t),prime=t): > Y:=X(t) + L1*sin(theta[1](t))+ L2*sin(theta[1](t)-theta[2](t)); > Y := X + L1 sin(θ1) + L2 sin(θ1 - θ2) > diff(Y,t); > X' + L1 cos(θ1) θ1' + L2 cos(θ1 - θ2) (θ1' - θ2') > > now, I can not find the partial differentiation!! (such as diff(Y,theta[1]). > > Do you have any suggestion? > > Thanks Roozbeh

Thanks Alec Mihailovs, I received your answer: PDEtools[declare](X(t),theta(t),prime=t): Y:=X(t) + L1*sin(theta[1](t))+ L2*sin(theta[1](t)-theta[2](t)); Y := X + L1 sin(θ1) + L2 sin(θ1 - θ2) diff(Y,t); X' + L1 cos(θ1) θ1' + L2 cos(θ1 - θ2) (θ1' - θ2') Roozbeh

Hello, I hope someone could help me with my problem. assume: Y:=X + L1 sin(theta1)+ L2 sin(theta1-theta2) where X,theta1, and theta2 are function of time. I know how I could find the partial differential such as diff(Y, theta2); but I can not find the derivative of Y in respect to time. Roozbeh