Derving piston position and instant velocity

[VooDoo]

hey guys,

Im doing a question where it asks to derive the instantaneous velocity of the piston. We are given the potion of the piston as a function of instantaneous angular displacement of the crank.
Now I checked the answer...and i m not sure how the \omega gets into the answer. if I take the derivative with respect to time of both sides, i don't end up with an \omega.

tau = ratio of crank radius to connecting rod length R/Lx = R[(1-cos\theta) + \frac{\tau}{4}(1-cos2\theta)]

\dot{x} = R\omega(sin\theta + \frac{\tau}{2}sin2\theta)

 

[Integral]

You have x( \theta ) and you are finding \frac {dx} {dt} so you must use the chain rule, what will show up is \frac {d \theta } {dt} = \omega. So you need to finish your derivitve.