Why is there a d\theta/dt at the end of the derivative?

[TheDoorsOfMe]

Homework Statement

r = r(t)
\theta = \theta(t)

x = r cos(\theta)

dx/dt =dr/dt cos(\theta) - r sin(\theta) d\theta/dt

The Attempt at a Solution

Where does the d\theta/dt come from at the end of the derivative? I know I'm using product rule here because r and theta are both functions of t. But, the derivative of cos is just -sin. Why would there be a d\theta/dt at the end?

 

[gabbagabbahey]

. But, the derivative of cos is just -sin. Why would there be a d\theta/dt at the end?

No, \frac{d}{d\theta}\cos\theta=-\sin\theta but \frac{d}{dt}\cos\theta=\left(\frac{d}{d\theta}\cos\theta\right)\left(\frac{d\theta}{dt}\right) via the chain rule.

 

[Mark44]

Homework Statement

r = r(t)
\theta = \theta(t)

x = r cos(\theta)

dx/dt =dr/dt cos(\theta) - r sin(\theta) d\theta/dt

The Attempt at a Solution

Where does the d\theta/dt come from at the end of the derivative? I know I'm using product rule here because r and theta are both functions of t. But, the derivative of cos is just -sin. Why would there be a d\theta/dt at the end?

Chain rule.
d/dt(cos(theta)) = -sin(theta)*d(theta)/dt

 

[TheDoorsOfMe]

oooooooooooooohhhhhhhhhhhh! man I'm kinda disappointed I didn't see that one : ( oh well. Thank very much guys!