# Product Rule of x = r cos()

• TheDoorsOfMe
In summary, the conversation discusses the use of the product rule and chain rule in finding the derivative of cos(theta) with respect to t. The chain rule is used because theta is a function of t, resulting in the inclusion of d\theta/dt in the final derivative.

## 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?

TheDoorsOfMe said:
. 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. TheDoorsOfMe said:

## 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

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