# Differentiating with respect to time

#### cabellos

How do you go about differentiating rcos$with respect to time....? In the book im studying from it says d$/dt d/d$(rcos$) is the process to find the answer... but what does this mean...?

#### cristo

Staff Emeritus
HAve you come across the chain rule? This states that $$\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}$$. Thus, applying this to the case you mention, we obtain $$\frac{d}{dt}(rcos\theta)=\frac{d}{d\theta}(rcos\theta)\frac{d\theta}{dt}$$

#### neutrino

The chain rule has been applied in arriving at the result.
http://mathworld.wolfram.com/ChainRule.html

$$\frac{df}{dt} = \frac{df}{d\theta}\frac{d\theta}{dt}$$, where f denotes the given function.

Edit: Since Cristo posted while I was playing with the typeset, I'm using theta instead of the more obvious x. Apparently, the '\$' symbol messes up Latex.

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#### HallsofIvy

If $\theta$ is a function of time but r does not, then it is true that $frac{df}{dt}= \frac{df}{d\theta}\frac{d\theta}{dt}$.
If both f depends on both r and $\theta$ and both r and $\theta$ depend upon time,
$$\frac{df}{dt}= \frac{\partial f}{\partial r}\frac{dr}{dt}+ \frac{\partial f}{\partial \theta}\frac{d\theta}{dt}$$