Find d/dx in terms of d/d(theta)

[maddogtheman]

If x=cos(theta) how do you find what d/dx is in terms of d/d(theta)?

 

[jostpuur]

By assuming that you have some test function \theta\mapsto f(\theta), defining a new function x\mapsto \bar{f}(x) by formula \bar{f}(x)=f(\theta(x)), where x\mapsto \theta(x) is some inverse of cosine, and then calculating

<br /> \frac{d}{dx}\bar{f}(x) = \cdots<br />

and then "thinking" that \bar{f} and f are somehow the same thing, and that you could cancel them out of the equation, so that you are left only with operators on the both sides.

 

[HallsofIvy]

Use the chain rule. If y is any function of x (and therefore of \theta),
\frac{dy}{dx}= \frac{dy}{d\theta}\frac{d\theta}{dx}

Since x= cos(\theta), then dx/d\theta= -sin(\theta) so that d\theta/dx= 1/(-sin(\theta)) and then
\frac{dy}{dx}= \frac{-1}{sin(\theta)}\frac{dy}{d\theta}