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

$$\frac{d}{dx}\bar{f}(x) = \cdots$$

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
$$\frac{dy}{dx}= \frac{-1}{sin(\theta)}\frac{dy}{d\theta}$$