Derivative of $f(\theta(t)))$ with respect to the derivative of $\theta$, i.e. $\frac{\partial f}{\partial \dot{\theta}}$?

Question

Suppose I have an angle $\theta(t)$ as a function of time, and I'm considering the expression

$$f(t) = \dot{\theta}\sin(\theta)$$

Is it true that $$ \frac{\partial f}{\partial \dot{\theta}} = \sin\theta $$

Or would we have something like $$ \frac{\partial f}{\partial \dot{\theta}} = \sin\theta + \dot{\theta} \frac{\partial}{\partial \dot{\theta}}\left(\sin\theta\right) $$

I guess I'm confused because both $\theta$ and $\dot{\theta}$ are functions of time, so I feel like the derivative should be more complicated than just $\sin\theta$. For example, could we write

$$ \frac{\partial f}{\partial \dot{\theta}} = \frac{\partial f}{\partial t}\frac{\partial t}{\partial \dot{\theta}} = \frac{\partial f}{\partial t}\left(\frac{\partial \dot{\theta}}{\partial t}\right)^{-1} = \frac{\partial f}{\partial t}\left(\ddot{\theta}\right)^{-1} $$

Does this answer your question? Why is $\frac{\operatorname dy'}{\operatorname dy}$ zero, since $y'$ depends on $y$? — Hans Lundmark, Apr 15 '22 at 15:16
@HansLundmark It certainly seems related but I haven't studied the Lagrangian, so I'm not sure if all of the information maps to this case? — nwsteg, Apr 15 '22 at 15:34
OK. But if I may ask just out of curiosity: if you're not studying Lagrangian mechanics (or calculus of variations), then where did you encounter derivatives with respect to $\dot\theta$? — Hans Lundmark, Apr 15 '22 at 18:30
I am studying a crank-slider mechanism and using a numerical method to simulate the movement of the linkages. The linkages are modeled as vectors — nwsteg, Apr 15 '22 at 19:18

score 0 · Accepted Answer · answered Apr 15 '22 at 15:49

Since $\dot\theta\sin\theta$ is a function of $\dot\theta$ and $\theta$ its partial derivatives w.r.t. $\dot\theta$ and $\theta$ are astonishingly simple, namely $\sin\theta$ and $\dot\theta\cos\theta$. I have deliberately not given that function a name, because writing $f(t)$ for it is when the confusion starts. Better to write $$ f(t)=\dot\theta(t)\sin(\theta(t))=g(\dot\theta(t),\theta(t))\,. $$ Then, by the chain rule, $$ \dot f(t)=(\partial_{\dot\theta}g)\frac{d\dot\theta}{dt}+(\partial_\theta g)\frac{d\theta}{dt}=\ddot\theta(t)\sin(\theta(t))+\dot\theta^2(t)\cos(\theta(t))\,. $$

Derivative of $f(\theta(t)))$ with respect to the derivative of $\theta$, i.e. $\frac{\partial f}{\partial \dot{\theta}}$?

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