differentiate by a differential

Question

This is a difficult question to phrase so I will show it mathematically. I am trying to differentiate something through chain rule but I am not sure if my steps are correct.

Let

$f(\theta) = \sin(\theta)$

Is it possible to obtain the answer to this:

$\frac{d }{d \dot{\theta}}f(\theta)$

I have tried to do the following:

Let $ z = \dot{\theta}$ so

$ \theta = \int{z} $

$\frac{d \sin{ \int{z} } }{dz} = z\cos{ \int{z} } $

Which gives me the answer as

$= \dot{\theta} \cos{\theta} $

Is this correct? Is it even possible to differentiate in this manner?

Answer 1

The symbolism $\frac{d f(\theta)}{d \dot{\theta}}$ is very strange (or crazy ?)

How can we understand it ?

$\dot{\theta}=\frac{d\theta}{dt}$

$d\dot{\theta}=d\left(\frac{d\theta}{dt}\right)=\frac{d^2\theta}{dt^2}dt=\ddot{\theta}dt$

$\frac{d f(\theta)}{d \dot{\theta}}=\frac{d f(\theta)}{\ddot{\theta}dt}=\frac{1}{\ddot{\theta}}\frac{df(\theta)}{dt}$

In case of $f(\theta)=\sin(\theta)$ this leads to : $\frac{d \sin(\theta)}{d \dot{\theta}}=\frac{1}{\ddot{\theta}}\frac{d\sin(\theta)}{dt}=\frac{1}{\ddot{\theta}}\cos(\theta)\frac{d\theta}{dt}=\frac{\dot{\theta}}{\ddot{\theta}}\:\cos{\theta}$

That is stretching a point to say the least.

.

In addition after the judicious orion's comment :

Of course if the typing was : $\quad \frac{\partial f(\theta \:,\:\dot{\theta}\:,\:\ddot{\theta},...) } {\partial \dot\theta}\quad$ there would be no ambiguity at all.

Answer 2

If this is something you get from Euler-Lagrange formalism, then you need to write a partial derivative, and there is no $\dot{\theta}$ in the expression, so the result is $0$.