# Partial derivative of position in respect to velocity [duplicate]

Out of interest What is the derivative of $$\frac{\partial x(t)}{ \partial v(t)}$$

Let me note that, since both $$x$$ and $$v$$ are functions of a single variable, we are dealing here not with partial derivatives but with ordinary derivatives which are equivalent to the corresponding differentials (partial derivatives are not). This may seem like a simple change of notation $$\frac{\partial x(t)}{\partial v(t)} \longrightarrow \frac{d x(t)}{d v(t)},$$ but mathematically these are two different things. Importantly, ordinary derivative is actually a true ratio of two differentials, $$dx = \dot{x}(t)dt, dv = \dot{v}dt$$, while a partial derivative is just a symbol.
Now we can consider $$x$$ as a function of $$v$$: $$x = f(v)$$, which is given to us in a parametric form, with $$t$$ being the parameter. The ordinary derivative of this function is then just the ratio of the two differentials, i.e. $$\frac{dx}{dv} = \frac{\dot{x}(t)}{\dot{v}(t)}$$ the answer is actually again given in a parametric form. If one wants it as a function of $$v$$; one needs to invert the function $$v(t)$$ and substitute it in the expression above. In other words $$f'(v) = \frac{\dot{x}(t)}{\dot{v}(t)}|_{v(t)=v}.$$