Leibniz rule for integrals and symmetry of second derivative

Question

Given two real valued functions $x(t), y(t)$, such that $\dot{y}(t) = \dot{x}(t)$, I need to compute $\frac{\partial y}{\partial x}$. One approach is:

\begin{equation} \dfrac{\partial y}{\partial x}(t) = \dfrac{\partial}{\partial x} \int \dot{x} \ dt = \dfrac{\partial}{\partial x} (x(t) + C) = 1 \end{equation}

On the other hand, using the symmetry of second derivative and taking $F(x,t) = \dot{x}(t)$, it follows that:

\begin{equation} \dfrac{\partial F(x,t)}{\partial x} = \dfrac{\partial \dot{x}(t)}{\partial x} = \dfrac{\partial }{\partial x} \dfrac{dx}{dt} = \dfrac{d}{dt} \dfrac{\partial x}{\partial x} = \dfrac{d}{dt} 1 = 0 \end{equation}

Therefore, using Leibniz rule for integrals,

\begin{equation} \dfrac{\partial y}{\partial x}(t) = \int \dfrac{\partial \dot{x}(t)}{\partial x} \ dt = \int 0 \ dt = C \end{equation}

With the first approach, we get $\frac{\partial y}{\partial x}$ = 1. With the second, $\frac{\partial y}{\partial x}$ = C $\in \mathbb{R}$. Which of them is correct? Or neither?

Answer 1

Your first answer is correct. In your second answer you incorrectly wrote that

$$ \frac{\partial} {\partial x} \frac{dx}{dt} = \frac{d}{dt} \frac{\partial x}{\partial x}. $$

This is not true in general; consider $x(t) = t^{2}$. Then $\dot{x}(t) = 2t = 2 \sqrt{x}$ and so

$$ \frac{\partial} {\partial x} \frac{dx}{dt} = x^{-1/2} \ne 0 $$ while

$$ \frac{d}{dt} \frac{\partial x}{\partial x} = \frac{d}{dt} 1 = 0. $$

Answer 2

It seems to me like you're overthinking it. If $\dot x(t) = \dot y(t)$, then $y(t) = x(t) + C$ for some constant $C$. So $\frac{dy}{dx} = 1$.