Derivative of a partial derivative

Question

Assuming that $\dot{x}=f(x,t,C)$, where $C$ is a parameter and $f(\cdot)$ is continuous, how do we show that

$$\frac{d}{dt} \bigg[\frac{d x}{d C} \bigg]=\frac{\partial f}{\partial C}+ \bigg(\frac{\partial f}{\partial x} \bigg) \bigg(\frac{d x}{d C} \bigg)$$

Please prove it step by step.

Below is what I did.

Let $x=F(x_0,x,t,C)$. We have

\begin{align} \dot{x} &= f(x,t,C) \\ &= F_t+F_x\frac{d x}{d t} \end{align}

as the initial value $x_0$ and $C$ are time invariant.

$$\frac{d}{dt} \bigg[\frac{d x}{d C} \bigg]=\frac{d}{d t} \bigg[\frac{\partial x}{\partial C} \bigg]+\frac{d}{d x} \bigg[\frac{\partial x}{\partial C} \bigg]\frac{d x}{d t}$$

I get stuck here.

Answer 1

After thinking for some time, I worked out the problem myself. I made a mistake above because $x=F(x_0,t,C)$ instead of $F(x_0,x,t,C)$. Hence,

$$\dot{x}=\frac{d F}{d t}=f(x,t,C)$$

\begin{align}\frac{d}{dt}\bigg[\frac{d x}{d C}\bigg]=&\frac{d}{d t}\bigg[\frac{d x}{d C}\bigg] =\frac{d^2x}{d td C} \end{align} Exchanging the order, we obtain \begin{align} \frac{d^2x}{d Cd t}=&\frac{d}{d C}\bigg[\frac{d x}{d t}\bigg]\\ =&\frac{d}{d C}f(x,t,C)\\ =&\frac{\partial f}{\partial C}+\bigg(\frac{\partial f}{\partial x}\bigg)\bigg(\frac{d x}{d C}\bigg) \end{align}

Done!