Chain rule, time derivative and change of variables A simple calculus question. If I apply the chain rule to a composite function:
$$\frac{d}{dt}f(x(t))=\frac{\partial}{\partial x}f(x(t))\frac{dx}{dt}$$
Now, if I change variables, and define:
$$x=x_1+\lambda x_2$$
I can say: 
\begin{equation}
\frac{d}{dt}f(x_1(t),x_2(t))=\frac{\partial}{\partial x_1}f(x_1(t),x_2(t))\frac{dx_1}{dt}+\frac{\partial}{\partial x_2}f(x_1(t),x_2(t))\frac{dx_2}{dt}
\label{eq:1}
\end{equation}
but also
$$\frac{\partial}{\partial x}=\frac{\partial}{\partial x_1}+\lambda ^{-1}\frac{\partial}{\partial x_2}$$
so that, from the equation at the top
$$\frac{d}{dt}f(x(t))=\left[\frac{\partial}{\partial x_1}+\lambda ^{-1}\frac{\partial}{\partial x_2}\right]f(x_1(t)+\lambda x_2(t))\left(\frac{dx_1}{dt}+\lambda\frac{dx_2}{dt}\right)$$
How does this relate to the third equation?
 A: In the original equation you wrote, all you have is a function of one variable inside a function of one variable, so using $\partial$ isn't necessary.  Just use the chain rule from Calculus I:  $$ \frac{d}{dt} f(x(t)) = \frac{df}{dx} \cdot \frac{dx}{dt}$$
When you make the change of variables $x = x_1 + \lambda x_2$, what you really have is $$f(x(x_1(t), x_2(t))),$$ not $$f(x_1(t), x_2(t)).$$  So then you would have:
\begin{align}
  \frac{d}{dt}f(x(x_1(t), x_2(t))) 
    &= \frac{d}{dx} f(x(x_1(t), x_2(t))) \cdot \frac{dx}{dt}\\[0.3cm]
    &= \frac{d}{dx} f(x(x_1(t), x_2(t))) \cdot \left(\frac{\partial x}{\partial x_1} \frac{dx_1}{dt} + \frac{\partial x}{\partial x_2} \frac{dx_2}{dt}\right)\\[0.3cm]
    &= \underbrace{f'(x(x_1(t), x_2(t)))}_{\color{red}{df/dx}} \cdot \underbrace{\left(\frac{dx_1}{dt} + \lambda \frac{dx_2}{dt}\right)}_{\color{red}{dx/dt}}
\end{align}
A: \begin{equation}
\frac{df(x)}{dt}=\frac{df(x)}{dx}\frac{dx}{dt}=\frac{df(x)}{dx}(\frac{dx}{dx_1}\frac{dx_1}{dt}+\frac{dx}{dx_2}\frac{dx_2}{dt})=\frac{df(x)}{dx}(\frac{dx_1}{dt}+\lambda^{-1}\frac{dx_2}{dt})
\label{eq:1}
\end{equation}
