(Why) can we treat a function of a variable as another independent variable? I'm currently reading my numerical analysis textbook and something's bugging me. To get into it, let's take a look at the following differential equation;
$$u'(x) = f(x, u(x))$$
In order to determine the stability of the equation, one may calculate the Jacobian,
$$J(x, u(x)) = \frac{\partial f}{\partial u}|_{(x, u(x))}$$
Here is a specific differential equation:
$$u'(x) = -\alpha(u(x) - sin(x)) + cos(x)$$
For which the Jacobian is
$$J(x, u(x)) = -\alpha$$
Basically, we treated both $sin(x)$ and $cos(x)$ as constants with respect to $u$, but I don't really understand why. Most of the time, when we take a derivative the variables are independant, which is not the case here as they both depend on the same variable $x$.
This means that the "rate of change of $sin(x)$ with respect to $u(x)$" is zero, but the value of $u(x)$ only changes if the value of x itself changes, so shouldn't the value of $sin(x)$ change aswell?
Thank you!
 A: There is a difference between the partial derivative $\frac{\partial}{\partial x}$ and the total derivative $\frac{d}{dx}$. For example, if we have variables $(u,x)$ and the equation $f=f(x,u(x))=x^2+u^3$ and we take the partial derivative we get $\frac{\partial f}{\partial x}=2x$ but if we take the total derivative we get $\frac{d f}{dx}=2x+3u^2\frac{\partial u}{\partial x}$, applying the chain rule. This distintion is a key point in classical mechanics for example and captures essentially what you are asking.
See: What exactly is the difference between a derivative and a total derivative?
A: To make things simpler, imagine a very simple autonomous dynamical system
$$
\frac{{\rm d}u}{{\rm d}x} = \alpha (u - u_0) \tag{1}
$$
for some constants $\alpha$ and $u_0$. The solutions to this system are of the form
$$
u(x) - u_0 = ce^{\alpha x} \tag{2}
$$
The interesting thing to note here is that if $\alpha > 0$ then the distance between $u(x)$ and $u_0$ grows exponentially with increasing $x$, so $u = u_0$ is said to be unstable. Whereas if $\alpha < 0$ the opposite occurs, and the distance between $u$ and $u_0$ shrinks with increasing $x$, in this case $u = u_0$ is stable. 
Now let's make things a bit more general. Imagine a system of the form
$$
\frac{{\rm d}u}{{\rm d}x} = f(u) \tag{3}
$$
and suppose there exists a point $u_0$ such that $f(u_0) = 0$ (like in Eq. (1)), if you want to understand the local stability of (3) you could Taylor expand $f$ around $u_0$
$$
f(u) = f(u_0) + \left.\frac{{\rm d}f}{{\rm d}u}\right|_{u = u_0}(u - u_0) + \cdots \tag{4}
$$
Remember that $f(u_0) = 0$, so at first order $f(u)\approx f'(u_0)(u - u_0)$, and Eq. (3)
$$
\frac{{\rm d}u}{{\rm d}x} \approx \underbrace{f'(u_0)}_{\alpha}(u- u_0) \tag{5}
$$
Now compare this with Eq. (1) and you realize that in order to understand the stability of the system around the point $u_0$ you need to know the value of $f'(u_0) $ a.k.a the Jacobian. There's a lot of caveats here you should be aware of, probably your text talks about them (e.g. what happens if $f'(u_0) = 0$, ...)

EDIT
Now imagine a system in two dimensions, something like
$$
\frac{{\rm d}u}{{\rm d}x} = f(u, v) ~~~~ \frac{{\rm d}v}{{\rm d}x} = g(u, v)\tag{6}
$$
You can define vectors ${\bf z} = {u \choose v}$ and ${\bf F} = {f \choose g}$ so that the system above can be written as 
$$
\frac{{\rm d}{\bf z}}{{\rm d}x} = {\bf F}({\bf z}) \tag{7}
$$
In this case nothing changes much, you can repeat the same analysis as in the first part and realize that the stability of the system around a point ${\bf z} = {\bf z}_0$ is given by the eigenvalues of the Jacobian evaluated at that location. And as before we require that $\color{blue}{{\bf F}({\bf z}_0) = 0}$. I highlight this condition because it will become important later.
Now to the final part. Instead of an autonomous system, consider a system of the form
$$
\frac{{\rm d}u}{{\rm d}x} = f(x, u) \tag{8}
$$
You could rename $v = x$ (that is, create a new state), and note that
$$
\frac{{\rm d}u}{{\rm d}x} = f(u, v) ~~~~ \frac{{\rm d}v}{{\rm d}x} = 1 \tag{9}
$$
So in theory you could repeat the same analysis all over again, but, you can see from this that the resulting field ${\bf F}$ never vanishes (that is, the blue expression above can never be satisfied)
