Time series: variations as a feature
The derivative is a linear transform, and you're using a linear model. As you've demonstrated, nothing is gained by adding linear combination of other features (like the derivative) as a new feature in a linear model.

You may want a derivative coefficient to help interpret the model. You could use the normal model $x_{n+1} = a * x_n + b * n_{n-1}$ and then report the value $a-b$ to the client. A more general approach is to examine the frequency response of the model (the derivative is a high-pass filter, your model is a FIR filter).

Alternately you can rotate the basis of the input such that the derivative is a feature and the dimensions are still independent as follows:

You can think of the independent variables $x_n$ and $x_{n-1}$ as forming a standard basis for the two dimensions of your model's input, that is $\langle 1, 0\rangle$ and $\langle 0, 1\rangle$. 
Neither $a\langle 1, 0 \rangle$ or $b \langle 0, 1 \rangle$ are orthogonal to  c$\langle1, -1\rangle$.

Therefore, if you want to use $c(x_n - x_{n-1})$ as a feature, I would also use the orthogonal feature $d(x_n + x_{n-1})$. That is,

$$ x_{n+1} = c(x_n - x_{n-1}) + d(x_n + x_{n-1}) $$.

You could think of this model as the derivative plus the integral. 

The same logic applies to the choice of scaling function for the Haar wavelet.