Line Follower Transfer Function 
Let's assume that I have a long straight line, and I'm trying to control the distance from line, $\Delta x$.
Given a differential drive robot, we have $v=\frac{v_l+v_r}{2}$ and $w=\frac{v_r-v_l}{d}$ where $v_r$ (right-wheel velocity) and $d$ (wheel separation) are constant.
In this case, $\dot{\Delta x} = v sin(\theta)$, so $\dot{\Delta x} = \frac{v_l+v_r}{2}sin(\int{\frac{v_r-v_l}{d}dt})$ assuming $\theta_0=0$.
With small-angle approximations, we can state that $\dot{\Delta x} = \frac{v_l+v_r}{2}\int{\frac{v_r-v_l}{d}dt}$.
Now, I'd like to get a transfer function from $\mathcal{L}\{v_l(t)\}$ to $\mathcal{L}\{\Delta x(t)\}$.
Is this possible? If so, how should I derive the transfer function?
 A: Start from
$$\Delta \dot{x} = v \sin \theta \implies \theta =\arcsin\left [\frac{\Delta \dot{x}}{v} \right]$$
$$\implies \int_{0}^{t}\dfrac{v_r-v_l(\tau)}{d}d\tau =\arcsin\left [\frac{\Delta \dot{x}}{v} \right]$$
Now, differentiate the expression and use the fundamental theorem of calculus
$$\dfrac{d}{dt}\int_{a}^{t}f(\tau)d\tau =f(t)$$ 
to obtain
$$\dfrac{v_r-v_l(t)}{d}=\dfrac{d}{dt}\arcsin\left [\frac{\Delta \dot{x}}{v} \right]$$
$$\dfrac{v_r-v_l(t)}{d}=\dfrac{1}{\sqrt{1-\left(\dfrac{\Delta \dot{x}}{v}\right)^2}}\dfrac{d}{dt}\left[\dfrac{\Delta \dot{x}}{v}\right] $$
$$\dfrac{v_r-v_l(t)}{d}=\dfrac{1}{\sqrt{1-\left(\dfrac{\Delta \dot{x}}{v}\right)^2}}\left[\dfrac{\Delta \ddot{x}v-\Delta \dot{x}\dot{v}}{v^2}\right] $$
$$\Phi(\Delta \dot{x},\Delta \ddot{x},v_l(t))=\dfrac{v_r-v_l(t)}{d}-\dfrac{\left[\dfrac{\Delta \ddot{x}\left(\frac{v_r+v_l(t)}{2} \right)-\Delta \dot{x}\left(\frac{\dot{v}_l(t)}{2} \right)}{\left(\frac{v_r+v_l(t)}{2} \right)^2}\right]}{\sqrt{1-\left(\dfrac{\Delta \dot{x}}{\left(\frac{v_r+v_l(t)}{2} \right)}\right)^2}}= 0$$
The linearization is given by:
$$\Phi(\Delta \dot{x},\Delta \ddot{x},v_l(t))\approx \left.\Phi(\Delta \dot{x},\Delta \ddot{x},v_l(t))\right|_{\text{R}}+\left.\dfrac{\partial \Phi}{\partial \Delta \dot{x}}\right|_{\text{R}}\left[\Delta \dot{x}-\left.\Delta \dot{x}\right|_{\text{R}} \right]$$
$$+\left.\dfrac{\partial \Phi}{\partial \Delta \ddot{x}}\right|_{\text{R}}\left[\Delta \ddot{x}-\left.\Delta \ddot{x}\right|_{\text{R}} \right]+\left.\dfrac{\partial \Phi}{\partial v_l(t)}\right|_{\text{R}}\left[v_l(t)-\left.v_l(t)\right|_{\text{R}} \right]=0.$$
The index $\text{R}$ is denoting that all expressions with this index need to be evaluated at the reference point that you selected.
Can you carry on from here? I would suggest using a computer algebra system (sympy for Python, Maple, MATLAB, Mathematica, ...) such that you do not end up doing a mistake in the calculations. 
Note, that it might happen that you end up with a linearized equation that does not contain your control input. If that happens, then you can conclude that linearization is not an appropriate method.
