$T = \frac{1}{2}M_{w}\dot{x}^{2} + \frac{1}{2}I_{w}\frac{\dot{x}^2}{r^2} + \frac{1}{2}M_{b}((\dot{x} + l\dot{\theta}cos(\theta))^2 + (l\dot{\theta}sin(\theta))^2) + \frac{1}{2}I_{b}\dot{\theta}^{2}$
$\frac{d}{dt}\frac{\partial T}{\partial \dot{\theta}}$
So far I have
$\frac{1}{2}M_{w}\dot{x}^{2}$ has no $\dot{\theta}$ term so it goes to 0
$\frac{1}{2}I_{w}\frac{\dot{x}^2}{r^2}$ has no $\dot{\theta}$ term so it goes to 0
$\frac{1}{2}I_{b}\dot{\theta}^{2}$ goes to $I_{b}\dot{\theta}$ then $I_{b}\ddot{\theta}$
but $\frac{1}{2}M_{b}((\dot{x} + l\dot{\theta}cos(\theta))^2 + (l\dot{\theta}sin(\theta))^2)$ I am lost on.
Could someone please explain in the two steps $\frac{\partial T}{\partial \dot{\theta}}$ then $\frac{d}{dt}$
We treat $\dot{\theta}$ as being independent of $\theta$. As you pointed out only two terms depend on $\dot{\theta}$.
Differentiating the first term,
$$ \frac{\partial }{\partial \dot{\theta} } \left( \frac12 M_b (\dot{x} + l \dot{\theta}\cos(\theta) )^2 \right) = \frac12 M_b 2(\dot{x}+l\dot{\theta}\cos(\theta)) \frac{\partial}{\partial \dot{\theta}}\left( \dot{x} + l \dot{\theta} \cos(\theta) \right) $$ $$ = M_b \left (\dot{x}+l\dot{\theta}\cos(\theta)\right)\ l\cos(\theta), $$
and now we differentiate the second term,
$$ \frac{\partial }{\partial \dot{\theta} } (-1)\left( l\dot{\theta}\sin(\theta)\right)^2 = -2 l \dot{\theta}\sin(\theta) \frac{\partial}{\partial \dot{\theta}} ( l \dot{\theta} \sin(\theta) )= -2 l^2 \dot{\theta}\sin^2(\theta)$$
Combining these we get,
$$(*)= M_b\dot{x} l \cos(\theta) + l^2 \dot{\theta}( M_b \cos^2(\theta) - 2 \sin^2(\theta) ) = \color{blue}{M_b\dot{x} l \cos(\theta)} +\color{green}{ l^2 \dot{\theta}( M_b - (2+M_b) \sin^2(\theta) )} $$
Now differentiating with respect to $t$ we get,
$$ \frac{d (*) }{dt} = \color{blue}{M_bl ( \ddot{x} \cos(\theta) - \dot{x} \sin(\theta) \dot{\theta})} + \color{green}{l^2 \ddot{\theta}(M_b - (2+M_b) \sin^2(\theta)) + l^2\dot{\theta} ( -2(2+M_b) \sin(\theta) \cos(\theta) \dot{\theta})}$$