Directional derivative - need some help with $D_{X_u}X_v$

Question

Let's say we have a surface in $\mathbb{R}^3$ which is parameterized with only two variables $u$ and $v$ so it can be defined as $X(u,v)=(f_1 (u,v),f_2(u,v),f_3(u,v))$. We calculate the tangent vectors $$X_u=\left (\frac{\partial f_1}{\partial u},\frac{\partial f_2}{\partial u} ,\frac{\partial f_3}{\partial u} \right )$$ and $$X_v=\left (\frac{\partial f_1}{\partial v},\frac{\partial f_2}{\partial v} ,\frac{\partial f_3}{\partial v} \right ),$$ which are objects in $\mathbb{R}^3$. Now I want to calculate such derivative $D_{X_u} X_v$ and this should be according to my understanding $$D_{X_u} X_v= \left(\sum_i \frac{\partial f_i}{\partial u} \cdot \frac{\partial }{\partial x_i} \frac{\partial f_1}{\partial v},\sum_i \frac{\partial f_i}{\partial u} \cdot \frac{\partial }{\partial x_i} \frac{\partial f_2}{\partial v},\sum_i \frac{\partial f_i}{\partial u} \cdot \frac{\partial }{\partial x_i} \frac{\partial f_3}{\partial v}\right).$$ Now, according to some lecture notes this expression should be equal to $$\frac{\partial X}{\partial u \partial v}=\left (\frac{\partial f_1}{\partial u\partial v},\frac{\partial f_2}{\partial u\partial v} ,\frac{\partial f_3}{\partial u\partial v} \right )$$ I don't see a connection between those equations.