Gradient of dot product using summation convention

Question

I'm trying to show that $\nabla (\vec u \cdot \vec v)=(\vec u')^t \vec v + (\vec v')^t \vec u$

Solution so far: $[\nabla (\vec u \cdot \vec v)]_i= \partial_i u_j v_j = u_j \partial_i v_j + v_j \partial_i u_j$

What do I do from here?

Answer 1

Note that $[\nabla (\vec u \cdot \vec v)]_i= \frac{\partial}{\partial_{x_i} }(\sum_ju_j v_j) = \sum_j (u_j \frac{\partial}{\partial_{x_i}} v_j + v_j \frac{\partial}{\partial_{x_i}} u_j)=((\vec u')^t \vec v + (\vec v')^t \vec u)_i$
Because note that $(\vec u')_{ij}=\frac{\partial}{\partial_{x_j}}u_i$