Suppose I have a scalar function $f(u(w(v_k))$, where
$u=[u_1(w(v_k)),u_2(w(v_k))]$
w is a another scalar function, and $v_k$ is the independent variable. If I was interested in calculating $\frac{df}{dv_k}$, I would think that the derivative would be
$\frac{df}{dv_k} = \frac{\partial f}{ \partial u} \frac{\partial u }{ \partial w} \frac{\partial w}{ \partial v_k}$.
The two first terms seem to me to be two vectors, and the last is a scalar. I believe somehow the first two terms are an inner product somehow (I'm missing a transpose)?
Can Someone give me some insight on this?
Yes, you're right, there is an inner product there. $$\frac{df}{dv_k} = \frac{\partial f}{ \partial \vec u} \cdot\frac{\partial \vec u }{ \partial w} \frac{\partial w}{ \partial v_k}=\frac{\partial f}{\partial u_1}\frac{\partial u_1}{\partial w}\frac{\partial w}{\partial v_k}+\frac{\partial f}{\partial u_2}\frac{\partial u_2}{\partial w}\frac{\partial w}{\partial v_k}$$ Here, in the last equality, everything is now scalar.