Take partial derivative of squared dot product

Question

I want to take the partial derivative of $ z = (y⋅x)^2$ with respect to $x$, where $y$, $x$ are vectors. Should I be correct to assume that we can see this as $ z = g(x)^2$, where $g(x) = (y⋅x)$. So using chain rule

\begin{align} ∂z/∂x &= ∂z/∂g * ∂g/∂x, \\ ∂z/∂x &= 2(y⋅x)y. \end{align}

I'm probably butchering the notation though.

Answer 1

Your calculation is correct. Recall that the dot product is defined by $y \cdot x = \sum_{i = 1}^n y_i x_i$, so $(y \cdot x)^2 = (\sum_i y_i x_i)^2$. Now you want to calculate $\partial z / \partial x$, which is the gradient $(\partial z/\partial x_1, \dots , \partial z /\partial x_n)$. By the chain rule,

$$\frac{\partial z}{\partial x_i} = 2y_i \left (\sum_i x_i y_i \right)$$

for all $i$, hence

$$\frac{\partial z}{\partial x} = 2 (x \cdot y) y.$$