Computing the derivative of an inner product

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I want to differentiate (1) wrt the vector $x \in \mathbb{R}^n$ where $w(x)$ is a zero-one diagonal indicator matrix

\begin{align} \frac{1}{2} g(x)^T w(x) g(x) &&&& (1) \end{align}

Since

\begin{align} \frac{1}{2} g(x)^T w(x) g(x)= \langle g(x),g(x) \rangle \end{align} Thus, \begin{align} \frac{d}{dx} \langle g(x),g(x) \rangle = \langle g(x)',g(x) \rangle + \langle g(x),g(x)' \rangle = 2 \langle g(x)',g(x) \rangle = (g(x)')^T w(x)g(x) \end{align}

but the notes I study from says the outcome is $g(x)'w(x)g(x)$

Why is that?