The derivative of a single variable function becomes a gradient function?

Question

From Convex Optimization by Boyd & Vandenberghe:

Let $f: \Bbb R^n \to \Bbb R$ be a convex and twice continuously differentiable function. Define $$g(t) = f(x+t \Delta x)$$ Then $$g''(t) = \Delta x^T\nabla^2f(x + t \Delta x)\Delta x$$

I'm not sure how this notation is used in the above box.

Since $g$ is a function of $t$, then $g'(t) = \nabla f(x+t\Delta x)\Delta x$ but this doesn't make sense since they're both vectors.

Can someone explain what I am misunderstanding in this notation here?

Answer 1

$\Delta x$ is a vector, $\nabla f(x+t\nabla x)$ is also a vector.

The book meant to say $$g'(t)=\nabla f(x + t\nabla x) ^T\Delta x,$$

their dot product.

$\nabla^2 f(x+t\Delta x)$ is a matrix. $\Delta x^T\nabla^2 f(x+t\Delta x)\Delta x$ gives us a scalar.