Three-times differentiable function

Question

Let $\mathbf a:\mathbb{R}\longrightarrow\mathbb{R^3}$ be a three-times differentiable vector valued function. If $\mathbf b(x)=\mathbf a(x)\cdot[\mathbf a'(x) \times\mathbf a''(x)]$ (eq $1$),

Show that:

$$\mathbf b'(x)=\mathbf a(x)\cdot\bigl[\mathbf a'(x) \times\mathbf a'''(x)\bigr]$$

What I have tried:

I think that to differentiate the RHS of the eq $1$ I use the product rule. So :

$\mathbf b'(x)=\mathbf a''(x)\cdot[\mathbf a(x)\times \mathbf a'(x)] -\mathbf a'(x)\cdot[\mathbf a(x)\times \mathbf a''(x)]$

Answer 1

Hint: Apply the rules for the dot product and the cross product: \begin{align} \mathbf b(x)'&=\mathbf a'(x)\cdot\bigl[\mathbf a'(x) \times\mathbf a''(x)\bigr]+\mathbf a(x)\cdot\bigl[\bigl(\mathbf a'(x) \times\mathbf a''(x)\bigr)'\bigr]\\ &=\mathbf a'(x)\cdot\bigl[\mathbf a'(x) \times\mathbf a''(x)\bigr]+\mathbf a(x)\cdot\bigl[\mathbf a''(x) \times\mathbf a''(x)+\mathbf a'(x) \times\mathbf a'''(x) \bigr]. \end{align}