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Let say that we have the vector $V(\theta)=[-2\theta \hspace{0.2cm} \theta^2 \hspace{0.2cm} \theta^3]^T$, and the elements of $V$ are differentiable functions of $\theta$ and The norm of the vector $V$ equal to $\|V(\theta)\| = \sqrt{(-2\theta)^2 + (\theta^2)^2 + (\theta^3)^2}$.

Is there a direct expression of the derivative of the following formula w.r.t the parameter $\theta$:

$$ \frac{\partial}{\partial \theta} \left (\frac{V(\theta)}{\|V(\theta)\|} \right)$$

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    Yes, you merely have to take the ordinary derivative of each component, each being a function of the single variable $\theta$. So$\frac{\partial}{\partial \theta} \left (\frac{V(\theta)}{\|V(\theta)\|} \right)=\frac{d}{d \theta} \left (\frac{V(\theta)}{\|V(\theta)\|} \right)$. – 2017-02-08
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    In fact I have tried that procedure by using a simple example with $V=[1\, \theta]^T$ and I also used the product rule $\frac{V'(\theta) \|V(\theta)\|-V(\theta)\|V(\theta)\|'}{\|V(\theta)\|^2} $ and the result is the same. So Seems like product rule works in this case as well, the question : is the product rule can be applicable for any vector $V(\theta)$. – 2017-02-09

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Let $\,\lambda=\|v\|\,\,$ denote the length of the vector, and find its differential $$\eqalign{ \lambda^2 &= v^Tv \cr 2\lambda\,d\lambda &= 2v^Tdv \implies d\lambda=\frac{v^Tdv}{\lambda} \cr }$$ Express the unit vector in terms of $\lambda$, then find its differential and derivative $$\eqalign{ s &= \frac{v}{\lambda} \cr\cr ds &= \frac{\lambda\,dv-v\,d\lambda}{\lambda^2} \cr &= \frac{1}{\lambda}(dv-ss^Tdv) \cr &= \frac{1}{\lambda}(I-ss^T)\,dv \cr\cr \frac{ds}{d\theta} &= \Big(\frac{I-ss^T}{\lambda}\Big)\,\frac{dv}{d\theta} \cr }$$