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

The norm of the vector $V$ equal to $\|V\| = \sqrt{(-2\theta)^2 + (\theta^2)^2 + (\theta^3)^2}$

Is the derivative of the vector norm given by this?:

$$ \frac{\partial \|V\|}{\partial \theta} = \frac{\frac{\partial f}{\partial \theta}}{2 \sqrt{f}}$$ Where $f= (-2\theta)^2 + (\theta^2)^2 + (\theta^3)^2$

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    So, you are asking whether $$ \frac{\partial \sqrt{f}}{\partial \theta} = \frac{\frac{\partial f}{\partial \theta}}{2 \sqrt{f}}\ ?$$ – 2017-01-25
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    What really I want is there a direct expression of the derivative of the vector norm w.r.t a parameter $\frac{\partial \|V\|}{\partial \theta}$. Or we have to follow the steps that I mentioned in the question to calculate it? – 2017-01-25
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    The formula in my comment is the most direct route there is. – 2017-01-25

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Let $g(\theta)=||V(\theta)||=\sqrt{f(\theta)}$. Then, by the chain rule:

$$g'(\theta)=\frac{1}{2 \sqrt{f(\theta)}}f'(\theta).$$