Derivative of a vector norm w.r.t. a parameter
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Let $g(\theta)=||V(\theta)||=\sqrt{f(\theta)}$. Then, by the chain rule:
$$g'(\theta)=\frac{1}{2 \sqrt{f(\theta)}}f'(\theta).$$
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user137684
Updated on January 26, 2020Comments
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user137684 almost 4 years
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 $f= (-2\theta)^2 + (\theta^2)^2 + (\theta^3)^2$