computing the gradient $f(\vec{x}) = \vec{x}^T\vec{x}+c$

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Could a helpful soul tell me if this approach is correct?

I need to compute the gradient $\nabla f$ and find the result and deriviation of $f(\vec{x}) = \vec{x}^T\vec{x}+c$


To solve this, I'm using a table in my textbook (First course in Machine Learning, 2nd edition, p. 23)): enter image description here

If above table can be used as I imagine, I get:

$f(\vec{x}) = \vec{x}^T\vec{x}+c \leftrightarrow \frac{\partial f}{\partial \vec{x}}(2\vec{x}+c)$.

Calculating $ \frac{\partial f}{\partial \vec{x}}(2\vec{x}+c) = 2$


The result is 2 and the derivative is $ \frac{\partial f}{\partial \vec{x}}(2\vec{x}+c)$

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$f(x+h) = f(x)+ 2 x^T h + h^Th$ from which we can read off the derivative as $Df(x)h = 2 x^T h$, or $\nabla f(x) = 2 x$.