Imagine I have the following function
$$ \vec{f}(\vec{x}) = x \vec{x}, x = | \vec{x} |, \vec{x} \in R^3 $$
That is, the function is essentially a quadratic function, but contains a vector direction as well. Intuitively from single variable calculus I would expect the gradient $ \nabla \vec{f} = (\partial \vec{f}/ \partial x_1,\partial \vec{f}/ \partial x_2,\partial \vec{f}/ \partial x_3) $ to be proportional to $2x$, however I also would expect it to be a 3×3 matrix.
My most naive attempt would be to do
$$ \vec{f} = x_1^2 \vec{e}_1 + x_2^2 \vec{e}_2 + x_3^2 \vec{e}_3 $$
and say that
$$ \nabla \vec{f} = 2 x_1 \vec{e}_1 + 2 x_2 \vec{e}_2 + 2 x_3 \vec{e}_3 $$
But it would mean that every gradient w.r.t. a vector would always be a diagonal matrix, which seems wrong to me. What I really want to create is the Jacobian $ \partial \vec{f}_i / \partial x_j $ but I think I get a little bit confused about what I do with the base vectors $ \vec{e_i} $ during the partial derivative.
Best Answer
You sound like you just want the Jacobian but are trying to force it into a singular gradient vector, and I think the main part you are getting confused on is how to differentiate that initial function f. Don't bother thinking of them as vectors for a second and just go through them normally for multivariable, treating all variables which you are not differentiating in regard to as just constants:
$\vec{f} = [f_1, f_2, f_3]^T$
$f_1(\vec{x}) = (x_1^2 + x_2^2 + x_3^2)^{1/2} * x_1$
$\frac{\partial f_1(\vec{x})}{\partial x_1} = (x_1^2 + x_2^2 + x_3^2)^{1/2} +\frac{x_1}{2(x_1^2 + x_2^2 + x_3^2)^{1/2}} * 2*x_1 = (x_1^2 + x_2^2 + x_3^2)^{1/2} +\frac{x_1^2}{(x_1^2 + x_2^2 + x_3^2)^{1/2}}$ by the chain rule.
By analogy, you can get $\frac{\partial f_2(\vec{x})}{\partial x_2}$ and $\frac{\partial f_3(\vec{x})}{\partial x_3}$. For the others, let's look at $\frac{\partial f_1(\vec{x})}{\partial x_2}$ for an example:
$\frac{\partial f_1(\vec{x})}{\partial x_2} = \frac{x_1*x_2}{(x_1^2 + x_2^2 + x_3^2)^{1/2}}$
To summarize, if $i = j$:
$\frac{\partial f_i(\vec{x})}{\partial x_j} = x +\frac{x_j^2}{x}$
Else:
$\frac{\partial f_i(\vec{x})}{\partial x_j} = \frac{x_i*x_j}{x}$
You can then plug this into a Jacobian form