Derivative of $f(x)\cdot x$ where $f$ is a $\mathbb{R}^3$ function: how to compute it? Let $f:\mathbb{R}^3\to\mathbb{R}^3$. Could someone please help me to compute the the derivative of
$$f(x)\cdot x?$$
If $f$ would be a real valued function I think it should be the derivative of a product, but how to proceed in this case?
Thank you in advance.
 A: With $f=(f_1,f_2,f_3)$ and $x= (x_1,x_2,x_3)$ we have
$$f(x)\cdot x=f_1(x)x_1+f_2(x)x_2+f_3(x)x_3.$$
Can you proceed ?
A: I notice that you are still unsure about how to proceed with Fred's answer, so I will write out a complete answer.
As $f$ maps vectors in $\mathbb{R}^3$ to another vector in $\mathbb{R}^3$, the derivative is defined as follows:
$$Df = \begin{pmatrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3}\\\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3}\\\ \frac{\partial f_3}{\partial x_1} & \frac{\partial f_3}{\partial x_2} & \frac{\partial f_3}{\partial x_3}\end{pmatrix}$$
This definition can be extended to any "nice" function $g: \mathbb{R}^k \rightarrow \mathbb{R}^q$ for any positive, finite integers $p,q$ assuming that these partial derivatives all exist.
Using this we can now solve your problem.
Let's define a new function $h(x) = f(x) \cdot x $ and apply this definition above.
Using the definition of the derivative in general (above), we notice that $h: \mathbb{R}^3 \rightarrow \mathbb{R}$. Therefore, inputting this into the definition of the derivative gives us the following formula:
$$Dh(x) = \begin{pmatrix}\frac{\partial h}{\partial x_1} & \frac{\partial h}{\partial x_2} & \frac{\partial h}{\partial x_3}\end{pmatrix} $$
And so by calculating the partial derivatives of $h(x) = (f(x) \cdot x)$, we arrive at an answer for the derivative.
