I am stuck on the following problem:
Let $f\colon \Bbb R^3 \to \Bbb R^3 $ be defined by $f(x_1,x_2,x_3)=(x_2+x_3,x_3+x_1,x_1+x_2).$ Then the first derivative of $f$ is :
1.not invertible anywhere
2.invertible only at the origin
3.invertible everywhere except at the origin
4.invertible everywhere.
My problem is I do not know how to calculate the derivative of $f$. Can someone point me in the right direction?
HINT: the derivative is a linear transformation $f':\mathbb{R^3}\rightarrow\mathbb{R}^3$ such that its matrix is
$$[f']=\left( \begin{array}{ccc} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3} \\ \frac{\partial f_3}{\partial x_1} & \frac{\partial f_3}{\partial x_2} & \frac{\partial f_3}{\partial x_3} \end{array} \right)$$
The question is most likely looking for the Jacobian:
$$J=f'(x_1,x_2,x_3)=\begin{bmatrix}\frac{\partial f_1}{\partial{x_1}} & \frac{\partial f_1}{\partial{x_2}} & \frac{\partial f_1}{\partial{x_3}}\\\frac{\partial f_2}{\partial{x_1}} & \frac{\partial f_2}{\partial{x_2}} &\frac{\partial f_2}{\partial{x_1}}\\\frac{\partial f_3}{\partial{x_1}} & \frac{\partial f_3}{\partial{x_2}} &\frac{\partial f_3}{\partial{x_3}}\end{bmatrix}$$
This seems like a lot of work, but since this is a linear transformation, the Jacobian will be identical to the transformation matrix itself.
The derivative of $f$ will be a three-by-three matrix, where the mnth entry is $\frac{\partial f_m}{\partial x_n}$.
This means row 1 column 1 entry will be 0, row 1 column 2 will be 1. row 1 column 3 entry will be 1 etc.
