I'm having difficulties with this exercise (from Elon LIMA's Curso de Análise, Vol. 2):
$f:U\longrightarrow\mathbb{R}$ is function, differentiable on the open set $U\subset\mathbb{R}^n$. Let $\{v_1,...,v_n\}$ be an arbitrary basis of $\mathbb{R}^n$ and $g^{ij}:=\left
$ . Show that grad$f(x)$ in this basis is $\textrm{grad} f(x) = \sum_i(\sum_j g^{ij}\frac{\partial f}{\partial v_j})v_i$ where $\frac{\partial f}{\partial v_j}$ is the directional derivative of $f$ along the vector $v_j$.
I tried starting with the RHS with $\frac{\partial f}{\partial v_j}=\left<\textrm{grad}f(x),v_j\right>$ and arriving at $\sum_i \left<\textrm{grad}f(x),e_i\right>e_i $ which is the "canonical" (see Note below) expression for the grad; writing each $v_i$ as $v_i=\sum_j \left
Am I approaching it incorrectly? How must this problem be tackled? I'm obviously missing something, or misinterpreting things.
NOTE: In the textbook, grad$f(x)$ is defined as $\sum_i \frac{\partial f}{\partial x_i}(x)e_i$ where $\frac{\partial f}{\partial x_i}(x)$ is the usual i-th partial derivative at the point $x$.