Find all first-order partial derivative of the given function defined on $R^n$.

Question

$$f(x)=\sum_{i=1}^n \sum_{j=1}^n a_{ij} x_i x_j$$ given $a_{ij}=a_{ji}$ my atempt. let $g(t)=f(x+te_k)$ where $e_k=$the k-th unite coordinate vector. Then computed $g'(0)=f'(x;e_k)=\sum_{i=1}^n \sum_{j=1}^n a_{ij}(x_i+x_j)$. But the ans is given as $2 \sum_{j=1}^n a_{kj}x_j$. how to solve? need a short hint.

Answer 1

Note first that the $i$th component of $x+te_k$ is $x_i$ when $i\ne k$ and $x_i+t$ when $i=k$. We have (using symmetry of $a_{ij}$ in the second term)

$$g(t) = f(x+te_k) = \sum\limits_{i,j\ne k} a_{ij}x_ix_j + 2\sum_{j\ne k} a_{kj}(x_k+t)x_j + a_{kk}(x_k+t)^2.$$ Therefore, $g'(t) = 2\sum_{j\ne k} a_{kj}x_j + 2a_{kk}(x_k+t)$, and so $$f'(x;e_k) = g'(0) = 2\sum_{j\ne k} a_{kj}x_j + 2a_{kk}x_k = 2\sum_j a_{kj}x_j.$$