The $j$th column of the <matrix> of $Df(a)$ in the standard bases is $Df(a)e_j$. Its $i$th component is $\partial f_i/\partial x_j(a)$, found by taking increments $te_j$. Thus \b[the <matrix> of the <total derivative> is the <Jacobian matrix>], with entry $(i,j)$ equal to $\partial f_i/\partial x_j(a)$.

Solution

Past exam of the options I course of the University of Oxford / 2005 / F / 1 / iii / Solution

 Solution (source code)