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

The total derivative is a linear map $Df(a):{\mathbb{R}}^n\to {\mathbb{R}}^n$ such thatIt is unique: the difference of two such linear maps, evaluated on $h=tv$, must vanish on every fixed vector $v$ after division by $t$ and passage to the limit.

The directional derivative in the specified vector direction isThe equality follows from the total-derivative remainder estimate; for the zero direction the result is zero. The convention keeps the supplied vector and does not replace it by a unit vector.

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 $t{e}_j$. Thus the matrix of the total derivative is the Jacobian matrix, with entry $(i,j)$ equal to $\partial f_i/\partial x_j(a)$.

Applying the chain rule to both inverse identities givesFor the unnumbered calculation, the requested point is $f(c,c)$. The Jacobian there isTo apply its inverse to $(1/2,1/2)$, solve $e^c(v_1+v_2)=1/2$, $e^c{v}_1-e^{-c}{v}_2=1/2$. Subtraction gives $v_2=0$, and henceFor precision, the displayed example maps bijectively onto $V=\{(s,t):s>0,t>0,st>1\}$, rather than the entire positive quadrant. Indeed writing $q=e^x$, the equations are $(s-q)(t-q)=1$ with $0<q<\min (s,t)$. This left side decreases continuously from $st$ to zero, giving exactly one $q$ if and only if $st>1$, after which $e^y=s-q$ is unique. Its Jacobian never vanishes, so the inverse is differentiable on $V$. The requested point lies in $V$, making the calculation well-defined.