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

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.