Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 12 / a

Assume $F$ is continuously differentiable near the point of interest on $F=0$, and that the relevant denominators below are nonzero. The total differential on the surface isHolding $z$ fixed gives $F_{x}dx+F_{y}dy=0$, so the implicit function theorem givesWhen $F_x{F}_y\ne 0$, their product proves the reciprocity rule for constrained partial derivatives:Similarly, holding $y$ or $x$ fixed gives $(\partial x/\partial z{)}_y=-F_z/F_x$ and $(\partial z/\partial y{)}_x=-F_y/F_z$. With $F_x{F}_y{F}_z\ne 0$, multiply the three ratios:This is the cyclic partial derivative identity. Nonzero denominators and local differentiable coordinate representations are substantive assumptions: at a singular coordinate representation, a formal cancellation of zero or infinite factors is not a valid proof.

Differentiate the defining function with the other coordinates independent. Put $C=\cosh (x+z)$ for brevity. ThenThe implicit differentiation ratios therefore giveEach expression applies on the implicit surface wherever its own denominator is nonzero. The hyperbolic cosine appears because the derivative of $\sinh (x+z)$ with respect to either $x$ or $z$ is $\cosh (x+z)$.