Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 18

Regard $u=u(s,t(s,v))$. Applying the multivariable chain rule while holding $v$ fixed givesEach subscript records the variable held fixed. A differentiable local coordinate change from $(s,t)$ to $(s,v)$ requires $(\partial v/\partial t{)}_s\ne 0$; the question's notation $t=t(s,v)$ presupposes such a branch.

For an exact differential $Pdx+Qdy$, equality of mixed partial derivatives requires $P_y=Q_x$. Here this givesOn an open two-dimensional domain the coefficient relation is $x{\mu }^{\prime }+3\mu =0$, so $\mu =C{x}^{-3}$ on a connected interval avoiding zero. The normalization gives $C=1$ on the positive interval through $x=1$. Integrating $f_x=2x^{-3}\sin y$ and checking $f_y$ yieldsThere is no smooth extension of this normalized integrating factor across $x=0$; an independently specified factor on $x<0$ is not fixed by data at one.

Let $a=f(b,c)$ and work where both $f_b$ and $f_c$ are nonzero, so the required inverse partial derivatives exist. At fixed $a$, the total differential gives $0=f_{b}db+f_{c}dc$, hence $(\partial b/\partial c{)}_a=-f_c/f_b$. At fixed $b$, $(\partial c/\partial a{)}_b=1/f_c$. Their product with $f_b$ is $-1$: this is the cyclic partial derivative identity.

For explicit verification in this example, put $P=b^4-6b^2{c}^2+c^4$ and $H=(b^2+c^2{)}^3$. Direct differentiation givesThereforeThe identity is local to regular branches: for these explicit formulas require $bcP\ne 0$. At singular points one cannot justify cancellation of nonexistent inverse derivatives merely by writing a formal product.