Derivative of a local inverse map ($Dg(y)=[Df(g(y)){]}^{-1}$)

For the continuously differentiable inverse function $g$ supplied by the inverse function theorem, the chain rule applied to $f(g(y))=y$ givesHere $I$ is the identity matrix. The matrix inverse is taken at the recovered input $g(y)$, not at the output $y$. In one dimension this is the inverse function rule $g^{\prime }(y)=1/f^{\prime }(g(y))$.

The formula also describes sensitivity. For a small output perturbation $h$, differentiability gives $g(y+h)-g(y)=Dg(y)h+o(\Vert h\Vert )$. A large operator norm of $Dg(y)$ permits large amplification in some directions. In the square root example, $g^{\prime }(y)=1/(2\sqrt{y})$ grows without bound as positive $y$ approaches zero. Existence of a local inverse function therefore does not imply that recovery is insensitive to measurement error.

This map is a diffeomorphism from ${\mathbb{R}}^2$ onto $V=\{(s,t):s>0,t>0,st>1\}$. Writing $q=e^x$, the inverse problem becomes $(s-q)(t-q)=1$ with $0<q<\min (s,t)$. The left side decreases strictly from $st$ to zero, proving existence and uniqueness exactly on $V$. Its Jacobian determinant is $-e^x(e^y+e^{-y})\ne 0$, so the inverse function theorem supplies a differentiable inverse. Along $(x,y)=(c,c)$, applying the inverse Jacobian to $(1/2,1/2)$ gives $(e^{-c}/2,0)$. Establishing the actual image matters before using a global inverse, even when a local inverse already exists at the point of interest.

Let a three-times continuously differentiable real function $f(g)$ have $df/dg\ne 0$. The inverse function theorem gives a local inverse $g(f)$. Writing $p=df/dg$, $q=d^{2}f/d{g}^2$ and $r=d^{3}f/d{g}^3$, its derivatives areIndeed $d/df=p^{-1}d/dg$ by the chain rule. Apply this operator successively to $g$, $p^{-1}$ and $-q{p}^{-3}$. Every forward derivative on the right is evaluated at the inverse image $g(f)$.