Implicit differentiation and adjoint formulas under a nonsingular residual Jacobian

If $m=0$, the state space is a singleton, its unique linear map is its own two-sided inverse, and the sole branch has zero derivative, as does the displayed formula. If $m\ge 1$ and $p=0$, the parameter space is a singleton. Write $A=R_z({\theta }_0,z_0)$. Differentiability gives $R({\theta }_0,z_0+v)=Av+o(\mathrm{\parallel }v{\mathrm{\parallel }}_2)$; since $\mathrm{\parallel }Av{\mathrm{\parallel }}_2\ge \mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{/}\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2$, this has no other zero for sufficiently small nonzero $v$. Thus the singleton branch is locally unique and has zero derivative, again the displayed formula. If $p,m\ge 1$, apply [F1] to obtain the local $C^1$ branch and $D\varphi =-R_z^{-1}{R}_{\theta }$.

The map $\theta \mapsto J(\theta ,\varphi (\theta ))$ is $C^1$. By [F2], its derivative at ${\theta }_0$ is $J_{\theta }+J_{z}D\varphi$. Substituting step 1.1 yields $J_{\theta }-J_z{R}_z^{-1}{R}_{\theta }$.

Since $R_z$ is invertible, $R_z^T$ is invertible. The adjoint equation has the unique solution satisfying ${\lambda }^T=J_z{R}_z^{-1}$, by transposing $R_z^T\lambda =J_z^T$. Therefore $J_z{R}_z^{-1}{R}_{\theta }={\lambda }^T{R}_{\theta }$, giving the stated adjoint formula.