Let $M$ be a manifold and fix a Lagrangian $L\in C^\infty(T M )$. Let $x_1,\dots x_n$ be local coordinates for $M$ and equip the tangent bundle and cotangent bundle with standard coordinates $(x_1,\dots, x_n, v_1,\dots, v_n)$ and $(x_1,\dots, x_n,\xi_1,\dots, \xi_n)$ respectively. The Legendre transform associated to $L$ is $$\Phi_L: T M\to T^\ast M \ , \ (p,X_p)\mapsto \Phi_{L_p}(p,X_p):=(p,\frac{\partial L}{\partial v}(p,X_p))$$We also define the dual function associated to $L$ which is $$L^\ast:T^\ast M\to\mathbb{R} \, \ (p,\alpha_p)\mapsto \min \left\{\alpha_p(X_p)L(p,X_p) \ , \ (p,X_p)\in T M\right\} $$ Under certain conditions on $L$, we get that $\Phi_L$ is a diffeomorphism with inverse $$\Phi_L^{1}(p,\alpha_p) = (p,X_p) \text{ where $(p,X_p)$ minimizes } \alpha_pL$$It can be shown that $$\Phi_L^{1}=\Phi_{L^\ast}$$My question is about the following claim. $$\text{If } \Phi_L(p,X_p)=(p,\alpha_p)\text{ then } \frac{\partial L}{\partial x}(p,X_p)=\frac{\partial L^\ast}{\partial x}(p,\alpha_p)$$ This is proved in Da Silva's book "Lectures on Symplectic Geometry" in the chapter on Hamiltonian mechanics. I do not understand her proof. Letting $H=L^\ast$, the definition of $H$ is $H(x_1,\dots x_n,\xi_1,\dots,\xi_n)=H(x,\xi)=\xi(x,W)L(x,W)$ where $(x,W)=\Phi_L^{1}(x,\xi)=\Phi_H(x,\xi)$ or equivalently $\xi=\Phi_L(W)$ so that $\xi$ is a function of the $x_i's$ and $v_i's$. She takes the total derivative of both sides of this equation with respect to $x$ to get $$\frac{\partial H}{\partial x}+\frac{\partial H}{\partial \xi}\frac{\partial \xi}{\partial x}=\frac{\partial\xi}{\partial x}v\frac{\partial L}{\partial x}$$ The left hand side is just the chain rule, right? But how is the right hand side obtained? If $\xi$ is dependent on $x$ and $v$ then isn't $v$ dependent on $x$ and $\xi$? I believe that once we have this formula the hypothesis gives us the result as $\frac{\partial H}{\partial \xi_i}(p,\alpha_p)=v^i(p,W_p)$. If someone could walk me through this calculation it would be greatly appreciated, I have spent more time on this then I would like to admit.

1$\begingroup$ The claim is fairly easy to prove when $M=\mathbb{R}$; maybe that case will help you sort things out. $\endgroup$– Brendan MurphyJun 29 '14 at 0:48
I actually was interpreting what she was saying wrong. The coordinates $\xi$ on $T^\ast M$ are of the form $\xi=\Phi_L(x,v)$ so that $\xi_i=\frac{\partial L}{\partial v_i}$ and the Legendre transform is involutive so $\Phi_L^{1}(\xi)=\Phi_{L^\ast}(\xi)=v$. This makes the computation fairly straightforward.