Work-energy balance $\frac{d}{dt}(K+U(t,x(t)))={\mathrm{\partial }}_{t}U(t,x(t))+{\sum }_i⟨d_i(t),x_i^{\mathrm{\prime }}(t)⟩$ for forces $m_i{x}_i^{\mathrm{\prime }\mathrm{\prime }}=-{\mathrm{\nabla }}_{x_i}U+d_i$

Each $x_i^{\mathrm{\prime }}$ is differentiable on $I$ because $x$ is twice differentiable, so [L1] gives differentiability of $t\mapsto \parallel x_i^{\mathrm{\prime }}(t){\parallel }_2^2$ with derivative $2⟨x_i^{\mathrm{\prime }\mathrm{\prime }}(t),x_i^{\mathrm{\prime }}(t)⟩$; by the sum and scalar rules of [L1] and linearity of the finite sum [L4], $K$ is differentiable with $K^{\mathrm{\prime }}=\frac{1}{2}{\sum }_i{m}_i\cdot 2⟨x_i^{\mathrm{\prime }\mathrm{\prime }},x_i^{\mathrm{\prime }}⟩={\sum }_i⟨m_i{x}_i^{\mathrm{\prime }\mathrm{\prime }},x_i^{\mathrm{\prime }}⟩,$ which is clause 1.

The curve $\gamma :I\to \mathbb{R}\times Q$, $\gamma (t):=(t,x(t))$, is differentiable (its components are the identity and the components of $x$; The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral), with ${\gamma }^{\mathrm{\prime }}(t)=(1,x^{\mathrm{\prime }}(t))$ under the identification $\mathbb{R}\times {\mathbb{R}}^{3N}={\mathbb{R}}^{1+3N}$. Since $\widehat{U}$ is $C^1$ on the open neighborhood $J\times Q$ of $\gamma (I)$, [L2] gives the usual chain rule at interior points. At an included endpoint, $\gamma (t+h)=\gamma (t)+h{\gamma }^{\mathrm{\prime }}(t)+o(h)$ as $h\to 0$ relative to $I$, and total differentiability gives $\widehat{U}(\gamma (t)+v)=\widehat{U}(\gamma (t))+D\widehat{U}(\gamma (t))v+o(\mathrm{\mid }v\mathrm{\mid })$. Substituting the first expansion into the second proves the same relative derivative there. Thus $(U\circ \gamma {)}^{\mathrm{\prime }}(t)=D\widehat{U}(\gamma (t)){\gamma }^{\mathrm{\prime }}(t)$, and [L3] evaluates the right side as the matrix product of the Jacobian of $U$ at $\gamma (t)$ with $(1,x^{\mathrm{\prime }}(t))$, that is $\frac{d}{dt}U(t,x(t))={\mathrm{\partial }}_{t}U(t,x(t))+{\sum }_i{\sum }_{k<3}{\mathrm{\partial }}_{i,k}U(t,x(t))x_{i,k}^{\mathrm{\prime }}(t)={\mathrm{\partial }}_{t}U(t,x(t))+{\sum }_i⟨{\mathrm{\nabla }}_{x_i}U(t,x(t)),x_i^{\mathrm{\prime }}(t)⟩,$ the last step regrouping the coordinates of $Q$ into the $N$ blocks ${\mathbb{R}}^3$ and using the coordinate formula for the inner product [L4]; this is clause 2.

Adding the identities of clauses 1 and 2 and substituting $m_i{x}_i^{\mathrm{\prime }\mathrm{\prime }}=-{\mathrm{\nabla }}_{x_i}U(t,x(t))+d_i$, the two occurrences of ${\sum }_i⟨{\mathrm{\nabla }}_{x_i}U(t,x(t)),x_i^{\mathrm{\prime }}(t)⟩$ cancel, leaving clause 3. If $d_i\equiv 0$ the derivative is ${\mathrm{\partial }}_{t}U(t,x(t))$, and if also ${\mathrm{\partial }}_{t}U\equiv 0$ then the derivative of $K+U(t,x(t))$ vanishes identically on the interval $I$, so that function is constant by [L5].