いくつかの方法で導く。

１ ニュートンの第２法則から

Newton's 2nd Low  $${F(x)=m\ddot{x}}$$　（$${F=F(x)}$$とする）
両辺×$${\dfrac{dx}{dt}}$$
　$${\dfrac{dx}{dt}F(x)=\dfrac{dx}{dt}m\ddot{x}}$$
　$${\dfrac{dx}{dt}\Big(\dfrac{d}{dx}\displaystyle\int_0^x F(x)dx\Big)=m\dot{x}\dfrac{d\dot{x}}{dt}}$$
　　　　　　↑ 積分して微分するので元に戻る
　$${\dfrac{d}{dx}\displaystyle\int_0^x F(x)dx=\dfrac{d}{dt}\Big(\dfrac{1}{2}m\dot{x}^2\Big)}$$
よって
　$${\dfrac{d}{dx}\Big(\dfrac{1}{2}m\dot{x}^2-\displaystyle\int_0^x F(x)dx\Big)=0}$$
積分すると
　$${\dfrac{1}{2}m\dot{x}^2-\displaystyle\int_0^x F(x)dx=\bm{const}}$$
　$${\dfrac{1}{2}m\dot{x}^2+\Big(-\displaystyle\int_0^x F(x)dx\Big)=\bm{const}}$$
$${\dfrac{1}{2}m\dot{x}^2=T}$$､$${-\displaystyle\int_0^x F(x)dx=U}$$を代入すると　$${T+U=\bm{const}}$$

２ 保存力から

保存力$${\displaystyle\sum_{j=1}^nF_i=\displaystyle\sum_{j=i}^n\Big(-\dfrac{\partial U}{\partial x_i}\Big)}$$
　$${\displaystyle\sum_{j=1}^nm_j\ddot{x_j}=\displaystyle\sum_{j=i}^n\Big(-\dfrac{\partial U}{\partial x_i}\Big)}$$　← $${F_i=m_i\ddot{x_j}}$$
両辺×$${\dot{x_j}}$$
　$${\displaystyle\sum_{j=1}^nm_j\ddot{x_j}\dot{x_j}=\displaystyle\sum_{j=i}^n\Big(-\dfrac{\partial U}{\partial x_i}\dot{x_j}\Big)}$$
　$${\displaystyle\sum_{j=1}^n\dfrac{d}{dt}\Big(\dfrac{1}{2}m_j\dot{x_j}^2\Big)=\displaystyle\sum_{j=i}^n\Big(-\dfrac{\partial U}{\partial x_i}\dfrac{dx_j}{dt}\Big)}$$
　$${\dfrac{d}{dt}\displaystyle\sum_{j=1}^n\Big(\dfrac{1}{2}m_j\dot{x_j}^2\Big)=\dfrac{1}{dt}\displaystyle\sum_{j=i}^n\Big(-\dfrac{\partial U}{\partial x_i}dx_j\Big)}$$
　$${\dfrac{d}{dt}T=-\dfrac{d}{dt}U}$$　
　$${\dfrac{d}{dt}(T+U)=0}$$　　積分すると　$${T+U=\bm{const}}$$

３ Lagrangeの運動方程式から

$${\dfrac{d}{dt}\Big(\dfrac{\partial\mathscr{L}}{\partial \dot{q_i}}\Big)-\dfrac{\partial\mathscr{L}}{\partial q_i}=0}$$（Lagrande's eq）より　$${\dfrac{\partial\mathscr{L}}{\partial q_i}=\dfrac{d}{dt}\Big(\dfrac{\partial\mathscr{L}}{\partial \dot{q_i}}\Big)}$$　(1)
$${\mathscr{L}(\{q_i\},\{\dot{q_i}\})}$$を全微分で表わすと
　$${d\mathscr{L}=\displaystyle\sum_{i=1}^f\dfrac{\partial\mathscr{L}}{\partial q_i}dq_i+\displaystyle\sum_{i=1}^f\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}d\dot{q_i}}$$　
$${t}$$での微分は
　$${\dfrac{d\mathscr{L}}{dt}=\displaystyle\sum_{i=1}^f\dfrac{\partial\mathscr{L}}{\partial q_i}\dot{q_i}+\displaystyle\sum_{i=1}^f\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}\ddot{q_i}}$$　　← (1)を代入
　　　$${=\displaystyle\sum_{i=1}^f\dfrac{d}{dt}\Big(\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}} \Big)\dot{q_i}+\displaystyle\sum_{i=1}^f\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}\ddot{q_i}=\displaystyle\sum_{i=1}^f\Big\{\dot{q_i}\dfrac{d}{dt}\Big(\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}} \Big)+\ddot{q_i}\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}\Big\}}$$
　　　$${=\displaystyle\sum_{i=1}^f\dfrac{d}{dt}\Big(\dot{q_i}\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}} \Big)=\dfrac{d}{dt}\Big(\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}} \Big)}$$
すなわち
　$${\dfrac{d}{dt}\Big(\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}-\mathscr{L} \Big)=0}$$
積分すると
　$${\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}-\mathscr{L}=\mathbf{const}}$$
$${左辺=\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial\mathscr{L}}{\partial\dot{q_i}}-\mathscr{L}=\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial(T-U)}{\partial\dot{q_i}}-(T-U)}$$
　　$${=\displaystyle\sum_{i=1}^f\dot{q_i}\Big(\dfrac{\partial T}{\partial\dot{q_i}}-\dfrac{\partial U}{\partial\dot{q_i}}\Big)-(T-U)}$$　　← $${\dfrac{\partial U}{\partial\dot{q_i}}=0}$$　
　　$${=\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial}{\partial\dot{q_i}}T-(T-U)=\displaystyle\sum_{i=1}^f\dot{q_i}\dfrac{\partial}{\partial\dot{q_i}}\displaystyle\sum_{i=1}^f\Big(\dfrac{1}{2}m_i\dot{q_i}^2\Big)-(T-U)}$$
　　 $${=\displaystyle\sum_{i=1}^f\dot{q_i}m_i\dot{q_i}-(T-U)=\displaystyle\sum_{i=1}^fm_i\dot{q_i}^2-(T-U)}$$
　　$${=2\displaystyle\sum_{i=1}^f\Big(\dfrac{1}{2}m_i\dot{q_i}^2\Big)-(T-U)=2T-(T-U)=T+U}$$
よって　$${T+U=\bm{const}}$$