# Ex 1.25 Properties of Dt, 3

Structure and Interpretation of Classical Mechanics

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Demonstrate that …

d . $\displaystyle{D_t (H \circ G) = (DH \circ G) D_t G}$

~~~ \displaystyle{ \begin{aligned} &D_t (HG) \circ \Gamma[q] (t) \\ \end{aligned} } \displaystyle{ \begin{aligned} &= \partial_0 \left[ (H \circ G) (t, q, v, a, ...) \right] \\ &+ \partial_1 \left[ (H \circ G) (t, q, v, a, ...) \right] v(t) \\ &+ \partial_2 \left[ (H \circ G) (t, q, v, a, ...) \right] a(t) + ... \\ \end{aligned} } \displaystyle{ \begin{aligned} &= \partial_0 \left[ H (G (t, q, v, a, ...)) \right] \\ &+ \partial_1 \left[ H (G (t, q, v, a, ...)) \right] v(t) \\ &+ \partial_2 \left[ H (G (t, q, v, a, ...)) \right] a(t) + ... \\ \end{aligned} } \displaystyle{ \begin{aligned} &= D (H (G (t, q, v, a, ...))) \partial_0 G (t, q, v, a, ...) \\ &+ D (H (G (t, q, v, a, ...))) \partial_1 G (t, q, v, a, ...) v(t) \\ &+ D (H (G (t, q, v, a, ...))) \partial_2 G (t, q, v, a, ...) a(t) + ... \\ \end{aligned} } \displaystyle{ \begin{aligned} &= D (H (G (t, q, v, a, ...))) \left[ \partial_0 G (t, q, ...) + \partial_1 G (t, q, ...) v(t) + \partial_2 G (t, q, ...) a(t) + ... \right] \\ \end{aligned} } \displaystyle{ \begin{aligned} &= D ( (H \circ G) \circ \Gamma[q] (t) ) D_t G \circ \Gamma[q] (t) \\ \end{aligned} } \displaystyle{ \begin{aligned} &= D_G (H \circ G) \circ \Gamma[q] (t) D_t G \circ \Gamma[q] (t) \\ \end{aligned} } $\displaystyle{ (fg) [q] \equiv f[q] g[q]}$ \displaystyle{ \begin{aligned} &= \{ [D_G (H \circ G)] D_t G \} \circ \Gamma[q] (t) \\ \end{aligned} }

— Me@2022-05-23 03:18:43 PM

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