Page:Annales de mathématiques pures et appliquées, 1829-1830, Tome 20.djvu/282

${\frac {\operatorname {d} S}{\operatorname {d} x}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} x}},\quad {\frac {\operatorname {d} S}{\operatorname {d} y}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} y}}\quad {\frac {\operatorname {d} S}{\operatorname {d} z}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} z}}.\quad$(59)

Si $P$ et $Q$ étant des fonctions de $x,y,z,\ S$ était fonction de $P$ et $Q$, on trouverait

$\left.{\begin{aligned}&{\frac {\operatorname {d} S}{\operatorname {d} x}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} x}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} x}},\\\\&{\frac {\operatorname {d} S}{\operatorname {d} y}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} y}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} y}},\\\\&{\frac {\operatorname {d} S}{\operatorname {d} z}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} z}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} z}}.\end{aligned}}\right\}$(60)

Si $P,Q,R$ étant des fonctions de $x,y,z,\ S$ était une fonction de $P,Q,R,$ on trouverait pareillement

$\left.{\begin{aligned}&{\frac {\operatorname {d} S}{\operatorname {d} x}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} x}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} x}}+{\frac {\operatorname {d} S}{\operatorname {d} R}}{\frac {\operatorname {d} R}{\operatorname {d} x}},\\\\&{\frac {\operatorname {d} S}{\operatorname {d} y}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} y}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} y}}+{\frac {\operatorname {d} S}{\operatorname {d} R}}{\frac {\operatorname {d} R}{\operatorname {d} y}}\,;\\\\&{\frac {\operatorname {d} S}{\operatorname {d} z}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} z}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} z}}+{\frac {\operatorname {d} S}{\operatorname {d} R}}{\frac {\operatorname {d} R}{\operatorname {d} z}}\,;\end{aligned}}\right\}$(61)

et ainsi de suite ; de sorte qu’en général, pour obtenir les coefficiens différentiels partiels d’une fonction de tant de quantités qu’on voudra, qui sont elles-mêmes fonctions d’un nombre quelconque de variables, il suffit d’opérer tour à tour, par rapport à chaque variable, comme si toutes les autres étaient des constantes.

Si, dans la formule (54), on suppose $Q=x,$ elle deviendra

${\frac {\operatorname {d} S}{\operatorname {d} x}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} x}}+{\frac {\operatorname {d} S}{\operatorname {d} x}}\,;\qquad$(62)