![{\displaystyle {\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c2b77520f6ae0587e0544e002307e4f1387a239)
(59)
Si
et
étant des fonctions de
était fonction de
et
, on trouverait
![{\displaystyle \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\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18b76a9660d52ecc1379fbf3ea9222339c0993c1)
(60)
Si
étant des fonctions de
était une fonction de
on trouverait pareillement
![{\displaystyle \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\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85b38d20763402fb7f965ca92934aa7023fc2005)
(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
elle deviendra
![{\displaystyle {\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/852bdfd7ebaeba4df316ae4a2ccc677f57a3dc38)
(62)