![{\displaystyle G={\frac {\operatorname {d} P}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+\ldots \,;\quad H={\frac {\operatorname {d} Q}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}Q}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13d9a370ec736334d729525bbd0b1544eebd2982)
mais
et
devenant respectivement
et
doit devenir
![{\displaystyle {\begin{alignedat}{1}S+{\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {G}{1}}&+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P^{2}}}{\frac {G^{2}}{1.2}}+\ldots \\\\+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {H}{1}}&+2{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P\operatorname {d} Q}}{\frac {GH}{1.2}}+\ldots \\\\&+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} Q^{2}}}{\frac {H^{2}}{1.2}}+\ldots \end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e14ab59f2c176601d1d518f2ccc39fd66f9d859)
ou bien, en substituant pour
et
leurs valeurs, développant et ordonnant,
![{\displaystyle S+\left({\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}}\right){\frac {g}{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb6492e26f8d55a83cf49d7ed670dbb175df3635)
![{\displaystyle +\left\{{\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} ^{2}Q}{\operatorname {d} x^{2}}}+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P^{2}}}\left({\frac {\operatorname {d} P}{\operatorname {d} x}}\right)^{2}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddf2f05b532e741bdcf4e7ef6650cd99c1820a46)
![{\displaystyle \left.+2{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P\operatorname {d} Q}}{\frac {\operatorname {d} P}{\operatorname {d} x}}{\frac {\operatorname {d} Q}{\operatorname {d} x}}+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} Q^{2}}}\left({\frac {\operatorname {d} Q}{\operatorname {d} x}}\right)^{2}\right\}{\frac {g^{2}}{1.2}}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0482135d344fda5ad4fed9809ed50bb2d52697e1)
mais, d’un autre côté, par l’intermédiaire de
et
étant aussi fonction de
lorsque
se change en
doit devenir
![{\displaystyle S+{\frac {\operatorname {d} S}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f68520e05fee8c00e1449ef6c3a9f77edaea13c0)
et ce dernier développement doit, quel que soit
être identique avec celui qui le précède. On doit donc avoir
![{\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} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} x}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/acb25b6114b7fd5bb3e08c556719c16cb6ce80c0)
(54)
Si
était fonction de trois quantités
qui fussent elles-mêmes des fonctions de
, on trouverait semblablement