Page:Annales de mathématiques pures et appliquées, 1829-1830, Tome 20.djvu/277

$G={\frac {\operatorname {d} P}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+\ldots \,;\quad H={\frac {\operatorname {d} Q}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}Q}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+\ldots \,;$

mais $P$ et $Q$ devenant respectivement $P+G$ et $Q+H,\ S$ doit devenir

${\begin{alignedat}{1}S+{\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {G}{1}}&+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P^{2}}}{\frac {G^{2}}{1.2}}+\ldots \\\\+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {H}{1}}&+2{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P\operatorname {d} Q}}{\frac {GH}{1.2}}+\ldots \\\\&+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} Q^{2}}}{\frac {H^{2}}{1.2}}+\ldots \end{alignedat}}$

ou bien, en substituant pour $G$ et $H$ leurs valeurs, développant et ordonnant,

$S+\left({\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} x}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} x}}\right){\frac {g}{1}}$

$+\left\{{\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} ^{2}P}{\operatorname {d} x^{2}}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} ^{2}Q}{\operatorname {d} x^{2}}}+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P^{2}}}\left({\frac {\operatorname {d} P}{\operatorname {d} x}}\right)^{2}\right.$

$\left.+2{\frac {\operatorname {d} ^{2}S}{\operatorname {d} P\operatorname {d} Q}}{\frac {\operatorname {d} P}{\operatorname {d} x}}{\frac {\operatorname {d} Q}{\operatorname {d} x}}+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} Q^{2}}}\left({\frac {\operatorname {d} Q}{\operatorname {d} x}}\right)^{2}\right\}{\frac {g^{2}}{1.2}}+\ldots \,;$

mais, d’un autre côté, par l’intermédiaire de $P$ et $Q,\ S$ étant aussi fonction de $x\,;$ lorsque $x$ se change en $x+g,\ S$ doit devenir

$S+{\frac {\operatorname {d} S}{\operatorname {d} x}}{\frac {g}{1}}+{\frac {\operatorname {d} ^{2}S}{\operatorname {d} x^{2}}}{\frac {g^{2}}{1.2}}+\ldots ,$

et ce dernier développement doit, quel que soit $g,$ être identique avec celui qui le précède. On doit donc avoir

${\frac {\operatorname {d} S}{\operatorname {d} x}}={\frac {\operatorname {d} S}{\operatorname {d} P}}{\frac {\operatorname {d} P}{\operatorname {d} x}}+{\frac {\operatorname {d} S}{\operatorname {d} Q}}{\frac {\operatorname {d} Q}{\operatorname {d} x}}.\qquad$(54)

Si $S$ était fonction de trois quantités $P,Q,R$ qui fussent elles-mêmes des fonctions de $x$, on trouverait semblablement