Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/254

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DÉVELOPPEMENT

${\frac {\operatorname {d} U}{\operatorname {d} a}}{\frac {\operatorname {d} V}{\operatorname {d} b}}={\frac {\operatorname {d} U}{\operatorname {d} b}}{\frac {\operatorname {d} V}{\operatorname {d} a}}.\qquad (3)$

L’on a encore

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}=V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} b\operatorname {d} a}}+{\frac {\operatorname {d} V}{\operatorname {d} b}}{\frac {\operatorname {d} U}{\operatorname {d} a}},$

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} b}}}{\operatorname {d} a}}=V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} b}}+{\frac {\operatorname {d} V}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} b}}\,;$

mais, en vertu de l’équation (3),

$V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} b}}=V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} b\operatorname {d} a}},\quad$et$\quad {\frac {\operatorname {d} V}{\operatorname {d} b}}{\frac {\operatorname {d} U}{\operatorname {d} a}}={\frac {\operatorname {d} V}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} b}}\,;$

donc finalement

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} b}}}{\operatorname {d} a}}\,;\qquad$(4)

et l’on aurait semblablement

${\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} b}}}{\operatorname {d} a}}.$

Si l’on avait

$x=\operatorname {f} (a,b,x,x'),\qquad x_{1}=\operatorname {f_{1}} (a_{1},b_{1},x,x_{1})\,;$

et que $U,V$ fussent des fonctions quelconques de $x,x_{1}$, sans $a,b,a_{1},b_{1},$ on trouverait, comme dans ce qui précède,