Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/258

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DÉVELOPPEMENT

${\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}=k{\frac {\operatorname {d} x}{\operatorname {d} a}}\,;$

d’où on conclut, au moyen de la première,

${\frac {\operatorname {d} x}{\operatorname {d} a}}={\frac {\operatorname {f} '(a+bz)}{1-b\left({\frac {\operatorname {d} z}{\operatorname {d} x}}+k{\frac {\operatorname {d} z}{\operatorname {d} x_{1}}}\right)\operatorname {f} '(a+bz)}}.$

On trouvera semblablement

${\frac {\operatorname {d} x_{1}}{\operatorname {d} b}}=k{\frac {\operatorname {d} x}{\operatorname {d} b}},$

${\frac {\operatorname {d} x}{\operatorname {d} b}}={\frac {z\operatorname {f} '(a+bz)}{1-b\left({\frac {\operatorname {d} z}{\operatorname {d} x}}+k{\frac {\operatorname {d} z}{\operatorname {d} x_{1}}}\right)\operatorname {f} '(a+bz)}}.$

Ces équations, comparées aux précédentes, donnent

$\left.{\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} b}}=z{\frac {\operatorname {d} x}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} b}}=z{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}.\end{aligned}}\right\}\quad$(12)

En suivant la marche qui a conduit à celles-ci, nous trouverons semblablement

$\left.{\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} b_{1}}}=z_{1}{\frac {\operatorname {d} x}{\operatorname {d} a_{1}}},\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} b_{1}}}=z_{1}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a_{1}}}.\end{aligned}}\right\}\quad$(13)

En observant donc que