Page:Annales de mathématiques pures et appliquées, 1814-1815, Tome 5.djvu/144

138

ESSAI SUR LES PRINCIPES

${\frac {\operatorname {d} }{u}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{m}\right\}=(\varphi p)^{m}.{\frac {\operatorname {d} }{x}}{\frac {\operatorname {d} }{u}}\operatorname {F} p+{\frac {\operatorname {d} }{x}}\operatorname {F} p.{\frac {\operatorname {d} }{u}}(\varphi p)^{m}\,;$

et comme, d’après (125), (126),

${\frac {\operatorname {d} }{x}}{\frac {\operatorname {d} }{u}}\operatorname {F} p=\varphi p.{\frac {\operatorname {d} ^{2}}{x}}\operatorname {F} p+{\frac {\operatorname {d} }{x}}\operatorname {F} p.{\frac {\operatorname {d} }{x}}\varphi p\,;$

${\frac {\operatorname {d} }{u}}(\varphi p)^{m}=m(\varphi p)^{m-1}.{\frac {\operatorname {d} }{u}}\varphi p=m(\varphi p)^{m}.{\frac {\operatorname {d} }{x}}\varphi p\,;$

Il viendra, en réduisant,

${\frac {\operatorname {d} }{u}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{m}\right\}={\frac {\operatorname {d} }{x}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{m+1}\right\}.\qquad$(128)

On trouvera, de la même manière

${\frac {\operatorname {d} }{v}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{m}.(\psi p)^{n}\right\}={\frac {\operatorname {d} }{x}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{m}(\psi p)^{n+1}\right\}.\qquad$(129)

Cela étant, en différenciant successivement, par rapport à $u,$ la première (125), on aura, eu égard à (128),

${\frac {\operatorname {d} ^{2}}{u}}\operatorname {F} p={\frac {\operatorname {d} }{u}}\left({\frac {\operatorname {d} }{x}}\operatorname {F} p.\varphi p\right)={\frac {\operatorname {d} }{x}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{2}\right\},$

${\frac {\operatorname {d} ^{3}}{u}}\operatorname {F} p={\frac {\operatorname {d} }{x}}{\frac {\operatorname {d} }{u}}\left[{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{2}\right]={\frac {\operatorname {d} ^{2}}{x}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{3}\right\}\,;$

et, en général

${\frac {\operatorname {d} ^{m}}{u}}\operatorname {F} p={\frac {\operatorname {d} ^{m-1}}{x}}\left\{{\frac {\operatorname {d} }{x}}\operatorname {F} p.(\varphi p)^{n}\right\}.\qquad$(130)