Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/85

81

INDÉTERMINÉES.

${\begin{array}{llr}\left({\frac {\operatorname {d} V}{\operatorname {d} x}}\right)&=0,\qquad \left({\frac {\operatorname {d} V}{\operatorname {d} x'}}\right)={\frac {x'}{\sqrt {x'^{2}+y'^{2}+z'^{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)=0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} x'}}\right)'&={\frac {x''\left(x'^{2}+y'^{2}+z'^{2}\right)-x'(x'x''+y'y''+z'z'')}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)'=0,\ldots \\\\&&\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)''=0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} y}}\right)&=0,\qquad \left({\frac {\operatorname {d} V}{\operatorname {d} y'}}\right)={\frac {y'}{\sqrt {x'^{2}+y'^{2}+z'^{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)=0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} y'}}\right)'&={\frac {y''\left(x'^{2}+y'^{2}+z'^{2}\right)-y'(x'x''+y'y''+z'z'')}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)'=0,\ldots \\\\&&\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)''=0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} z}}\right)&=0,\qquad \left({\frac {\operatorname {d} V}{\operatorname {d} z'}}\right)={\frac {z'}{\sqrt {x'^{2}+y'^{2}+z'^{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} z''}}\right)=0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} z'}}\right)'&={\frac {z''\left(x'^{2}+y'^{2}+z'^{2}\right)-z'(x'x''+y'y''+z'z'')}{\left(x'^{2}+y'^{2}+z'^{2}\right)^{\frac {3}{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} z''}}\right)'=0,\ldots \\\\&&\left({\frac {\operatorname {d} V}{\operatorname {d} z''}}\right)''=0,\ldots \end{array}}$

En conséquence, l’équation (14) sera