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QUESTIONS RÉSOLUES.
Le cas le plus simple que puisse présenter le problème général que nous venons de résoudre, est celui où l’on veut passer de l’hypothèse où
est fonction de
et
à celle où, par exemple,
est fonction de
et
; on peut poser alors
![{\displaystyle x=x,\quad y=u,\quad z=v~;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cba38f43907eb894a523f83f91d9a073a0081189)
d’où
![{\displaystyle {\begin{array}{lllll}{\tfrac {\mathrm {d} x}{\mathrm {d} u}}={\tfrac {\mathrm {d} x}{\mathrm {d} y}}&,{\tfrac {\mathrm {d} x}{\mathrm {d} v}}={\tfrac {\mathrm {d} x}{\mathrm {d} z}}&,{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} u^{2}}}={\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y^{2}}}&,{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} u\mathrm {d} v}}={\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y\mathrm {d} z}}&,{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} v^{2}}}={\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}},\\{\tfrac {\mathrm {d} y}{\mathrm {d} u}}=1&,{\tfrac {\mathrm {d} y}{\mathrm {d} v}}=0&,{\tfrac {\mathrm {d} ^{2}y}{\mathrm {d} u^{2}}}=0&,{\tfrac {\mathrm {d} ^{2}y}{\mathrm {d} u\mathrm {d} v}}=0&,{\tfrac {\mathrm {d} ^{2}y}{\mathrm {d} v^{2}}}=0,\\{\tfrac {\mathrm {d} z}{\mathrm {d} u}}=0&,{\tfrac {\mathrm {d} z}{\mathrm {d} v}}=1&,{\tfrac {\mathrm {d} ^{2}z}{\mathrm {d} u^{2}}}=0&,{\tfrac {\mathrm {d} ^{2}z}{\mathrm {d} u\mathrm {d} v}}=0&,{\tfrac {\mathrm {d} ^{2}z}{\mathrm {d} v^{2}}}=0~;\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abf4ec2571b2a2869c07a5a968b878c1da21a988)
par suite de quoi les valeurs générales de
deviennent
![{\displaystyle {\begin{aligned}p&={\tfrac {1}{\tfrac {\mathrm {d} x}{\mathrm {d} z}}},\qquad \qquad q=-{\tfrac {\tfrac {\mathrm {d} x}{\mathrm {d} y}}{\tfrac {\mathrm {d} x}{\mathrm {d} z}}},\\r&=-{\tfrac {\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}}{\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{3}}},\qquad s={\tfrac {{\tfrac {\mathrm {d} x}{\mathrm {d} z}}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y\mathrm {d} z}}-{\tfrac {\mathrm {d} x}{\mathrm {d} y}}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}}}{\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{3}}},\\t&=-{\tfrac {\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{2}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y^{2}}}-2{\tfrac {\mathrm {d} x}{\mathrm {d} z}}{\tfrac {\mathrm {d} x}{\mathrm {d} y}}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} y\mathrm {d} z}}+\left({\tfrac {\mathrm {d} x}{\mathrm {d} y}}\right)^{2}{\tfrac {\mathrm {d} ^{2}x}{\mathrm {d} z^{2}}}}{\left({\tfrac {\mathrm {d} x}{\mathrm {d} z}}\right)^{3}}}.&\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d819e88734ceedca4941f4853af5303a849b6b5)