Soient
de semblables fonctions de
que
sont de
et posons
![{\displaystyle \mathrm {Q} =q+\mathrm {A+B+C+D} +\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/151cc673f36d201832c2ee9ca802095b835cdfdc)
où
sont des fonctions déterminées de
Il sera donc
![{\displaystyle \mathrm {Q} '=q'+{\frac {d\mathrm {A} }{dq}}+{\frac {d\mathrm {B} }{dq}}+{\frac {d\mathrm {C} }{dq}}+\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dc6f4d5705e34b88437dea9810428ed58d93fea)
car, en supposant
![{\displaystyle \mathrm {Q} '=q'+\mathrm {A'+B'+C'+D'} +\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/381c54ecd18d992c58a6ee2d5e069516efd9187c)
on a
![{\displaystyle {\frac {d\mathrm {A} }{dq}}=\mathrm {A} ',\qquad {\frac {d\mathrm {B} }{dq}}=\mathrm {B} ',\qquad \ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c67b0c78ad63051515be3acb7fcb47a0d9a90739)
mais, en prenant le différentiel complet de
il résulte
![{\displaystyle {\frac {dx}{dt}}\mathrm {Q} '=q'+{\frac {d\mathrm {A} }{dq}}+{\frac {d\mathrm {B} }{dq}}+{\frac {d\mathrm {C} }{dq}}+\ldots +{\frac {d\mathrm {A} }{dt}}+{\frac {d\mathrm {B} }{dt}}+{\frac {d\mathrm {C} }{dt}}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49685c765fb3f54fa6afc7de3f96e460f3b018c7)
d’où l’on tire
![{\displaystyle {\frac {dx}{dt}}\mathrm {Q'-Q'} ={\frac {dx}{dt}}\mathrm {R} '={\frac {d\mathrm {A} }{dt}}+{\frac {d\mathrm {B} }{dt}}+{\frac {d\mathrm {C} }{dt}}+\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/723467092f9a9d5620f4ff5849c0bdb875190579)
Posons maintenant
![{\displaystyle \mathrm {R} =r+{\mathfrak {A+B+C+D}}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dd94371896557297eecab5f252bc2992bd71666)
il s’ensuivra
![{\displaystyle {\frac {dx}{dt}}\mathrm {R} '=r'+{\frac {d{\mathfrak {A}}}{dt}}+{\frac {d{\mathfrak {B}}}{dt}}+{\frac {d{\mathfrak {C}}}{dt}}+{\frac {d{\mathfrak {D}}}{dt}}+\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37e7f09192fb922f4c19c85756c989e4d091f42d)
par conséquent,
![{\displaystyle r'+{\frac {d{\mathfrak {A}}}{dt}}+{\frac {d{\mathfrak {B}}}{dt}}+{\frac {d{\mathfrak {C}}}{dt}}+{\frac {d{\mathfrak {D}}}{dt}}+\ldots ={\frac {d\mathrm {A} }{dt}}+{\frac {d\mathrm {B} }{dt}}+{\frac {d\mathrm {C} }{dt}}+\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e9fb028b82f5291f53302f1c23333d828c8ed58)
En comparant les membres :
1o
|
|
|
donc
![{\displaystyle \mathrm {A} =pq'\qquad {\text{et}}\qquad {\mathfrak {A}}=pr'=pp'q'\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c8c9bfa741415324bbe55cf4407e8a0842e8867)