248
DÉVELOPPEMENT
![{\displaystyle {\frac {\operatorname {d} x}{\operatorname {d} a}}={\frac {\operatorname {f} '(a+bz)}{1-b{\frac {\operatorname {d} z}{\operatorname {d} x}}\operatorname {f} '(a+bz)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e00db89d61f0ccb5d1522b239e46018408bcf5b3)
on trouverait de la même manière
![{\displaystyle {\frac {\operatorname {d} x}{\operatorname {d} b}}={\frac {z\operatorname {f} '(a+bz)}{1-b{\frac {\operatorname {d} z}{\operatorname {d} x}}\operatorname {f} '(a+bz)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8507ad70240862f376ed7dc33565b810d7dd47)
La comparaison de ces deux valeurs donne
![{\displaystyle {\frac {\operatorname {d} x}{\operatorname {d} b}}=z{\frac {\operatorname {d} x}{\operatorname {d} a}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16e142035dbea7bc0677f3445fb1839d7b75e791)
comparant ensuite ce résultat avec les équations (1), il viendra
![{\displaystyle {\frac {\operatorname {d} U}{\operatorname {d} b}}=z{\frac {\operatorname {d} U}{\operatorname {d} a}}\,;\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/01046aea6c4e98b58a7f1ea0f2573c0d06e1a2c1)
(8)
ce qui fera devenir l’équation (4)
![{\displaystyle {\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .zV{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a}}\,;\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e748f664e26bb775bc110c5a196fc1fe97cd205)
(9)
changeant ensuite
en
on aura, pour ce cas particulier,
![{\displaystyle {\frac {\operatorname {d} .z^{n}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .z^{n+1}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/04ac08aaebf51d3e16a168b4811d4ea3616af969)
(10)
L’équation (8) donnera donc, en vertu de cette dernière,
![{\displaystyle {\frac {\operatorname {d} ^{2}U}{\operatorname {d} b^{2}}}={\frac {\operatorname {d} .z{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .z^{2}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/786ce60c523670d2e7c38be81a726a26d1ce83ef)
d’où, on conclura, de la même manière,