249
DES FONCTIONS.
![{\displaystyle {\frac {\operatorname {d} ^{3}U}{\operatorname {d} b^{3}}}={\frac {\operatorname {d} ^{2}.z^{2}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a\operatorname {d} b}}={\frac {\operatorname {d} ^{2}.z^{3}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/184113df869bdf9219ac1592484048dc1e2a4de4)
et de celle-là
![{\displaystyle {\frac {\operatorname {d} ^{4}U}{\operatorname {d} b^{4}}}={\frac {\operatorname {d} ^{3}.z^{3}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{2}\operatorname {d} b}}={\frac {\operatorname {d} ^{3}.z^{4}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{3}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d17ca1742a0c012bb30c97e66d2484b0c4f5abb6)
et ainsi de suite ; de manière qu’on aura, en général
![{\displaystyle {\frac {\operatorname {d} ^{m}U}{\operatorname {d} b^{m}}}={\frac {\operatorname {d} ^{m-1}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{m-1}}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cd88b16147819596b99a9c819d03e76203f9eb0)
(11)
Une fois parvenu à la formule (11), le développement de
suivant les puissances de
ne présente plus de difficulté.
Soit encore
![{\displaystyle x=\operatorname {f} (a+bz)\,;\qquad x_{1}=\operatorname {f_{1}} (a_{1}+b_{1}z_{1})\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/017f9aae670c60360f8b3d728ef14dcfd07572a3)
étant des fonctions quelconques de
sans
On trouvera, par la différentiation,
![{\displaystyle {\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} a}}=\left(1+b{\frac {\operatorname {d} z}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+b{\frac {\operatorname {d} z}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}\right)\operatorname {f} '(a+bz),\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}=\left(b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}\right)\operatorname {f_{1}} '(a_{1}+b_{1}z_{1}).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/674d08c35d5e0e3043d71226fcc85c78717b7403)
En posant, pour abréger,
![{\displaystyle k={\frac {b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}\operatorname {f_{1}} '(a_{1}+b_{1}z_{1})}{1-b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x_{1}}}\operatorname {f_{1}} '(a_{1}+b_{1}z_{1})}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5c872f1c21358af8ab6eb7e41e1b192c0d20eb2)
la dernière de ces équations donne