Passons présentement aux fonctions circulaires. On sait que
étant une fonction quelconque de
on a
![{\displaystyle \operatorname {Ang} .(\operatorname {Tang} .=X)={\frac {X}{1}}-{\frac {X^{3}}{3}}+{\frac {X^{5}}{5}}-{\frac {X^{7}}{7}}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05ee4e49c45b1b0eadf137ffc992f65ed1823bdc)
donc (2)
![{\displaystyle \operatorname {d} .\operatorname {Ang} .(\operatorname {Tang} .=X)=\operatorname {d} .{\frac {X}{1}}-\operatorname {d} .{\frac {X^{3}}{3}}+\operatorname {d} .{\frac {X^{5}}{5}}-\operatorname {d} .{\frac {X^{7}}{7}}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67cae1c94960c58768ffc53868def2bbe2b35dbf)
mais on a généralement, (16) et (18),
![{\displaystyle \operatorname {d} .{\frac {X^{n}}{n}}=X^{n-1}\operatorname {d} X\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba4f1298582f771fa22425190a9e5da597d6c410)
donc, en substituant,
c’est-à-dire, comme ci-dessus,
![{\displaystyle \operatorname {d} .e^{X}=e^{X}.\operatorname {d} X.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55b82267450d8e2dd957f990a77ba6a6dffda5fb)
On aurait de même
![{\displaystyle \operatorname {d} .e^{X'}=e^{X'}.\operatorname {d} X'\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c36829ad8f044f76a5ac17c9cccbaa6558a2545d)
posant alors
![{\displaystyle e^{X'}=X,\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d47055b9c09a572b42275c31caddebd4560fa432)
d’où
![{\displaystyle \quad X'=\operatorname {Log} .X\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/79971df4f418eadf594aced77e5d551fd80bf8b0)
et
![{\displaystyle \quad \operatorname {d} X'=\operatorname {d} .\operatorname {Log} .X,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b27d688c2b03f40eeda8838000793d8a3c800948)
il viendrait, en substituant,
![{\displaystyle \operatorname {d} X=X\operatorname {d} .\operatorname {Log} .X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/101ac9783d6a59239a32ef18000b274a21c3fb83)
d’où l’on conclurait, comme nous l’avons trouvé ci-dessus,
![{\displaystyle \operatorname {d} .\operatorname {Log} .X={\frac {\operatorname {d} X}{X}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a882cded3099346e5746544f3beab319ca277a98)