Pourquoi la formule de dérivation fonctionne

Démonstration

Pour $\big[\ln(u(x))\big]' = \dfrac{u'(x)}{u(x)}$

$$f'(x) = \big[\ln(u(x))\big]' = \big[g(u(x))\big]' = u'(x)\times g'(u(x)) = u'(x)\times\dfrac{1}{u(x)}$$

car $g'(x) = (\ln x)' = \dfrac{1}{x}$.

Conclusion : $\big[\ln(u(x))\big]' = \dfrac{u'(x)}{u(x)}$.

Pour $\big[\ln(|u(x)|)\big]' = \dfrac{u'(x)}{u(x)}$

1er cas : $|u(x)| = u(x)$, déjà démontré ci-dessus.

2ème cas : $|u(x)| = -u(x)$

$$\big[\ln(|u(x)|)\big]' = \big[\ln(-u(x))\big]' = (-u(x))' \times \ln'(-u(x)) = -u'(x) \times \dfrac{1}{-u(x)} = \dfrac{u'(x)}{u(x)}$$

Conclusion : $\big[\ln(|u(x)|)\big]' = \dfrac{u'(x)}{u(x)}$.