Dérivées usuelles et règles de dérivation

Tableau des dérivées usuelles

Les résultats suivants sont à connaître par cœur :

Fonction $f(x)$	Ensemble de dérivabilité	Dérivée $$(ku)' = ku' \qquad (k \in \mathbb{R})$$0
$$(ku)' = ku' \qquad (k \in \mathbb{R})$$1 (constante)	$$(ku)' = ku' \qquad (k \in \mathbb{R})$$2	$$(ku)' = ku' \qquad (k \in \mathbb{R})$$3
$$(ku)' = ku' \qquad (k \in \mathbb{R})$$4	$$(ku)' = ku' \qquad (k \in \mathbb{R})$$5	$$(ku)' = ku' \qquad (k \in \mathbb{R})$$6
$$(ku)' = ku' \qquad (k \in \mathbb{R})$$7	$$(ku)' = ku' \qquad (k \in \mathbb{R})$$8	$$(ku)' = ku' \qquad (k \in \mathbb{R})$$9
$$(uv)' = u'v + uv'$$0	$$(uv)' = u'v + uv'$$1	$$(uv)' = u'v + uv'$$2
$$(uv)' = u'v + uv'$$3 ($$(uv)' = u'v + uv'$$4)	$$(uv)' = u'v + uv'$$5	$$(uv)' = u'v + uv'$$6
$$(uv)' = u'v + uv'$$7	$$(uv)' = u'v + uv'$$8	$$(uv)' = u'v + uv'$$9
$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$0	$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$1	$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$2

---

Opérations sur les dérivées

Somme

$$(u + v)' = u' + v'$$

Multiplication par un scalaire

$$(ku)' = ku' \qquad (k \in \mathbb{R})$$

Produit

$$(uv)' = u'v + uv'$$

Moyen mnémotechnique : « la dérivée du premier fois le second, plus le premier fois la dérivée du second ».

Quotient

$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$

---

Exemples d'application

Exemple 1 : dérivée d'un polynôme

Soit $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$3.

$$f'(x) = 12x^3 - 6x^2 + 5$$

Exemple 2 : dérivée d'un produit

Soit $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$4.

On pose $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$5 et $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$6.

• $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$7, $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$8

$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$

Exemple 3 : dérivée d'un quotient

Soit $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \qquad (v \neq 0)$$9.

• $$f'(x) = 12x^3 - 6x^2 + 5$$0, $$f'(x) = 12x^3 - 6x^2 + 5$$1
• $$f'(x) = 12x^3 - 6x^2 + 5$$2, $$f'(x) = 12x^3 - 6x^2 + 5$$3

$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$

---

Dérivées composées

La dérivée de $$f'(x) = 12x^3 - 6x^2 + 5$$4 (« $$f'(x) = 12x^3 - 6x^2 + 5$$5 de $$f'(x) = 12x^3 - 6x^2 + 5$$6 ») est :

$$(g \circ u)'(x) = u'(x) \cdot g'(u(x))$$

Applications fréquentes

Fonction	Dérivée
$$f'(x) = 12x^3 - 6x^2 + 5$$7	$$f'(x) = 12x^3 - 6x^2 + 5$$8
$$f'(x) = 12x^3 - 6x^2 + 5$$9	$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$0
$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$1	$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$2

Exemple

Soit $$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$3.

On pose $$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$4, donc $$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$5.

$$f'(x) = 4 \times 3 \times (3x+1)^3 = 12(3x+1)^3$$

---

Tableau récapitulatif des règles

Opération	Formule
$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$6	$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$7
$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$8	$$f'(x) = 2(x^2-3) + (2x+1) \cdot 2x = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$$9
$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$0	$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$1
$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$2	$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$3
$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$4	$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$5
$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$6	$$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$7

---

À retenir

• Connaître le tableau des dérivées usuelles ($$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$8, $$f'(x) = \frac{1 \cdot (x^2+1) - (x+1) \cdot 2x}{(x^2+1)^2} = \frac{x^2+1-2x^2-2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2}$$9, $$(g \circ u)'(x) = u'(x) \cdot g'(u(x))$$0).
• Maîtriser les quatre opérations : somme, produit, quotient, composée.
• La formule du quotient nécessite que $$(g \circ u)'(x) = u'(x) \cdot g'(u(x))$$1 : $$(g \circ u)'(x) = u'(x) \cdot g'(u(x))$$2.