derivative of e^{19x}

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for expand factor rationalize

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x²+6x+8=0

Solution

x=−2, x=−4

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Solution steps

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x²+2(x+2)²(x+3)

Créer le modèle d'équation partielle en utilisant le dénominateur (x+2)²(x+3)

x²+2(x+2)²(x+3) =a₀x+2 +a₁(x+2)² +a₂x+3

Multiplier l'équation par le dénominateur

(x²+2)(x+2)²(x+3)(x+2)²(x+3) =a₀(x+2)²(x+3)x+2 +a₁(x+2)²(x+3)(x+2)² +a₂(x+2)²(x+3)x+3

Simplifier

x²+2=a₀(x+2)(x+3)+a₁(x+3)+a₂(x+2)²

Résoudre les paramètres inconnus en intégrant les vraies racines du dénominateur : −2, −3

Pour la racine du dénominateur −2: a₁=6

Pour la racine du dénominateur −3: a₂=11

a₁=6, a₂=11

Intégrer les solutions aux paramètres connus

x²+2=a₀(x+2)(x+3)+6(x+3)+11(x+2)²

Développer

x²+2=a₀x²+11x²+5a₀x+50x+6a₀+62

Grouper les éléments en fonction des puissances de x

1· x²+2=x²(a₀+11)+x(5a₀+50)+(6a₀+62)

Résoudre 6a₀+62=2 pour a₀

a₀=−10

Intégrer les solutions aux paramètres de la fraction partielle pour obtenir le résultat final

(−10)x+2 +6(x+2)² +11x+3

Simplifier

−10x+2 +6(x+2)² +11x+3

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

derivative of e^{19x}

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