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Matemáticas 5 · Tema 5

Regla del producto (P1)

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Video — Regla del producto

Apuntes

La regla del producto se usa para derivar el producto de dos funciones. Si f=uvf = uv, su derivada es la derivada de la primera parte por la segunda sin derivar, más la primera sin derivar por la derivada de la segunda:

f=uv  ⟹  f′=u′v+uv′f = uv \quad \implies \quad f' = u'v + uv'

Ejemplo

En f(x)=(5x2−4x)(3x2−x)f(x) = (5x^2 - 4x)(3x^2 - x) identificamos u=(5x2−4x)u = (5x^2 - 4x) y v=(3x2−x)v = (3x^2 - x).

f(x)=(5x2−4x)(3x2−x)f′(x)=(10x−4)(3x2−x)+(5x2−4x)(6x−1)f′(x)=30x3−10x2−12x2+4x+30x3−5x2−24x2+4xf′(x)=60x3−51x2+8x\begin{aligned} f(x) &= (5x^2 - 4x)(3x^2 - x) \\ f'(x) &= (10x - 4)(3x^2 - x) + (5x^2 - 4x)(6x - 1) \\ f'(x) &= 30x^3 - 10x^2 - 12x^2 + 4x + 30x^3 - 5x^2 - 24x^2 + 4x \\ \mathbf{f'(x)} &= \mathbf{60x^3 - 51x^2 + 8x} \end{aligned}

Ejercicios

Bloque único
f(x)=(8x2−2x)(6x2+x)f(x) = (8x^2 - 2x)(6x^2 + x)
f(x)=(10x3−4x2)(5x−3x2)f(x) = (10x^3 - 4x^2)(5x - 3x^2)
f(x)=(5x+7x2)(4x+2x2)f(x) = (5x + 7x^2)(4x + 2x^2)
f(x)=(8x−9x2)(x+2x3)f(x) = (8x - 9x^2)(x + 2x^3)
f(x)=(3x2−5x)(2x−6x2)f(x) = (3x^2 - 5x)(2x - 6x^2)
f(x)=(5x3+4x)(6x2−8)f(x) = (5x^3 + 4x)(6x^2 - 8)

Respuestas

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Bloque único
f(x)=(8x2−2x)(6x2+x)f′(x)=(16x−2)(6x2+x)+(8x2−2x)(12x+1)f′(x)=96x3+16x2−12x2−2x+96x3+8x2−24x2−2xf′(x)=192x3−12x2−4x\begin{aligned} f(x) &= (8x^2 - 2x)(6x^2 + x) \\ f'(x) &= (16x - 2)(6x^2 + x) + (8x^2 - 2x)(12x + 1) \\ f'(x) &= 96x^3 + 16x^2 - 12x^2 - 2x + 96x^3 + 8x^2 - 24x^2 - 2x \\ \mathbf{f'(x)} &= \mathbf{192x^3 - 12x^2 - 4x} \end{aligned}
f(x)=(10x3−4x2)(5x−3x2)f′(x)=(30x2−8x)(5x−3x2)+(10x3−4x2)(5−6x)f′(x)=150x3−90x4−40x2+24x3+50x3−60x4−20x2+24x3f′(x)=−150x4+248x3−60x2\begin{aligned} f(x) &= (10x^3 - 4x^2)(5x - 3x^2) \\ f'(x) &= (30x^2 - 8x)(5x - 3x^2) + (10x^3 - 4x^2)(5 - 6x) \\ f'(x) &= 150x^3 - 90x^4 - 40x^2 + 24x^3 + 50x^3 - 60x^4 - 20x^2 + 24x^3 \\ \mathbf{f'(x)} &= \mathbf{-150x^4 + 248x^3 - 60x^2} \end{aligned}
f(x)=(5x+7x2)(4x+2x2)f′(x)=(5+14x)(4x+2x2)+(5x+7x2)(4+4x)f′(x)=20x+10x2+56x2+28x3+20x+20x2+28x2+28x3f′(x)=56x3+114x2+40x\begin{aligned} f(x) &= (5x + 7x^2)(4x + 2x^2) \\ f'(x) &= (5 + 14x)(4x + 2x^2) + (5x + 7x^2)(4 + 4x) \\ f'(x) &= 20x + 10x^2 + 56x^2 + 28x^3 + 20x + 20x^2 + 28x^2 + 28x^3 \\ \mathbf{f'(x)} &= \mathbf{56x^3 + 114x^2 + 40x} \end{aligned}
f(x)=(8x−9x2)(x+2x3)f′(x)=(8−18x)(x+2x3)+(8x−9x2)(1+6x2)f′(x)=8x+16x3−18x2−36x4+8x+48x3−9x2−54x4f′(x)=−90x4+64x3−27x2+16x\begin{aligned} f(x) &= (8x - 9x^2)(x + 2x^3) \\ f'(x) &= (8 - 18x)(x + 2x^3) + (8x - 9x^2)(1 + 6x^2) \\ f'(x) &= 8x + 16x^3 - 18x^2 - 36x^4 + 8x + 48x^3 - 9x^2 - 54x^4 \\ \mathbf{f'(x)} &= \mathbf{-90x^4 + 64x^3 - 27x^2 + 16x} \end{aligned}
f(x)=(3x2−5x)(2x−6x2)f′(x)=(6x−5)(2x−6x2)+(3x2−5x)(2−12x)f′(x)=12x2−36x3−10x+30x2+6x2−36x3−10x+60x2f′(x)=−72x3+108x2−20x\begin{aligned} f(x) &= (3x^2 - 5x)(2x - 6x^2) \\ f'(x) &= (6x - 5)(2x - 6x^2) + (3x^2 - 5x)(2 - 12x) \\ f'(x) &= 12x^2 - 36x^3 - 10x + 30x^2 + 6x^2 - 36x^3 - 10x + 60x^2 \\ \mathbf{f'(x)} &= \mathbf{-72x^3 + 108x^2 - 20x} \end{aligned}
f(x)=(5x3+4x)(6x2−8)f′(x)=(15x2+4)(6x2−8)+(5x3+4x)(12x)f′(x)=90x4−120x2+24x2−32+60x4+48x2f′(x)=150x4−48x2−32\begin{aligned} f(x) &= (5x^3 + 4x)(6x^2 - 8) \\ f'(x) &= (15x^2 + 4)(6x^2 - 8) + (5x^3 + 4x)(12x) \\ f'(x) &= 90x^4 - 120x^2 + 24x^2 - 32 + 60x^4 + 48x^2 \\ \mathbf{f'(x)} &= \mathbf{150x^4 - 48x^2 - 32} \end{aligned}

Ejercicios extra 1

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Bloque A
f(x)=(2x2+3x)(x2−5)f(x) = (2x^2 + 3x)(x^2 - 5)
f(x)=(4x3−x)(2x2+6x)f(x) = (4x^3 - x)(2x^2 + 6x)
f(x)=(x−7x2)(3x2+4)f(x) = (x - 7x^2)(3x^2 + 4)
f(x)=(6x2−2x)(4x−1)f(x) = (6x^2 - 2x)(4x - 1)
f(x)=(3x3+5x2)(x2−2x)f(x) = (3x^3 + 5x^2)(x^2 - 2x)
f(x)=(9x+2x2)(5x2−x)f(x) = (9x + 2x^2)(5x^2 - x)

Ejercicios extra 2

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Bloque B
f(x)=(x3−4x)(2x2+3x)f(x) = (x^3 - 4x)(2x^2 + 3x)
f(x)=(5x2+2x)(7x−x2)f(x) = (5x^2 + 2x)(7x - x^2)
f(x)=(8x2−3)(x3+x)f(x) = (8x^2 - 3)(x^3 + x)
f(x)=(12x−4x2)(x2+5)f(x) = (12x - 4x^2)(x^2 + 5)
f(x)=(2x3−6x)(3x2−2)f(x) = (2x^3 - 6x)(3x^2 - 2)
f(x)=(4x2+1)(x2−9x)f(x) = (4x^2 + 1)(x^2 - 9x)
← Derivadas de raíces Regla del producto (P2) →