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Matemáticas 5 · Tema 7

Regla del cociente (P1)

Apuntes

La regla del cociente sirve para derivar una función que es el cociente de otras dos. Si la función tiene la forma:

f(x)=uvf(x) = \dfrac{u}{v}

su derivada está dada por:

f′(x)=u′v−uv′v2f'(x) = \dfrac{u'v - uv'}{v^2}

donde uu es el numerador y vv el denominador.

Ejemplo

Sea la función f(x)=5x2−3x4x2+xf(x) = \dfrac{5x^2 - 3x}{4x^2 + x}

f′(x)=(10x−3)(4x2+x)−(5x2−3x)(8x+1)(4x2+x)2f′(x)=40x3+10x2−12x2−3x−(40x3+5x2−24x2−3x)(4x2)2+2(4x2)(x)+(x)2f′(x)=40x3+10x2−12x2−3x−40x3−5x2+24x2+3x16x4+8x3+x2f′(x)=17x216x4+8x3+x2\begin{aligned} f'(x) &= \frac{(10x - 3)(4x^2 + x) - (5x^2 - 3x)(8x + 1)}{(4x^2 + x)^2} \\ f'(x) &= \frac{40x^3 + 10x^2 - 12x^2 - 3x - (40x^3 + 5x^2 - 24x^2 - 3x)}{(4x^2)^2 + 2(4x^2)(x) + (x)^2} \\ f'(x) &= \frac{40x^3 + 10x^2 - 12x^2 - 3x - 40x^3 - 5x^2 + 24x^2 + 3x}{16x^4 + 8x^3 + x^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{17x^2}{16x^4 + 8x^3 + x^2}} \end{aligned}

Ejercicios

Bloque 1
f(x)=4x2−5x6x2+3f(x) = \dfrac{4x^2 - 5x}{6x^2 + 3}
f(x)=5x4−2x3x2+8f(x) = \dfrac{5x^4 - 2x}{3x^2 + 8}
f(x)=2x3+4x5x3−1f(x) = \dfrac{2x^3 + 4x}{5x^3 - 1}
Bloque 2
f(x)=x5−6x22x3+4xf(x) = \dfrac{x^5 - 6x^2}{2x^3 + 4x}
f(x)=3x4−5x4x3−2xf(x) = \dfrac{3x^4 - 5x}{4x^3 - 2x}
f(x)=4x2−7x5x4+3x2f(x) = \dfrac{4x^2 - 7x}{5x^4 + 3x^2}

Respuestas

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Bloque 1
f(x)=4x2−5x6x2+3f′(x)=(6x2+3)(8x−5)−(4x2−5x)(12x)(6x2+3)2f′(x)=48x3−30x2+24x−15−(48x3−60x2)(6x2+3)2f′(x)=48x3−30x2+24x−15−48x3+60x2(6x2+3)2f′(x)=30x2+24x−15(6x2+3)2\begin{aligned} f(x) &= \frac{4x^2 - 5x}{6x^2 + 3} \\ f'(x) &= \frac{(6x^2 + 3)(8x - 5) - (4x^2 - 5x)(12x)}{(6x^2 + 3)^2} \\ f'(x) &= \frac{48x^3 - 30x^2 + 24x - 15 - (48x^3 - 60x^2)}{(6x^2 + 3)^2} \\ f'(x) &= \frac{48x^3 - 30x^2 + 24x - 15 - 48x^3 + 60x^2}{(6x^2 + 3)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{30x^2 + 24x - 15}{(6x^2 + 3)^2}} \end{aligned}
f(x)=5x4−2x3x2+8f′(x)=(3x2+8)(20x3−2)−(5x4−2x)(6x)(3x2+8)2f′(x)=60x5−6x2+160x3−16−(30x5−12x2)(3x2+8)2f′(x)=60x5−6x2+160x3−16−30x5+12x2(3x2+8)2f′(x)=30x5+160x3+6x2−16(3x2+8)2\begin{aligned} f(x) &= \frac{5x^4 - 2x}{3x^2 + 8} \\ f'(x) &= \frac{(3x^2 + 8)(20x^3 - 2) - (5x^4 - 2x)(6x)}{(3x^2 + 8)^2} \\ f'(x) &= \frac{60x^5 - 6x^2 + 160x^3 - 16 - (30x^5 - 12x^2)}{(3x^2 + 8)^2} \\ f'(x) &= \frac{60x^5 - 6x^2 + 160x^3 - 16 - 30x^5 + 12x^2}{(3x^2 + 8)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{30x^5 + 160x^3 + 6x^2 - 16}{(3x^2 + 8)^2}} \end{aligned}
f(x)=2x3+4x5x3−1f′(x)=(5x3−1)(6x2+4)−(2x3+4x)(15x2)(5x3−1)2f′(x)=30x5+20x3−6x2−4−(30x5+60x3)(5x3−1)2f′(x)=30x5+20x3−6x2−4−30x5−60x3(5x3−1)2f′(x)=−40x3−6x2−4(5x3−1)2\begin{aligned} f(x) &= \frac{2x^3 + 4x}{5x^3 - 1} \\ f'(x) &= \frac{(5x^3 - 1)(6x^2 + 4) - (2x^3 + 4x)(15x^2)}{(5x^3 - 1)^2} \\ f'(x) &= \frac{30x^5 + 20x^3 - 6x^2 - 4 - (30x^5 + 60x^3)}{(5x^3 - 1)^2} \\ f'(x) &= \frac{30x^5 + 20x^3 - 6x^2 - 4 - 30x^5 - 60x^3}{(5x^3 - 1)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{-40x^3 - 6x^2 - 4}{(5x^3 - 1)^2}} \end{aligned}
Bloque 2
f(x)=x5−6x22x3+4xf′(x)=(2x3+4x)(5x4−12x)−(x5−6x2)(6x2+4)(2x3+4x)2f′(x)=10x7−24x4+20x5−48x2−(6x7+4x5−36x4−24x2)(2x3+4x)2f′(x)=10x7−24x4+20x5−48x2−6x7−4x5+36x4+24x2(2x3+4x)2f′(x)=4x7+16x5+12x4−24x2(2x3+4x)2\begin{aligned} f(x) &= \frac{x^5 - 6x^2}{2x^3 + 4x} \\ f'(x) &= \frac{(2x^3 + 4x)(5x^4 - 12x) - (x^5 - 6x^2)(6x^2 + 4)}{(2x^3 + 4x)^2} \\ f'(x) &= \frac{10x^7 - 24x^4 + 20x^5 - 48x^2 - (6x^7 + 4x^5 - 36x^4 - 24x^2)}{(2x^3 + 4x)^2} \\ f'(x) &= \frac{10x^7 - 24x^4 + 20x^5 - 48x^2 - 6x^7 - 4x^5 + 36x^4 + 24x^2}{(2x^3 + 4x)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{4x^7 + 16x^5 + 12x^4 - 24x^2}{(2x^3 + 4x)^2}} \end{aligned}
f(x)=3x4−5x4x3−2xf′(x)=(4x3−2x)(12x3−5)−(3x4−5x)(12x2−2)(4x3−2x)2f′(x)=48x6−20x3−24x4+10x−(36x6−6x4−60x3+10x)(4x3−2x)2f′(x)=48x6−20x3−24x4+10x−36x6+6x4+60x3−10x(4x3−2x)2f′(x)=12x6−18x4+40x3(4x3−2x)2\begin{aligned} f(x) &= \frac{3x^4 - 5x}{4x^3 - 2x} \\ f'(x) &= \frac{(4x^3 - 2x)(12x^3 - 5) - (3x^4 - 5x)(12x^2 - 2)}{(4x^3 - 2x)^2} \\ f'(x) &= \frac{48x^6 - 20x^3 - 24x^4 + 10x - (36x^6 - 6x^4 - 60x^3 + 10x)}{(4x^3 - 2x)^2} \\ f'(x) &= \frac{48x^6 - 20x^3 - 24x^4 + 10x - 36x^6 + 6x^4 + 60x^3 - 10x}{(4x^3 - 2x)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{12x^6 - 18x^4 + 40x^3}{(4x^3 - 2x)^2}} \end{aligned}
f(x)=4x2−7x5x4+3x2f′(x)=(5x4+3x2)(8x−7)−(4x2−7x)(20x3+6x)(5x4+3x2)2f′(x)=40x5−35x4+24x3−21x2−(80x5+24x3−140x4−42x2)(5x4+3x2)2f′(x)=40x5−35x4+24x3−21x2−80x5−24x3+140x4+42x2(5x4+3x2)2f′(x)=−40x5+105x4+21x2(5x4+3x2)2\begin{aligned} f(x) &= \frac{4x^2 - 7x}{5x^4 + 3x^2} \\ f'(x) &= \frac{(5x^4 + 3x^2)(8x - 7) - (4x^2 - 7x)(20x^3 + 6x)}{(5x^4 + 3x^2)^2} \\ f'(x) &= \frac{40x^5 - 35x^4 + 24x^3 - 21x^2 - (80x^5 + 24x^3 - 140x^4 - 42x^2)}{(5x^4 + 3x^2)^2} \\ f'(x) &= \frac{40x^5 - 35x^4 + 24x^3 - 21x^2 - 80x^5 - 24x^3 + 140x^4 + 42x^2}{(5x^4 + 3x^2)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{-40x^5 + 105x^4 + 21x^2}{(5x^4 + 3x^2)^2}} \end{aligned}

Ejercicios extra 1

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Bloque A
f(x)=7x2−4x5x+2f(x) = \dfrac{7x^2 - 4x}{5x + 2}
f(x)=6x3−2x4x2+5f(x) = \dfrac{6x^3 - 2x}{4x^2 + 5}
f(x)=2x5−4x25x3+6f(x) = \dfrac{2x^5 - 4x^2}{5x^3 + 6}
Bloque B
g(x)=−8x2−3xx−2x3g(x) = \dfrac{-8x^2 - 3x}{x - 2x^3}
h(x)=4x2−57x+2x3h(x) = \dfrac{4x^2 - 5}{7x + 2x^3}
h(x)=3x2−58x+5x2h(x) = \dfrac{3x^2 - 5}{8x + 5x^2}

Ejercicios extra 2

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Bloque C
f(x)=4x3+2x6x2−7f(x) = \dfrac{4x^3 + 2x}{6x^2 - 7}
g(x)=−5x2−x2x3+3g(x) = \dfrac{-5x^2 - x}{2x^3 + 3}
f(x)=8x2−5x3x4+2f(x) = \dfrac{8x^2 - 5x}{3x^4 + 2}
Bloque D
g(x)=7x2−5x6x+10x3g(x) = \dfrac{7x^2 - 5x}{6x + 10x^3}
h(x)=4−8x27x+x2h(x) = \dfrac{4 - 8x^2}{7x + x^2}
g(x)=x4−8x3x3−2xg(x) = \dfrac{x^4 - 8x}{3x^3 - 2x}
Bloque E
h(x)=5x2−92x4+xh(x) = \dfrac{5x^2 - 9}{2x^4 + x}
g(x)=7x2−3xx4−5xg(x) = \dfrac{7x^2 - 3x}{x^4 - 5x}
h(x)=3x4−18x2+4xh(x) = \dfrac{3x^4 - 1}{8x^2 + 4x}
Bloque F
h(x)=x6−2x24x3−xh(x) = \dfrac{x^6 - 2x^2}{4x^3 - x}
g(x)=2x4−6x5x2−3xg(x) = \dfrac{2x^4 - 6x}{5x^2 - 3x}
h(x)=9x3−4x5+7xh(x) = \dfrac{9x^3 - 4}{x^5 + 7x}
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