Inicio › Matemáticas 5 › Regla de la cadena (P1)

Matemáticas 5 · Tema 9

Regla de la cadena (P1)

Apuntes

La regla de la cadena se utiliza para derivar funciones compuestas:

h(x)=f(g(x))⟶h′(x)=[f′(g(x))][g′(x)]h(x) = f(g(x)) \longrightarrow h'(x) = [f'(g(x))][g'(x)]

Ejemplos

h(x)=−2(4−5x2)3h′(x)=−6(4−5x2)2(−10x)h′(x)=60x[16−40x2+25x4]h′(x)=960x−2400x3+1500x5\begin{aligned} h(x) &= -2(4 - 5x^2)^3 \\ h'(x) &= -6(4 - 5x^2)^2(-10x) \\ h'(x) &= 60x[16 - 40x^2 + 25x^4] \\ \mathbf{h'(x)} &= \mathbf{960x - 2400x^3 + 1500x^5} \end{aligned}
g(x)=3(2x3+5x2)2g′(x)=6(2x3+5x2)1(6x2+10x)g′(x)=6(12x5+30x4+20x4+50x3)g′(x)=72x5+300x4+300x3\begin{aligned} g(x) &= 3(2x^3 + 5x^2)^2 \\ g'(x) &= 6(2x^3 + 5x^2)^1(6x^2 + 10x) \\ g'(x) &= 6(12x^5 + 30x^4 + 20x^4 + 50x^3) \\ \mathbf{g'(x)} &= \mathbf{72x^5 + 300x^4 + 300x^3} \end{aligned}
h(x)=4x2−53=(4x2−5)13h′(x)=13(4x2−5)−23(8x)h′(x)=8x3(4x2−5)23h′(x)=8x3(4x2−53)2\begin{aligned} h(x) &= \sqrt[3]{4x^2 - 5} = (4x^2 - 5)^{\frac{1}{3}} \\ h'(x) &= \frac{1}{3}(4x^2 - 5)^{-\frac{2}{3}}(8x) \\ h'(x) &= \frac{8x}{3(4x^2 - 5)^{\frac{2}{3}}} \\ \mathbf{h'(x)} &= \mathbf{\frac{8x}{3\left(\sqrt[3]{4x^2 - 5}\right)^2}} \end{aligned}

Ejercicios

Bloque 1
h(x)=8(3−4x2)3h(x) = 8(3 - 4x^2)^3
g(x)=4(5x2−2)3g(x) = 4(5x^2 - 2)^3
h(x)=3(2−5x2)3h(x) = 3(2 - 5x^2)^3
g(x)=−2(3x2+4)3g(x) = -2(3x^2 + 4)^3
Bloque 2
h(x)=4(5x−2x2)2h(x) = 4(5x - 2x^2)^2
g(x)=−5(2x3−6x)2g(x) = -5(2x^3 - 6x)^2
h(x)=3(4x−x2)2h(x) = 3(4x - x^2)^2
g(x)=−2(3x3−5x)2g(x) = -2(3x^3 - 5x)^2
Bloque 3
h(x)=3x2−45h(x) = \sqrt[5]{3x^2 - 4}
g(x)=6−7x23g(x) = \sqrt[3]{6 - 7x^2}
h(x)=8x2+13h(x) = \sqrt[3]{8x^2 + 1}
g(x)=4−7x25g(x) = \sqrt[5]{4 - 7x^2}

Respuestas

▼
Bloque 1
h(x)=8(3−4x2)3h′(x)=24(3−4x2)2(−8x)h′(x)=−192x(9−24x2+16x4)h′(x)=−1728x+4608x3−3072x5\begin{aligned} h(x) &= 8(3 - 4x^2)^3 \\ h'(x) &= 24(3 - 4x^2)^2(-8x) \\ h'(x) &= -192x(9 - 24x^2 + 16x^4) \\ \mathbf{h'(x)} &= \mathbf{-1728x + 4608x^3 - 3072x^5} \end{aligned}
g(x)=4(5x2−2)3g′(x)=12(5x2−2)2(10x)g′(x)=120x(25x4−20x2+4)g′(x)=3000x5−2400x3+480x\begin{aligned} g(x) &= 4(5x^2 - 2)^3 \\ g'(x) &= 12(5x^2 - 2)^2(10x) \\ g'(x) &= 120x(25x^4 - 20x^2 + 4) \\ \mathbf{g'(x)} &= \mathbf{3000x^5 - 2400x^3 + 480x} \end{aligned}
h(x)=3(2−5x2)3h′(x)=9(2−5x2)2(−10x)h′(x)=−90x(4−20x2+25x4)h′(x)=−360x+1800x3−2250x5\begin{aligned} h(x) &= 3(2 - 5x^2)^3 \\ h'(x) &= 9(2 - 5x^2)^2(-10x) \\ h'(x) &= -90x(4 - 20x^2 + 25x^4) \\ \mathbf{h'(x)} &= \mathbf{-360x + 1800x^3 - 2250x^5} \end{aligned}
g(x)=−2(3x2+4)3g′(x)=−6(3x2+4)2(6x)g′(x)=−36x(9x4+24x2+16)g′(x)=−324x5−864x3−576x\begin{aligned} g(x) &= -2(3x^2 + 4)^3 \\ g'(x) &= -6(3x^2 + 4)^2(6x) \\ g'(x) &= -36x(9x^4 + 24x^2 + 16) \\ \mathbf{g'(x)} &= \mathbf{-324x^5 - 864x^3 - 576x} \end{aligned}
Bloque 2
h(x)=4(5x−2x2)2h′(x)=8(5x−2x2)1(5−4x)h′(x)=8(25x−20x2−10x2+8x3)h′(x)=64x3−240x2+200x\begin{aligned} h(x) &= 4(5x - 2x^2)^2 \\ h'(x) &= 8(5x - 2x^2)^1(5 - 4x) \\ h'(x) &= 8(25x - 20x^2 - 10x^2 + 8x^3) \\ \mathbf{h'(x)} &= \mathbf{64x^3 - 240x^2 + 200x} \end{aligned}
g(x)=−5(2x3−6x)2g′(x)=−10(2x3−6x)1(6x2−6)g′(x)=−10(12x5−12x3−36x3+36x)g′(x)=−120x5+480x3−360x\begin{aligned} g(x) &= -5(2x^3 - 6x)^2 \\ g'(x) &= -10(2x^3 - 6x)^1(6x^2 - 6) \\ g'(x) &= -10(12x^5 - 12x^3 - 36x^3 + 36x) \\ \mathbf{g'(x)} &= \mathbf{-120x^5 + 480x^3 - 360x} \end{aligned}
h(x)=3(4x−x2)2h′(x)=6(4x−x2)1(4−2x)h′(x)=6(16x−8x2−4x2+2x3)h′(x)=12x3−72x2+96x\begin{aligned} h(x) &= 3(4x - x^2)^2 \\ h'(x) &= 6(4x - x^2)^1(4 - 2x) \\ h'(x) &= 6(16x - 8x^2 - 4x^2 + 2x^3) \\ \mathbf{h'(x)} &= \mathbf{12x^3 - 72x^2 + 96x} \end{aligned}
g(x)=−2(3x3−5x)2g′(x)=−4(3x3−5x)1(9x2−5)g′(x)=−4(27x5−15x3−45x3+25x)g′(x)=−108x5+240x3−100x\begin{aligned} g(x) &= -2(3x^3 - 5x)^2 \\ g'(x) &= -4(3x^3 - 5x)^1(9x^2 - 5) \\ g'(x) &= -4(27x^5 - 15x^3 - 45x^3 + 25x) \\ \mathbf{g'(x)} &= \mathbf{-108x^5 + 240x^3 - 100x} \end{aligned}
Bloque 3
h(x)=3x2−45=(3x2−4)15h′(x)=15(3x2−4)−45(6x)h′(x)=6x5(3x2−45)4\begin{aligned} h(x) &= \sqrt[5]{3x^2 - 4} = (3x^2 - 4)^{\frac{1}{5}} \\ h'(x) &= \frac{1}{5}(3x^2 - 4)^{-\frac{4}{5}}(6x) \\ \mathbf{h'(x)} &= \mathbf{\frac{6x}{5\left(\sqrt[5]{3x^2 - 4}\right)^4}} \end{aligned}
g(x)=6−7x23=(6−7x2)13g′(x)=13(6−7x2)−23(−14x)g′(x)=−14x3(6−7x23)2\begin{aligned} g(x) &= \sqrt[3]{6 - 7x^2} = (6 - 7x^2)^{\frac{1}{3}} \\ g'(x) &= \frac{1}{3}(6 - 7x^2)^{-\frac{2}{3}}(-14x) \\ \mathbf{g'(x)} &= \mathbf{\frac{-14x}{3\left(\sqrt[3]{6 - 7x^2}\right)^2}} \end{aligned}
h(x)=8x2+13=(8x2+1)13h′(x)=13(8x2+1)−23(16x)h′(x)=16x3(8x2+13)2\begin{aligned} h(x) &= \sqrt[3]{8x^2 + 1} = (8x^2 + 1)^{\frac{1}{3}} \\ h'(x) &= \frac{1}{3}(8x^2 + 1)^{-\frac{2}{3}}(16x) \\ \mathbf{h'(x)} &= \mathbf{\frac{16x}{3\left(\sqrt[3]{8x^2 + 1}\right)^2}} \end{aligned}
g(x)=4−7x25=(4−7x2)15g′(x)=15(4−7x2)−45(−14x)g′(x)=−14x5(4−7x25)4\begin{aligned} g(x) &= \sqrt[5]{4 - 7x^2} = (4 - 7x^2)^{\frac{1}{5}} \\ g'(x) &= \frac{1}{5}(4 - 7x^2)^{-\frac{4}{5}}(-14x) \\ \mathbf{g'(x)} &= \mathbf{\frac{-14x}{5\left(\sqrt[5]{4 - 7x^2}\right)^4}} \end{aligned}

Ejercicios extra 1

▼
Bloque A
h(x)=−4(3−2x2)3h(x) = -4(3 - 2x^2)^3
g(x)=3(4x2+1)3g(x) = 3(4x^2 + 1)^3
Bloque B
g(x)=6(x3−4x)2g(x) = 6(x^3 - 4x)^2
h(x)=−2(x3−x)2h(x) = -2(x^3 - x)^2
Bloque C
h(x)=2x2−54h(x) = \sqrt[4]{2x^2 - 5}
g(x)=5x2−23g(x) = \sqrt[3]{5x^2 - 2}

Ejercicios extra 2

▼
Bloque D
h(x)=5(1−x2)3h(x) = 5(1 - x^2)^3
g(x)=7(2x2−3)3g(x) = 7(2x^2 - 3)^3
h(x)=−2(5−3x2)3h(x) = -2(5 - 3x^2)^3
g(x)=2(6x2−5)3g(x) = 2(6x^2 - 5)^3
h(x)=9(2−x2)3h(x) = 9(2 - x^2)^3
g(x)=−4(x2+8)3g(x) = -4(x^2 + 8)^3
Bloque E
g(x)=−3(2x3+5x)2g(x) = -3(2x^3 + 5x)^2
h(x)=4(3x3−2x)2h(x) = 4(3x^3 - 2x)^2
g(x)=5(2x3−7x)2g(x) = 5(2x^3 - 7x)^2
h(x)=−6(x3+4x)2h(x) = -6(x^3 + 4x)^2
g(x)=4(3x3+2x)2g(x) = 4(3x^3 + 2x)^2
h(x)=3(5x3−4x)2h(x) = 3(5x^3 - 4x)^2
Bloque F
h(x)=3x2+46h(x) = \sqrt[6]{3x^2 + 4}
g(x)=x2−65g(x) = \sqrt[5]{x^2 - 6}
h(x)=4x2−13h(x) = \sqrt[3]{4x^2 - 1}
g(x)=7x2+24g(x) = \sqrt[4]{7x^2 + 2}
h(x)=6x2−35h(x) = \sqrt[5]{6x^2 - 3}
g(x)=9−2x23g(x) = \sqrt[3]{9 - 2x^2}
← Regla del cociente (P2) Regla de la cadena (P2) →