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Matemáticas 5 · Tema 8

Regla del cociente (P2)

Apuntes

La misma regla del cociente, ahora con senos, cosenos y tangentes:

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

Ejemplos

f(x)=3x25cos⁡xf′(x)=(6x)(5cos⁡x)−(3x2)(−5sen⁡x)(5cos⁡x)2f′(x)=30xcos⁡x+15x2sen⁡x25cos⁡2x\begin{aligned} f(x) &= \frac{3x^2}{5\cos x} \\ f'(x) &= \frac{(6x)(5\cos x) - (3x^2)(-5\operatorname{sen} x)}{(5\cos x)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{30x\cos x + 15x^2\operatorname{sen} x}{25\cos^2 x}} \end{aligned}
f(x)=−6x2+3cos⁡xsen⁡xf′(x)=(−12x−3sen⁡x)(sen⁡x)−(−6x2+3cos⁡x)(cos⁡x)sen⁡2xf′(x)=−12xsen⁡x−3sen⁡2x+6x2cos⁡x−3cos⁡2xsen⁡2x\begin{aligned} f(x) &= \frac{-6x^2 + 3\cos x}{\operatorname{sen} x} \\ f'(x) &= \frac{(-12x - 3\operatorname{sen} x)(\operatorname{sen} x) - (-6x^2 + 3\cos x)(\cos x)}{\operatorname{sen}^2 x} \\ \mathbf{f'(x)} &= \mathbf{\frac{-12x\operatorname{sen} x - 3\operatorname{sen}^2 x + 6x^2\cos x - 3\cos^2 x}{\operatorname{sen}^2 x}} \end{aligned}

Ejercicios

Bloque 1
f(x)=4x25cos⁡xf(x) = \dfrac{4x^2}{5\cos x}
h(x)=2tan⁡x8x2h(x) = \dfrac{2\tan x}{8x^2}
Bloque 2
f(x)=5x2−3sen⁡x2cos⁡xf(x) = \dfrac{5x^2 - 3\operatorname{sen} x}{2\cos x}
f(x)=4x3+2cos⁡xsen⁡xf(x) = \dfrac{4x^3 + 2\cos x}{\operatorname{sen} x}
Bloque 3
g(x)=7cos⁡x2x2−4sen⁡xg(x) = \dfrac{7\cos x}{2x^2 - 4\operatorname{sen} x}
f(x)=2cos⁡x3x2−5sen⁡xf(x) = \dfrac{2\cos x}{3x^2 - 5\operatorname{sen} x}

Respuestas

▼
Bloque 1
f(x)=4x25cos⁡xf′(x)=(8x)(5cos⁡x)−(4x2)(−5sen⁡x)(5cos⁡x)2f′(x)=40xcos⁡x+20x2sen⁡x25cos⁡2x\begin{aligned} f(x) &= \frac{4x^2}{5\cos x} \\ f'(x) &= \frac{(8x)(5\cos x) - (4x^2)(-5\operatorname{sen} x)}{(5\cos x)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{40x\cos x + 20x^2\operatorname{sen} x}{25\cos^2 x}} \end{aligned}
h(x)=2tan⁡x8x2h′(x)=(2sec⁡2x)(8x2)−(2tan⁡x)(16x)(8x2)2h′(x)=16x2sec⁡2x−32xtan⁡x64x4\begin{aligned} h(x) &= \frac{2\tan x}{8x^2} \\ h'(x) &= \frac{(2\sec^2 x)(8x^2) - (2\tan x)(16x)}{(8x^2)^2} \\ \mathbf{h'(x)} &= \mathbf{\frac{16x^2\sec^2 x - 32x\tan x}{64x^4}} \end{aligned}
Bloque 2
f(x)=5x2−3sen⁡x2cos⁡xf′(x)=(10x−3cos⁡x)(2cos⁡x)−(5x2−3sen⁡x)(−2sen⁡x)(2cos⁡x)2f′(x)=20xcos⁡x−6cos⁡2x+10x2sen⁡x−6sen⁡2x4cos⁡2x\begin{aligned} f(x) &= \frac{5x^2 - 3\operatorname{sen} x}{2\cos x} \\ f'(x) &= \frac{(10x - 3\cos x)(2\cos x) - (5x^2 - 3\operatorname{sen} x)(-2\operatorname{sen} x)}{(2\cos x)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{20x\cos x - 6\cos^2 x + 10x^2\operatorname{sen} x - 6\operatorname{sen}^2 x}{4\cos^2 x}} \end{aligned}
f(x)=4x3+2cos⁡xsen⁡xf′(x)=(12x2−2sen⁡x)(sen⁡x)−(4x3+2cos⁡x)(cos⁡x)sen⁡2xf′(x)=12x2sen⁡x−2sen⁡2x−4x3cos⁡x−2cos⁡2xsen⁡2x\begin{aligned} f(x) &= \frac{4x^3 + 2\cos x}{\operatorname{sen} x} \\ f'(x) &= \frac{(12x^2 - 2\operatorname{sen} x)(\operatorname{sen} x) - (4x^3 + 2\cos x)(\cos x)}{\operatorname{sen}^2 x} \\ \mathbf{f'(x)} &= \mathbf{\frac{12x^2\operatorname{sen} x - 2\operatorname{sen}^2 x - 4x^3\cos x - 2\cos^2 x}{\operatorname{sen}^2 x}} \end{aligned}
Bloque 3
g(x)=7cos⁡x2x2−4sen⁡xg′(x)=(−7sen⁡x)(2x2−4sen⁡x)−(7cos⁡x)(4x−4cos⁡x)(2x2−4sen⁡x)2g′(x)=−14x2sen⁡x+28sen⁡2x−28xcos⁡x+28cos⁡2x(2x2−4sen⁡x)2\begin{aligned} g(x) &= \frac{7\cos x}{2x^2 - 4\operatorname{sen} x} \\ g'(x) &= \frac{(-7\operatorname{sen} x)(2x^2 - 4\operatorname{sen} x) - (7\cos x)(4x - 4\cos x)}{(2x^2 - 4\operatorname{sen} x)^2} \\ \mathbf{g'(x)} &= \mathbf{\frac{-14x^2\operatorname{sen} x + 28\operatorname{sen}^2 x - 28x\cos x + 28\cos^2 x}{(2x^2 - 4\operatorname{sen} x)^2}} \end{aligned}
f(x)=2cos⁡x3x2−5sen⁡xf′(x)=(−2sen⁡x)(3x2−5sen⁡x)−(2cos⁡x)(6x−5cos⁡x)(3x2−5sen⁡x)2f′(x)=−6x2sen⁡x+10sen⁡2x−12xcos⁡x+10cos⁡2x(3x2−5sen⁡x)2\begin{aligned} f(x) &= \frac{2\cos x}{3x^2 - 5\operatorname{sen} x} \\ f'(x) &= \frac{(-2\operatorname{sen} x)(3x^2 - 5\operatorname{sen} x) - (2\cos x)(6x - 5\cos x)}{(3x^2 - 5\operatorname{sen} x)^2} \\ \mathbf{f'(x)} &= \mathbf{\frac{-6x^2\operatorname{sen} x + 10\operatorname{sen}^2 x - 12x\cos x + 10\cos^2 x}{(3x^2 - 5\operatorname{sen} x)^2}} \end{aligned}

Ejercicios extra 1

▼
Bloque A
f(x)=3x44cos⁡xf(x) = \dfrac{3x^4}{4\cos x}
g(x)=−2tan⁡x5x3g(x) = \dfrac{-2\tan x}{5x^3}
h(x)=6x25cos⁡xh(x) = \dfrac{6x^2}{5\cos x}
g(x)=4tan⁡x3x4g(x) = \dfrac{4\tan x}{3x^4}
Bloque B
g(x)=5x2−3cos⁡xsen⁡xg(x) = \dfrac{5x^2 - 3\cos x}{\operatorname{sen} x}
g(x)=2x2−4cos⁡xsen⁡xg(x) = \dfrac{2x^2 - 4\cos x}{\operatorname{sen} x}
f(x)=6x3−5sen⁡x3cos⁡xf(x) = \dfrac{6x^3 - 5\operatorname{sen} x}{3\cos x}
h(x)=3x2+4cos⁡x2sen⁡xh(x) = \dfrac{3x^2 + 4\cos x}{2\operatorname{sen} x}
Bloque C
f(x)=4cos⁡x2x3−3sen⁡xf(x) = \dfrac{4\cos x}{2x^3 - 3\operatorname{sen} x}
h(x)=4cos⁡x2x2−5sen⁡xh(x) = \dfrac{4\cos x}{2x^2 - 5\operatorname{sen} x}
f(x)=3cos⁡x4x2−2sen⁡xf(x) = \dfrac{3\cos x}{4x^2 - 2\operatorname{sen} x}
h(x)=8cos⁡x5x2−3sen⁡xh(x) = \dfrac{8\cos x}{5x^2 - 3\operatorname{sen} x}

Ejercicios extra 2

▼
Bloque D
g(x)=7x36sen⁡xg(x) = \dfrac{7x^3}{6\operatorname{sen} x}
f(x)=−4cos⁡x6x2f(x) = \dfrac{-4\cos x}{6x^2}
f(x)=2x43cos⁡xf(x) = \dfrac{2x^4}{3\cos x}
g(x)=−5tan⁡x2x3g(x) = \dfrac{-5\tan x}{2x^3}
h(x)=5x32cos⁡xh(x) = \dfrac{5x^3}{2\cos x}
f(x)=6x32cos⁡xf(x) = \dfrac{6x^3}{2\cos x}
h(x)=−3x24cos⁡xh(x) = \dfrac{-3x^2}{4\cos x}
Bloque E
h(x)=3x2−2sen⁡x4cos⁡xh(x) = \dfrac{3x^2 - 2\operatorname{sen} x}{4\cos x}
g(x)=4x2−2cos⁡xsen⁡xg(x) = \dfrac{4x^2 - 2\cos x}{\operatorname{sen} x}
h(x)=7x2+2sen⁡x5cos⁡xh(x) = \dfrac{7x^2 + 2\operatorname{sen} x}{5\cos x}
f(x)=x3−6cos⁡x4sen⁡xf(x) = \dfrac{x^3 - 6\cos x}{4\operatorname{sen} x}
Bloque F
h(x)=6cos⁡x3x2−2sen⁡xh(x) = \dfrac{6\cos x}{3x^2 - 2\operatorname{sen} x}
g(x)=5sen⁡x4x2+3cos⁡xg(x) = \dfrac{5\operatorname{sen} x}{4x^2 + 3\cos x}
f(x)=2cos⁡xx3−4sen⁡xf(x) = \dfrac{2\cos x}{x^3 - 4\operatorname{sen} x}
h(x)=9sen⁡x2x2−7cos⁡xh(x) = \dfrac{9\operatorname{sen} x}{2x^2 - 7\cos x}
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