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Matemáticas 5 · Tema 3

Derivadas: senos, cosenos…

Apuntes — Tabla de derivadas

Seno
f(x)=sen⁡x  →  f′(x)=cos⁡xf(x) = \operatorname{sen} x \;\rightarrow\; f'(x) = \cos x
Coseno
f(x)=cos⁡x  →  f′(x)=−sen⁡xf(x) = \cos x \;\rightarrow\; f'(x) = -\operatorname{sen} x
Tangente
f(x)=tan⁡x  →  f′(x)=sec⁡2xf(x) = \tan x \;\rightarrow\; f'(x) = \sec^2 x
Logaritmo natural
f(x)=ln⁡x  →  f′(x)=1xf(x) = \ln x \;\rightarrow\; f'(x) = \dfrac{1}{x}
Exponencial exe^x
f(x)=ex  →  f′(x)=exf(x) = e^x \;\rightarrow\; f'(x) = e^x
Exponencial axa^x
f(x)=ax  →  f′(x)=axln⁡(a)f(x) = a^x \;\rightarrow\; f'(x) = a^x \ln(a)

Ejemplo

Deriva la siguiente función usando la tabla. El término con xx en el denominador se convierte primero a exponente negativo:

f(x)=−5x−4x3+5ln⁡x+6cos⁡x−3sen⁡x−4ex+6x−5tan⁡x−1f(x) = -5x - \dfrac{4}{x^3} + 5 \ln x + 6 \cos x - 3 \operatorname{sen} x - 4e^x + 6^x - 5 \tan x - 1
−4x3=−4x−3derivando12x−4=12x4-\dfrac{4}{x^3} = -4x^{-3} \quad \text{derivando} \quad 12x^{-4} = \dfrac{12}{x^4}
f′(x)=−5+12x4+5x−6sen⁡x−3cos⁡x−4ex+6xln⁡6−5sec⁡2xf'(x) = -5 + \dfrac{12}{x^4} + \dfrac{5}{x} - 6 \operatorname{sen} x - 3 \cos x - 4e^x + 6^x \ln 6 - 5 \sec^2 x

Ejercicios

Bloque único
f(x)=−5x−4x3+2ln⁡x+7cos⁡x−4sen⁡x+7ex−2x+10f(x) = -5x - \dfrac{4}{x^3} + 2 \ln x + 7 \cos x - 4 \operatorname{sen} x + 7e^x - 2^x + 10
f(x)=3tan⁡x−7+8x3−2x4+8sen⁡x−9cos⁡x−6x+4x−3f(x) = 3 \tan x - 7 + \dfrac{8}{x^3} - 2x^4 + 8 \operatorname{sen} x - 9 \cos x - \dfrac{6}{x} + 4^x - 3
f(x)=8sen⁡x−ln⁡x+6x4−3x2−7x+2ex+x−tan⁡xf(x) = 8 \operatorname{sen} x - \ln x + 6x^4 - \dfrac{3}{x^2} - 7^x + 2e^x + x - \tan x
f(x)=−4sen⁡x+6tan⁡x−8cos⁡x−4x7−2x3+7−9xf(x) = -4 \operatorname{sen} x + 6 \tan x - 8 \cos x - \dfrac{4}{x^7} - 2x^3 + 7 - 9^x
f(x)=−9cos⁡x−8+4x6+4x8−6tan⁡x−4sen⁡x−3x+8x−2f(x) = -9 \cos x - 8 + \dfrac{4}{x^6} + 4x^8 - 6 \tan x - 4 \operatorname{sen} x - 3x + 8^x - 2
f(x)=6tan⁡x−2x5+5sen⁡x−3x4+ln⁡x−5x+4ex−9cos⁡x+3f(x) = 6 \tan x - \dfrac{2}{x^5} + 5 \operatorname{sen} x - 3x^4 + \ln x - 5^x + 4e^x - 9 \cos x + 3

Respuestas

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Bloque único
f(x)=−5x−4x3+2ln⁡x+7cos⁡x−4sen⁡x+7ex−2x+10f(x) = -5x - \frac{4}{x^3} + 2 \ln x + 7 \cos x - 4 \operatorname{sen} x + 7e^x - 2^x + 10
−4x3=−4x−3derivando12x−4=12x4-\dfrac{4}{x^3} = -4x^{-3} \quad \text{derivando} \quad 12x^{-4} = \dfrac{12}{x^4}
f′(x)=−5+12x4+2x−7sen⁡x−4cos⁡x+7ex−2xln⁡2\mathbf{f'(x) = -5 + \frac{12}{x^4} + \frac{2}{x} - 7 \operatorname{sen} x - 4 \cos x + 7e^x - 2^x \ln 2}
f(x)=3tan⁡x+8x3−2x4+8sen⁡x−9cos⁡x−6x+4x−10f(x) = 3 \tan x + \frac{8}{x^3} - 2x^4 + 8 \operatorname{sen} x - 9 \cos x - \frac{6}{x} + 4^x - 10
8x3=8x−3derivando−24x−4=−24x4\dfrac{8}{x^3} = 8x^{-3} \quad \text{derivando} \quad -24x^{-4} = -\dfrac{24}{x^4}
−6x=−6x−1derivando6x−2=6x2-\dfrac{6}{x} = -6x^{-1} \quad \text{derivando} \quad 6x^{-2} = \dfrac{6}{x^2}
f′(x)=3sec⁡2x−24x4−8x3+8cos⁡x+9sen⁡x+6x2+4xln⁡4\mathbf{f'(x) = 3 \sec^2 x - \frac{24}{x^4} - 8x^3 + 8 \cos x + 9 \operatorname{sen} x + \frac{6}{x^2} + 4^x \ln 4}
f(x)=8sen⁡x−ln⁡x+6x4−3x2−7x+2ex+x−tan⁡xf(x) = 8 \operatorname{sen} x - \ln x + 6x^4 - \frac{3}{x^2} - 7^x + 2e^x + x - \tan x
−3x2=−3x−2derivando6x−3=6x3-\dfrac{3}{x^2} = -3x^{-2} \quad \text{derivando} \quad 6x^{-3} = \dfrac{6}{x^3}
f′(x)=8cos⁡x−1x+24x3+6x3−7xln⁡7+2ex+1−sec⁡2x\mathbf{f'(x) = 8 \cos x - \frac{1}{x} + 24x^3 + \frac{6}{x^3} - 7^x \ln 7 + 2e^x + 1 - \sec^2 x}
f(x)=−4sen⁡x+6tan⁡x−8cos⁡x−4x7−2x3+7−9xf(x) = -4 \operatorname{sen} x + 6 \tan x - 8 \cos x - \frac{4}{x^7} - 2x^3 + 7 - 9^x
−4x7=−4x−7derivando28x−8=28x8-\dfrac{4}{x^7} = -4x^{-7} \quad \text{derivando} \quad 28x^{-8} = \dfrac{28}{x^8}
f′(x)=−4cos⁡x+6sec⁡2x+8sen⁡x+28x8−6x2−9xln⁡9\mathbf{f'(x) = -4 \cos x + 6 \sec^2 x + 8 \operatorname{sen} x + \frac{28}{x^8} - 6x^2 - 9^x \ln 9}
f(x)=−9cos⁡x+4x6+4x8−6tan⁡x−4sen⁡x−3x+8x−10f(x) = -9 \cos x + \frac{4}{x^6} + 4x^8 - 6 \tan x - 4 \operatorname{sen} x - 3x + 8^x - 10
4x6=4x−6derivando−24x−7=−24x7\dfrac{4}{x^6} = 4x^{-6} \quad \text{derivando} \quad -24x^{-7} = -\dfrac{24}{x^7}
f′(x)=9sen⁡x−24x7+32x7−6sec⁡2x−4cos⁡x−3+8xln⁡8\mathbf{f'(x) = 9 \operatorname{sen} x - \frac{24}{x^7} + 32x^7 - 6 \sec^2 x - 4 \cos x - 3 + 8^x \ln 8}
f(x)=6tan⁡x−2x5+5sen⁡x−3x4+ln⁡x−5x+4ex−9cos⁡x+3f(x) = 6 \tan x - \frac{2}{x^5} + 5 \operatorname{sen} x - 3x^4 + \ln x - 5^x + 4e^x - 9 \cos x + 3
−2x5=−2x−5derivando10x−6=10x6-\dfrac{2}{x^5} = -2x^{-5} \quad \text{derivando} \quad 10x^{-6} = \dfrac{10}{x^6}
f′(x)=6sec⁡2x+10x6+5cos⁡x−12x3+1x−5xln⁡5+4ex+9sen⁡x\mathbf{f'(x) = 6 \sec^2 x + \frac{10}{x^6} + 5 \cos x - 12x^3 + \frac{1}{x} - 5^x \ln 5 + 4e^x + 9 \operatorname{sen} x}

Ejercicios extra 1

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Bloque A
f(x)=4sen⁡x−3x2+2x5−7cos⁡x+ln⁡x−6x+5ex+9−2xf(x) = 4\operatorname{sen} x - \dfrac{3}{x^2} + 2x^5 - 7\cos x + \ln x - 6^x + 5e^x + 9 - 2x
f(x)=−2tan⁡x+8x4−5x3+3cos⁡x−4ln⁡x+7x−6ex+x−1f(x) = -2\tan x + \dfrac{8}{x^4} - 5x^3 + 3\cos x - 4\ln x + 7^x - 6e^x + x - 1
f(x)=7cos⁡x+5x3−2x6+4sen⁡x+3ln⁡x−8x+2ex−3x+6f(x) = 7\cos x + \dfrac{5}{x^3} - 2x^6 + 4\operatorname{sen} x + 3\ln x - 8^x + 2e^x - 3x + 6
f(x)=−6sen⁡x−4x5+3x4+9tan⁡x−2ln⁡x+5x+4ex+2x−8f(x) = -6\operatorname{sen} x - \dfrac{4}{x^5} + 3x^4 + 9\tan x - 2\ln x + 5^x + 4e^x + 2x - 8
f(x)=5tan⁡x+7x2−4x7−3cos⁡x+6ln⁡x−2x+8ex−x+4f(x) = 5\tan x + \dfrac{7}{x^2} - 4x^7 - 3\cos x + 6\ln x - 2^x + 8e^x - x + 4
f(x)=−3cos⁡x+2x6+6x3−5sen⁡x+4ln⁡x+9x−7ex+5x−2f(x) = -3\cos x + \dfrac{2}{x^6} + 6x^3 - 5\operatorname{sen} x + 4\ln x + 9^x - 7e^x + 5x - 2
Bloque B
f(x)=8sen⁡x−5x4+2x8−6tan⁡x+7ln⁡x−3x+4ex−4x+1f(x) = 8\operatorname{sen} x - \dfrac{5}{x^4} + 2x^8 - 6\tan x + 7\ln x - 3^x + 4e^x - 4x + 1
f(x)=−4tan⁡x+3x7−7x2+2cos⁡x−5ln⁡x+6x+9ex+3x−5f(x) = -4\tan x + \dfrac{3}{x^7} - 7x^2 + 2\cos x - 5\ln x + 6^x + 9e^x + 3x - 5
f(x)=9cos⁡x+6x3−3x5+8sen⁡x−ln⁡x+4x−5ex+6x+2f(x) = 9\cos x + \dfrac{6}{x^3} - 3x^5 + 8\operatorname{sen} x - \ln x + 4^x - 5e^x + 6x + 2
f(x)=−7sen⁡x−2x2+5x4+3tan⁡x+8ln⁡x−9x+6ex−2x+3f(x) = -7\operatorname{sen} x - \dfrac{2}{x^2} + 5x^4 + 3\tan x + 8\ln x - 9^x + 6e^x - 2x + 3
f(x)=6tan⁡x+4x5−8x3−2cos⁡x+5ln⁡x+7x−3ex+x−6f(x) = 6\tan x + \dfrac{4}{x^5} - 8x^3 - 2\cos x + 5\ln x + 7^x - 3e^x + x - 6
f(x)=−5cos⁡x+9x4+4x6+7sen⁡x−3ln⁡x−2x+ex−5x+8f(x) = -5\cos x + \dfrac{9}{x^4} + 4x^6 + 7\operatorname{sen} x - 3\ln x - 2^x + e^x - 5x + 8

Ejercicios extra 2

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Bloque C
f(x)=6sen⁡x−2x4+3exf(x) = 6 \operatorname{sen} x - \dfrac{2}{x^4} + 3e^x
f(x)=−4cos⁡x+ln⁡x−5x2f(x) = -4 \cos x + \ln x - 5x^2
f(x)=7x−2tan⁡x+5xf(x) = 7^x - 2 \tan x + \dfrac{5}{x}
f(x)=1x2+8cos⁡x−4xf(x) = \dfrac{1}{x^2} + 8 \cos x - 4^x
f(x)=3ex−6sen⁡x+x5f(x) = 3e^x - 6 \operatorname{sen} x + x^5
f(x)=9tan⁡x+3x3−ln⁡xf(x) = 9 \tan x + \dfrac{3}{x^3} - \ln x
Bloque D
f(x)=−2x6+5ln⁡x−sen⁡xf(x) = -2x^6 + 5 \ln x - \operatorname{sen} x
f(x)=7x5+2x−3cos⁡xf(x) = \dfrac{7}{x^5} + 2^x - 3 \cos x
f(x)=4tan⁡x−ex+6x2f(x) = 4 \tan x - e^x + \dfrac{6}{x^2}
f(x)=10cos⁡x−4x+9xf(x) = 10 \cos x - \dfrac{4}{x} + 9^x
f(x)=ln⁡x+7sen⁡x−2x4f(x) = \ln x + 7 \operatorname{sen} x - 2x^4
f(x)=−3tan⁡x+8x6+5exf(x) = -3 \tan x + \dfrac{8}{x^6} + 5e^x
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