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Matemáticas 5 · Tema 18

Derivada por definición

Apuntes

La derivada de una función f(x)f(x) medida bajo la definición de límite se determina con la fórmula del cociente de diferencias:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Para resolverla se calcula f(x+h)f(x+h), se sustituye en el límite, se simplifica el numerador para poder factorizar y cancelar la hh que causa la indeterminación, y al final se evalúa el límite.

Ejemplos

Ejemplo 1

Dada f(x)=5x−3f(x) = 5x - 3, con f(x+h)=5(x+h)−3f(x+h) = 5(x+h) - 3:

f′(x)=lim⁡h→05(x+h)−3−(5x−3)h=lim⁡h→05x+5h−3−5x+3h=lim⁡h→05hh=lim⁡h→05f′(x)=5\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{5(x+h) - 3 - (5x - 3)}{h} \\ &= \lim_{h \to 0} \frac{5x + 5h - 3 - 5x + 3}{h} \\ &= \lim_{h \to 0} \frac{5h}{h} \\ &= \lim_{h \to 0} 5 \\ \mathbf{f'(x)} &= \mathbf{5} \end{aligned}
Ejemplo 2

Dada f(x)=2x2+4f(x) = 2x^2 + 4, con f(x+h)=2(x+h)2+4f(x+h) = 2(x+h)^2 + 4:

f′(x)=lim⁡h→02(x+h)2+4−(2x2+4)h=lim⁡h→02(x2+2xh+h2)+4−2x2−4h=lim⁡h→02x2+4xh+2h2−2x2h=lim⁡h→0h(4x+2h)h=lim⁡h→0(4x+2h)f′(x)=4x\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{2(x+h)^2 + 4 - (2x^2 + 4)}{h} \\ &= \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) + 4 - 2x^2 - 4}{h} \\ &= \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 - 2x^2}{h} \\ &= \lim_{h \to 0} \frac{h(4x + 2h)}{h} \\ &= \lim_{h \to 0} (4x + 2h) \\ \mathbf{f'(x)} &= \mathbf{4x} \end{aligned}

Ejercicios

Bloque 1
f(x)=4x+2f(x) = 4x + 2
f(x)=6x−5f(x) = 6x - 5
Bloque 2
f(x)=3x2f(x) = 3x^2
f(x)=−5x2f(x) = -5x^2
Bloque 3
f(x)=x2−3x−2f(x) = x^2 - 3x - 2
f(x)=x2−x+5f(x) = x^2 - x + 5

Respuestas

▼
Bloque 1
Dada f(x)=4x+2f(x) = 4x + 2
f′(x)=lim⁡h→04(x+h)+2−(4x+2)h=lim⁡h→04x+4h+2−4x−2h=lim⁡h→04hh=lim⁡h→04f′(x)=4\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{4(x+h) + 2 - (4x + 2)}{h} \\ &= \lim_{h \to 0} \frac{4x + 4h + 2 - 4x - 2}{h} \\ &= \lim_{h \to 0} \frac{4h}{h} \\ &= \lim_{h \to 0} 4 \\ \mathbf{f'(x)} &= \mathbf{4} \end{aligned}
Dada f(x)=6x−5f(x) = 6x - 5
f′(x)=lim⁡h→06(x+h)−5−(6x−5)h=lim⁡h→06x+6h−5−6x+5h=lim⁡h→06hh=lim⁡h→06f′(x)=6\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{6(x+h) - 5 - (6x - 5)}{h} \\ &= \lim_{h \to 0} \frac{6x + 6h - 5 - 6x + 5}{h} \\ &= \lim_{h \to 0} \frac{6h}{h} \\ &= \lim_{h \to 0} 6 \\ \mathbf{f'(x)} &= \mathbf{6} \end{aligned}
Bloque 2
Dada f(x)=3x2f(x) = 3x^2
f′(x)=lim⁡h→03(x+h)2−3x2h=lim⁡h→03(x2+2xh+h2)−3x2h=lim⁡h→03x2+6xh+3h2−3x2h=lim⁡h→0h(6x+3h)h=lim⁡h→0(6x+3h)f′(x)=6x\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{3(x+h)^2 - 3x^2}{h} \\ &= \lim_{h \to 0} \frac{3(x^2 + 2xh + h^2) - 3x^2}{h} \\ &= \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} \\ &= \lim_{h \to 0} \frac{h(6x + 3h)}{h} \\ &= \lim_{h \to 0} (6x + 3h) \\ \mathbf{f'(x)} &= \mathbf{6x} \end{aligned}
Dada f(x)=−5x2f(x) = -5x^2
f′(x)=lim⁡h→0−5(x+h)2−(−5x2)h=lim⁡h→0−5(x2+2xh+h2)+5x2h=lim⁡h→0−5x2−10xh−5h2+5x2h=lim⁡h→0h(−10x−5h)h=lim⁡h→0(−10x−5h)f′(x)=−10x\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{-5(x+h)^2 - (-5x^2)}{h} \\ &= \lim_{h \to 0} \frac{-5(x^2 + 2xh + h^2) + 5x^2}{h} \\ &= \lim_{h \to 0} \frac{-5x^2 - 10xh - 5h^2 + 5x^2}{h} \\ &= \lim_{h \to 0} \frac{h(-10x - 5h)}{h} \\ &= \lim_{h \to 0} (-10x - 5h) \\ \mathbf{f'(x)} &= \mathbf{-10x} \end{aligned}
Bloque 3
Dada f(x)=x2−3x−2f(x) = x^2 - 3x - 2
f′(x)=lim⁡h→0(x+h)2−3(x+h)−2−(x2−3x−2)h=lim⁡h→0x2+2xh+h2−3x−3h−2−x2+3x+2h=lim⁡h→02xh+h2−3hh=lim⁡h→0h(2x+h−3)h=lim⁡h→0(2x+h−3)f′(x)=2x−3\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{(x+h)^2 - 3(x+h) - 2 - (x^2 - 3x - 2)}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 3x - 3h - 2 - x^2 + 3x + 2}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 - 3h}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h - 3)}{h} \\ &= \lim_{h \to 0} (2x + h - 3) \\ \mathbf{f'(x)} &= \mathbf{2x - 3} \end{aligned}
Dada f(x)=x2−x+5f(x) = x^2 - x + 5
f′(x)=lim⁡h→0(x+h)2−(x+h)+5−(x2−x+5)h=lim⁡h→0x2+2xh+h2−x−h+5−x2+x−5h=lim⁡h→02xh+h2−hh=lim⁡h→0h(2x+h−1)h=lim⁡h→0(2x+h−1)f′(x)=2x−1\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{(x+h)^2 - (x+h) + 5 - (x^2 - x + 5)}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x - h + 5 - x^2 + x - 5}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 - h}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h - 1)}{h} \\ &= \lim_{h \to 0} (2x + h - 1) \\ \mathbf{f'(x)} &= \mathbf{2x - 1} \end{aligned}

Ejercicios extra 1

▼
Bloque A
f(x)=3x+7f(x) = 3x + 7
f(x)=−2x+9f(x) = -2x + 9
Bloque B
f(x)=2x2f(x) = 2x^2
f(x)=−4x2f(x) = -4x^2
Bloque C
f(x)=x2−5x+4f(x) = x^2 - 5x + 4
f(x)=2x2+3x−1f(x) = 2x^2 + 3x - 1

Ejercicios extra 2

▼
Bloque D
f(x)=5x−4f(x) = 5x - 4
f(x)=6x2f(x) = 6x^2
f(x)=−x2+4x+6f(x) = -x^2 + 4x + 6
Bloque E
f(x)=−7x−2f(x) = -7x - 2
f(x)=−3x2f(x) = -3x^2
f(x)=3x2−2x+5f(x) = 3x^2 - 2x + 5
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