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Lição 11 — Derivada: definição via limite

Derivada como limite do quociente incremental de Newton. Interpretação geométrica (reta tangente) e física (velocidade instantânea). Notações. Derivabilidade implica continuidade.

Used in: Cálculo 1 — Unidade 2 · USP MAC0105 · ITA MA-011

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

O quociente incremental f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} mede a taxa média de variação de ff no intervalo [a, a+h][a,\ a+h]. A derivada f′(a)f'(a) é o limite dessa taxa quando h→0h \to 0 — a taxa instantânea de variação. Geometricamente, é o coeficiente angular da reta tangente ao gráfico de ff em (a,f(a))(a,f(a)).

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Rigorous notation, full derivation, hypotheses

Definição rigorosa

A derivada — definição e primeiras propriedades

Exemplos resolvidos

Exercícios

Exercise list

30 exercises · 7 with worked solution (25%)

Application 24Understanding 6
  1. Ex. 11.1Application

    Calcule a inclinação da reta secante a f(x)=4x+7f(x) = 4x + 7 entre x1=2x_1 = 2 e x2=5x_2 = 5.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 1
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    A inclinação da reta secante é f(5)−f(2)5−2=27−153=4\frac{f(5)-f(2)}{5-2} = \frac{27-15}{3} = 4. Como f(x)=4x+7f(x)=4x+7 é linear, a secante tem sempre inclinação 4.
  2. Ex. 11.2ApplicationAnswer key

    Calcule a inclinação da reta secante a f(x)=8x−3f(x) = 8x - 3 entre x1=−1x_1 = -1 e x2=3x_2 = 3.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 2
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    f(3)−f(−1)3−(−1)=(24−3)−(−8−3)4=21−(−11)4=324=8\frac{f(3)-f(-1)}{3-(-1)} = \frac{(24-3)-(-8-3)}{4} = \frac{21-(-11)}{4} = \frac{32}{4} = 8. Novamente, ff é linear, então a inclinação da secante é sempre igual ao coeficiente angular.
  3. Ex. 11.3Application

    Calcule a inclinação da reta secante a f(x)=x2+2x+1f(x) = x^2 + 2x + 1 entre x1=3x_1 = 3 e x2=3,5x_2 = 3{,}5.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 3
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    f(3)=9+6+1=16f(3) = 9+6+1 = 16 e f(3,5)=12,25+7+1=20,25f(3{,}5) = 12{,}25+7+1 = 20{,}25. Inclinação: 20,25−160,5=4,250,5=8,5\frac{20{,}25 - 16}{0{,}5} = \frac{4{,}25}{0{,}5} = 8{,}5.
  4. Ex. 11.4Application

    Calcule a inclinação da reta secante a f(x)=−x2+x+2f(x) = -x^2 + x + 2 entre x1=0,5x_1 = 0{,}5 e x2=1,5x_2 = 1{,}5.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 4
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    f(0,5)=−0,25+0,5+2=2,25f(0{,}5) = -0{,}25+0{,}5+2 = 2{,}25 e f(1,5)=−2,25+1,5+2=1,25f(1{,}5) = -2{,}25+1{,}5+2 = 1{,}25. Inclinação: 1,25−2,251=−1\frac{1{,}25-2{,}25}{1} = -1.
    Show step-by-step (with the why)
    1. Calcule f(0,5)=−(0,5)2+0,5+2=−0,25+0,5+2=2,25f(0{,}5) = -(0{,}5)^2 + 0{,}5 + 2 = -0{,}25 + 0{,}5 + 2 = 2{,}25.
    2. Calcule f(1,5)=−(1,5)2+1,5+2=−2,25+1,5+2=1,25f(1{,}5) = -(1{,}5)^2 + 1{,}5 + 2 = -2{,}25 + 1{,}5 + 2 = 1{,}25.
    3. Inclinação: m=f(1,5)−f(0,5)1,5−0,5=1,25−2,251=−1m = \frac{f(1{,}5)-f(0{,}5)}{1{,}5-0{,}5} = \frac{1{,}25-2{,}25}{1} = -1.
  5. Ex. 11.5Application

    Calcule a inclinação da reta secante a f(x)=43x−1f(x) = \tfrac{4}{3}x - 1 entre x1=1x_1 = 1 e x2=3x_2 = 3.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 5
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    f(1)=43−1=13f(1) = \frac{4}{3}-1 = \frac{1}{3} e f(3)=4−1=3f(3) = 4-1 = 3. Inclinação: 3−1/33−1=8/32=43\frac{3 - 1/3}{3-1} = \frac{8/3}{2} = \frac{4}{3}.
  6. Ex. 11.6ApplicationAnswer key

    Calcule a inclinação da reta secante a f(x)=x−72x+1f(x) = \dfrac{x-7}{2x+1} entre x1=0x_1 = 0 e x2=2x_2 = 2.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 6
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    f(0)=0−70+1=−7f(0) = \frac{0-7}{0+1} = -7 e f(2)=2−74+1=−1f(2) = \frac{2-7}{4+1} = -1. Inclinação: −1−(−7)2−0=62=3\frac{-1-(-7)}{2-0} = \frac{6}{2} = 3.
  7. Ex. 11.7Application

    Calcule a inclinação da reta secante a f(x)=xf(x) = \sqrt{x} entre x1=1x_1 = 1 e x2=16x_2 = 16.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 7
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    f(1)=1f(1)=1 e f(16)=4f(16)=4. Inclinação: 4−116−1=315=15\frac{4-1}{16-1} = \frac{3}{15} = \frac{1}{5}.
    Show step-by-step (with the why)
    1. Calcule f(1)=1=1f(1) = \sqrt{1} = 1 e f(16)=16=4f(16) = \sqrt{16} = 4.
    2. Inclinação: m=4−116−1=315=15m = \frac{4 - 1}{16 - 1} = \frac{3}{15} = \frac{1}{5}.
  8. Ex. 11.8ApplicationAnswer key

    Calcule a inclinação da reta secante a f(x)=x−9f(x) = \sqrt{x-9} entre x1=10x_1 = 10 e x2=13x_2 = 13.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 8
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    f(10)=1=1f(10)=\sqrt{1}=1 e f(13)=4=2f(13)=\sqrt{4}=2. Inclinação: 2−113−10=13\frac{2-1}{13-10} = \frac{1}{3}.
  9. Ex. 11.9Application

    Calcule a inclinação da reta secante a f(x)=x1/3+1f(x) = x^{1/3} + 1 entre x1=0x_1 = 0 e x2=8x_2 = 8.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 9
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    f(0)=0+1=1f(0)=0+1=1 e f(8)=2+1=3f(8)=2+1=3. Inclinação: 3−18−0=28=14\frac{3-1}{8-0} = \frac{2}{8} = \frac{1}{4}.
  10. Ex. 11.10Application

    Use a definição para calcular f′(2)f'(2) sendo f(x)=3−4xf(x) = 3 - 4x.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 11
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    Pela definição: f′(2)=lim⁡h→0(3−4(2+h))−(3−8)h=lim⁡h→0−4hh=−4f'(2) = \lim_{h\to 0}\frac{(3-4(2+h))-(3-8)}{h} = \lim_{h\to 0}\frac{-4h}{h} = -4. Sendo ff linear, a derivada é constante e igual ao coeficiente angular.
  11. Ex. 11.11Application

    Use a definição para calcular f′(1)f'(1) sendo f(x)=x2+xf(x) = x^2 + x.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 13
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    f′(1)=lim⁡h→0(1+h)2+(1+h)−2h=lim⁡h→03h+h2h=3f'(1) = \lim_{h\to 0}\frac{(1+h)^2+(1+h)-2}{h} = \lim_{h\to 0}\frac{3h+h^2}{h} = 3. Portanto f′(1)=3f'(1)=3.
    Show step-by-step (with the why)
    1. Escreva f(1+h)=(1+h)2+(1+h)=1+2h+h2+1+h=2+3h+h2f(1+h) = (1+h)^2+(1+h) = 1+2h+h^2+1+h = 2+3h+h^2.
    2. Subtraia f(1)=1+1=2f(1) = 1+1 = 2: numerador =3h+h2= 3h+h^2.
    3. Divida por hh e tome o limite: lim⁡h→0(3+h)=3\lim_{h\to 0}(3+h) = 3.
  12. Ex. 11.12Application

    Use a definição para calcular f′(0)f'(0) sendo f(x)=1−x−x2f(x) = 1 - x - x^2.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 14
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    f′(0)=lim⁡h→0(1−h−h2)−1h=lim⁡h→0(−1−h)=−1f'(0) = \lim_{h\to 0}\frac{(1-h-h^2)-1}{h} = \lim_{h\to 0}(-1-h) = -1.
  13. Ex. 11.13Application

    Use a definição para calcular f′(4)f'(4) sendo f(x)=xf(x) = \sqrt{x}.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 25
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    Multiplicando pelo conjugado: 4+h−2h⋅4+h+24+h+2=hh(4+h+2)→14\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2} = \frac{h}{h(\sqrt{4+h}+2)} \to \frac{1}{4}.
  14. Ex. 11.14Application

    Use a definição para calcular f′(2)f'(2) sendo f(x)=1xf(x) = \dfrac{1}{x}.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 27
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    12+h−12h=2−(2+h)2h(2+h)=−12(2+h)→−14\frac{\frac{1}{2+h}-\frac{1}{2}}{h} = \frac{2-(2+h)}{2h(2+h)} = \frac{-1}{2(2+h)} \to -\frac{1}{4}.
    Show step-by-step (with the why)
    1. Escreva o quociente incremental: 1/(2+h)−1/2h\frac{1/(2+h)-1/2}{h}.
    2. Some as frações no numerador: =2−(2+h)2h(2+h)=−h2h(2+h)= \frac{2-(2+h)}{2h(2+h)} = \frac{-h}{2h(2+h)}.
    3. Cancele hh: =−12(2+h)= \frac{-1}{2(2+h)}.
    4. Tome o limite: lim⁡h→0−12(2+h)=−14\lim_{h\to 0} \frac{-1}{2(2+h)} = -\frac{1}{4}.
  15. Ex. 11.15Application

    Use a definição para calcular f′(2)f'(2) sendo f(x)=x2+9xf(x) = x^2 + 9x.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 23
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    f(2+h)=(2+h)2+9(2+h)=4+4h+h2+18+9hf(2+h) = (2+h)^2+9(2+h) = 4+4h+h^2+18+9h. Subtraindo f(2)=22f(2)=22: 13h+h2h→13\frac{13h+h^2}{h}\to 13.
  16. Ex. 11.16ApplicationAnswer key

    Use a definição para calcular f′(1)f'(1) sendo f(x)=3x2−x+2f(x) = 3x^2 - x + 2.

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    Solve onlineOpenStax Calculus Volume 1 · §3.1 · ex. 24
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    f(1+h)=3(1+h)2−(1+h)+2=3+6h+3h2−1−h+2=4+5h+3h2f(1+h) = 3(1+h)^2-(1+h)+2 = 3+6h+3h^2-1-h+2 = 4+5h+3h^2. f(1)=4f(1)=4. Portanto 5h+3h2h→5\frac{5h+3h^2}{h}\to 5.
  17. Ex. 11.17Understanding

    Use a definição para encontrar f′(x)f'(x) sendo f(x)=6f(x) = 6.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 54
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    Pela definição: f′(x)=lim⁡h→06−6h=0f'(x) = \lim_{h\to 0}\frac{6-6}{h} = 0. A derivada de toda constante é zero.
  18. Ex. 11.18Application

    Use a definição para encontrar f′(x)f'(x) sendo f(x)=2−3xf(x) = 2 - 3x.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 55
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    (2−3(x+h))−(2−3x)h=−3hh=−3\frac{(2-3(x+h))-(2-3x)}{h} = \frac{-3h}{h} = -3. Logo f′(x)=−3f'(x) = -3.
  19. Ex. 11.19ApplicationAnswer key

    Use a definição para encontrar f′(x)f'(x) sendo f(x)=4x2f(x) = 4x^2.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 57
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    4(x+h)2−4x2h=8xh+4h2h=8x+4h→8x\frac{4(x+h)^2-4x^2}{h} = \frac{8xh+4h^2}{h} = 8x+4h \to 8x. Logo f′(x)=8xf'(x) = 8x.
    Show step-by-step (with the why)
    1. Expanda: 4(x+h)2=4x2+8xh+4h24(x+h)^2 = 4x^2 + 8xh + 4h^2.
    2. Subtraia f(x)=4x2f(x) = 4x^2: numerador =8xh+4h2= 8xh + 4h^2.
    3. Divida por hh: =8x+4h= 8x + 4h.
    4. Tome o limite: f′(x)=8xf'(x) = 8x.
  20. Ex. 11.20Application

    Use a definição para encontrar f′(x)f'(x) sendo f(x)=5x−x2f(x) = 5x - x^2.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 58
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    5(x+h)−(x+h)2−(5x−x2)h=5h−2xh−h2h=5−2x−h→5−2x\frac{5(x+h)-(x+h)^2-(5x-x^2)}{h} = \frac{5h-2xh-h^2}{h} = 5-2x-h \to 5-2x.
  21. Ex. 11.21Application

    Use a definição para encontrar f′(x)f'(x) sendo f(x)=x+1xf(x) = x + \dfrac{1}{x}.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 62
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    (x+h+1x+h)−(x+1x)h=1+1h(1x+h−1x)=1−1x(x+h)→1−1x2\frac{(x+h+\frac{1}{x+h})-(x+\frac{1}{x})}{h} = 1 + \frac{1}{h}\left(\frac{1}{x+h}-\frac{1}{x}\right) = 1 - \frac{1}{x(x+h)} \to 1-\frac{1}{x^2}.
    Show step-by-step (with the why)
    1. Numerador: (x+h+1x+h)−(x+1x)=h+1x+h−1x(x+h+\frac{1}{x+h}) - (x+\frac{1}{x}) = h + \frac{1}{x+h} - \frac{1}{x}.
    2. Simplifique a parte racional: 1x+h−1x=−hx(x+h)\frac{1}{x+h}-\frac{1}{x} = \frac{-h}{x(x+h)}.
    3. Divida por hh: 1−1x(x+h)1 - \frac{1}{x(x+h)}.
    4. Limite: f′(x)=1−1x2f'(x) = 1 - \frac{1}{x^2}.
  22. Ex. 11.22ApplicationAnswer key

    Use a definição para encontrar f′(x)f'(x) sendo f(x)=1xf(x) = \dfrac{1}{x}.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 63
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    1x+h−1xh=−hhx(x+h)=−1x(x+h)→−1x2\frac{\frac{1}{x+h}-\frac{1}{x}}{h} = \frac{-h}{hx(x+h)} = \frac{-1}{x(x+h)} \to -\frac{1}{x^2}.
  23. Ex. 11.23Understanding

    O limite lim⁡h→0(1+h)2/3−1h\lim_{h\to 0}\dfrac{(1+h)^{2/3}-1}{h} representa f′(a)f'(a) para qual função ff e ponto aa?

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 68
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    O limite lim⁡h→0(1+h)2/3−1h\lim_{h\to 0}\frac{(1+h)^{2/3}-1}{h} tem a forma f(1+h)−f(1)h\frac{f(1+h)-f(1)}{h} com f(x)=x2/3f(x)=x^{2/3} e f(1)=1f(1)=1, logo a=1a=1.
  24. Ex. 11.24Understanding

    O limite lim⁡h→03(2+h)2+2−14h\lim_{h\to 0}\dfrac{3(2+h)^2+2-14}{h} representa f′(a)f'(a). Identifique f(x)f(x), aa e calcule o valor.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 69
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    O limite lim⁡h→03(2+h)2+2−14h\lim_{h\to 0}\frac{3(2+h)^2+2-14}{h} tem a forma f′(2)f'(2) com f(x)=3x2+2f(x)=3x^2+2 e f(2)=14f(2)=14. Logo f′(2)=6⋅2=12f'(2)=6\cdot 2 = 12.
    Show step-by-step (with the why)
    1. Identifique f(2+h)=3(2+h)2+2f(2+h) = 3(2+h)^2+2 e f(2)=3⋅4+2=14f(2) = 3\cdot 4 + 2 = 14.
    2. Expanda: 3(2+h)2+2=12+12h+3h2+2=14+12h+3h23(2+h)^2+2 = 12+12h+3h^2+2 = 14+12h+3h^2.
    3. Subtraia 14 e divida por hh: 12+3h→1212+3h\to 12.
  25. Ex. 11.25Understanding

    O limite lim⁡h→0(2+h)4−16h\lim_{h\to 0}\dfrac{(2+h)^4-16}{h} representa f′(a)f'(a). Identifique f(x)f(x), aa e calcule o valor.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 71
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    O limite lim⁡h→0(2+h)4−16h\lim_{h\to 0}\frac{(2+h)^4-16}{h} tem a forma f′(2)f'(2) com f(x)=x4f(x)=x^4 (pois 24=162^4=16). Derivada: f′(x)=4x3f'(x)=4x^3, logo f′(2)=32f'(2)=32.
  26. Ex. 11.26UnderstandingAnswer key

    O limite lim⁡h→02(3+h)2−(3+h)−15h\lim_{h\to 0}\dfrac{2(3+h)^2-(3+h)-15}{h} representa f′(a)f'(a). Identifique f(x)f(x), aa e calcule o valor.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 72
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    O numerador é f(3+h)−f(3)f(3+h)-f(3) com f(x)=2x2−xf(x)=2x^2-x e f(3)=18−3=15f(3)=18-3=15. Então f′(3)=4(3)−1=11f'(3)=4(3)-1=11.
  27. Ex. 11.27Understanding

    O limite lim⁡h→0eh−1h\lim_{h\to 0}\dfrac{e^h - 1}{h} representa f′(a)f'(a). Identifique f(x)f(x), aa e o valor do limite.

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    Solve onlineOpenStax Calculus Volume 1 · §3.2 · ex. 73
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    O limite lim⁡h→0eh−1h\lim_{h\to 0}\frac{e^h-1}{h} tem a forma e0+h−e0h\frac{e^{0+h}-e^0}{h}, ou seja, f′(0)f'(0) com f(x)=exf(x)=e^x. Como (ex)′=ex(e^x)' = e^x, temos f′(0)=1f'(0)=1.
  28. Ex. 11.28Application

    Use a definição para calcular f′(3)f'(3) sendo f(x)=2x2f(x) = \dfrac{2}{x^2}.

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    Solve onlineActive Calculus · §1.3 · ex. 3
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    Pela definição: f′(3)=lim⁡h→02(3+h)2−29h=lim⁡h→018−2(3+h)29h(3+h)2f'(3) = \lim_{h\to 0}\frac{\frac{2}{(3+h)^2}-\frac{2}{9}}{h} = \lim_{h\to 0}\frac{18-2(3+h)^2}{9h(3+h)^2}. Expandindo: 18−2(9+6h+h2)=−12h−2h218-2(9+6h+h^2) = -12h-2h^2. Logo f′(3)=−129⋅9=−1281=−427f'(3) = \frac{-12}{9\cdot 9} = -\frac{12}{81} = -\frac{4}{27}.
    Show step-by-step (with the why)
    1. Escreva o quociente: 2/(3+h)2−2/9h\frac{2/(3+h)^2 - 2/9}{h}.
    2. Some com denominador comum 9(3+h)29(3+h)^2: numerador =18−2(3+h)2=−12h−2h2= 18 - 2(3+h)^2 = -12h - 2h^2.
    3. Divida por hh: −12−2h9(3+h)2\frac{-12-2h}{9(3+h)^2}.
    4. Limite: −1281=−427\frac{-12}{81} = -\frac{4}{27}.
  29. Ex. 11.29Application

    Use a definição para encontrar f′(a)f'(a) sendo f(x)=7+11x−3x2f(x) = 7 + 11x - 3x^2.

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    Solve onlineActive Calculus · §1.3 · ex. 8
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    Pela definição: f(a+h)=7+11(a+h)−3(a+h)2=7+11a+11h−3a2−6ah−3h2f(a+h) = 7+11(a+h)-3(a+h)^2 = 7+11a+11h-3a^2-6ah-3h^2. Subtraindo f(a)f(a) e dividindo por hh: 11−6a−3h→11−6a11-6a-3h \to 11-6a.
  30. Ex. 11.30Application

    Seja f(x)=9x2−6x+2f(x) = 9x^2 - 6x + 2. O quociente f(1+h)−f(1)h\dfrac{f(1+h)-f(1)}{h} pode ser simplificado para ah+bah + b. Quais são os valores de aa e bb?

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    Solve onlineActive Calculus · §1.4 · ex. 6
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    f(1+h)=9(1+h)2−6(1+h)+2=9+18h+9h2−6−6h+2=5+12h+9h2f(1+h) = 9(1+h)^2-6(1+h)+2 = 9+18h+9h^2-6-6h+2 = 5+12h+9h^2. Como f(1)=5f(1)=5, temos f(1+h)−f(1)h=12+9h=9h+12\frac{f(1+h)-f(1)}{h} = 12+9h = 9h+12. Logo a=9a=9 e b=12b=12.

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Updated on 2026-05-21 · Author(s): Clube da Matemática

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