Achar y^' = underset(Delta x -> 0)(lim) (Delta y)/(Delta x) para as funções:

  1. y = x^3 ;
  2. y = (1)/(x^2) ;
  3. y = sqrt(x) ;
  4. y = cotg x .

Passo 1

Vamos começar lembrando que:

Delta y = f (x + Delta x) - f ( x )

Passo 2

  1. y = x^3

Assim, temos:

Delta y = (x + Delta x)^3 - (x)^3

Delta y = (x^2 + 2 x Delta x + Delta x^2) (x + Delta x) - x^3

Delta y = x^3 + x^2 Delta x + 2 x^2 Delta x + 2 x Delta x^2 + x Delta x^2 + Delta x^3 - x^3

Delta y = 3 x^2 Delta x + 3 x Delta x^2 + Delta x^3

Delta y = Delta x ( 3 x^2 + 3 x Delta x + Delta x^2 )

Aplicando o limite:

y^' = underset(Delta x -> 0)(lim) (Delta y)/(Delta x)

y^' = underset(Delta x -> 0)(lim) (Delta x ( 3 x^2 + 3 x Delta x + Delta x^2 ))/(Delta x)

y^' = underset(Delta x -> 0)(l i m) (3 x^2 + 3 x Delta x + Delta x^2)

y^' = 3 x^2 + 3 x (0) + (0)^2

y^' = 3 x^2

Passo 3

  1. y = (1)/(x^2)
  2. Logo temos:

    Delta y = (1)/((x + Delta x)^2) - (1)/(x^2)

    Delta y = (1)/(x^2 + 2 x Delta x + Delta x^2) - (1)/(x^2)

    Delta y = (x^2 - (x^2 + 2 x Delta x + Delta x^2))/((x^2 + 2 x Delta x + Delta x^2) (x^2))

    Delta y = (x^2 - x^2 - 2 x Delta x - Delta x^2)/(x^4 + 2 x^3 Delta x + x^2 Delta x^2)

    Delta y = (- Delta x ( 2 x + Delta x ))/(x^4 + 2 x^3 Delta x + x^2 Delta x^2)

    Aplicando o limite:

    y^' = underset(Delta x -> 0)(lim) (Delta y)/(Delta x)

    y^' = underset(Delta x -> 0)(lim) ((- Delta x (2 x + Delta x))/(x^4 + 2 x^3 Delta x + x^2 Delta x^2))/(Delta x)

    y^' = underset(Delta x -> 0)(lim) (- Delta x (2 x + Delta x))/(x^4 + 2 x^3 Delta x + x^2 Delta x^2) * (1)/(Delta x)

    y^' = underset(Delta x -> 0)(lim) (- (2 x + Delta x))/(x^4 + 2 x^3 Delta x + x^2 Delta x^2)

    y^' = (- 2 x - 0)/(x^4 + 2 x^3 (0) + x^2 (0)^2)

    y^' = - (2 x)/(x^4)

    y^' = - (2)/(x^3)

    Passo 4

  3. sqrt(x)
  4. Então:

    Delta y = sqrt(x + Delta x) - sqrt(x)

    Aplicando o limite:

    y^' = underset(Delta x -> 0)(lim) (sqrt(x + Delta x) - sqrt(x))/(Delta x)

    Vamos multiplicar o numerador e denominador por ( sqrt(x + Delta x) + sqrt(x) ) para facilitar nosso trabalho:

    y^' = underset(Delta x -> 0)(lim) ((sqrt(x + Delta x) - sqrt(x)) (sqrt(x + Delta x) - sqrt(x)))/(Delta x (sqrt(x + Delta x) - sqrt(x)))

    y^' = underset(Delta x -> 0)(lim) ((sqrt(x + Delta x) - sqrt(x)) (sqrt(x + Delta x) + sqrt(x)))/(Delta x (sqrt(x + Delta x) + sqrt(x)))

    y^' = underset(Delta x -> 0)(lim) ((sqrt(x + Delta x))^2 - (sqrt(x))^2)/(Delta x (sqrt(x + Delta x) + sqrt(x)))

    y^' = underset(Delta x -> 0)(lim) (x + Delta x - x)/(Delta x (sqrt(x + Delta x) + sqrt(x)))

    y^' = underset(Delta x -> 0)(lim) (Delta x)/(Delta x (sqrt(x + Delta x) + sqrt(x)))

    y^' = underset(Delta x -> 0)(lim) (1)/((sqrt(x + Delta x) + sqrt(x)))

    y^' = (1)/(sqrt(x + 0) + sqrt(x))

    y^' = (1)/(2 sqrt(x))

    Passo 5

  5. y = cotg x

Vamos direto para a definição do limite, onde teremos:

y^' = underset(Delta x -> 0)(lim) (f (x + Delta x) - f ( x ))/(Delta x)

y^' = underset(Delta x -> 0)(lim) (cotg ( x + Delta x ) - c o t g ( x ))/(Delta x)

Com as identidades trigonométricas, sabemos que cotg a = (cos a)/(sen a) . Então substituímos:

y^' = underset(Delta x -> 0)(lim) ((cos ( x + Delta x ))/(sen ( x + Delta x )) - (cos x)/(sen x))/(Delta x)

y^' = underset(Delta x -> 0)(lim) (sen x cos (x + Delta x) - cos x sen (x + Delta x))/(Delta x sen (x + Delta x) sen x)

Recorrendo às identidades trigonométricas novamente, temos que:

cos ( a + b ) = cos a cos b - sen a sen b

sen ( a + b ) = sen a cos b + cos a sen b

Então, aplicando, teremos:

y^' = underset(Delta x -> 0)(lim) (sen x (cos x cos Delta x - sen x sen Delta x) - cos x (sen x cos Delta x + cos x sen Delta x))/(Delta x sen (x + Delta x) sen x)

y^' = underset(Delta x -> 0)(lim) (sen x cos x cos Delta x - sen^2 x sen Delta x - cos x sen x cos Delta x - cos^2 x sen Delta x)/(Delta x sen (x + Delta x) sen x)

y^' = underset(Delta x -> 0)(lim) (- sen^2 x sen Delta x - cos^2 x sen Delta x)/(Delta x sen (x + Delta x) sen x)

y^' = underset(Delta x -> 0)(lim) (- sen^2 x sen Delta x - cos^2 x sen Delta x)/(Delta x sen (x + Delta x) sen x)

y^' = underset(Delta x -> 0)(lim) (- sen Delta x ( cos^2 x + sen^2 x ))/(Delta x sen (x + Delta x) sen x)

Lembrando também, que:

cos^2 a + sen^2 a = 1

Logo:

y^' = underset(Delta x -> 0)(l i m) - (sen Delta x)/(Delta x) * (1)/(sen (x + Delta x) sen x)

Aplicando o limite, teremos:

y^' = - (sen 0)/(0) * (1)/(sen (x + 0) sen x)

y^' = - (0)/(0) * (1)/(sen x * sen x)

Note que:

(A)/(A) = 1 e n t ã o (0)/(0) = 1

Ficamos com:

y^' = - (1)/(sen^2 x) = - cosec^2 x

Resposta

Deu certo aí também?

  1. y^' = 3 x^2
  2. y^' = - (2)/(x^3)
  3. y^' = (1)/(2 sqrt(x))
  4. y ' = - cosec^2 x

Qualquer dúvida, pode mandar pra gente!