Calcular f'(8) , se f(x) = raizCubica(x) .

Passo 1

Vamos derivar mais um pouco.

Lembrando que:

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

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

Passo 2

Então, aplicando:

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

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

Vamos lembrar que:

(a - b) (a^2 + a b + b^2) = a^3 - b^3

Tendo a = root(3)(x + Delta x) e b = root(3)(x) , podemos fazer:

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

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

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

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

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

Agora, já podemos substituir Delta x para encontrar o limite:

y^' = (1)/((root(3)(x + 0))^2 + (root(3)(x + 0)) (root(3)(x)) + (root(3)(x))^2)

y^' = (1)/((root(3)(x))^2 + (root(3)(x))^2 + (root(3)(x))^2)

y^' = (1)/(x^((2)/(3)) + x^((2)/(3)) + x^((2)/(3)))

y^' = (1)/(3 x^((2)/(3))) = (1)/(3 root(3)(x^2))

Então, f ' ( 8 ) será:

y^' (8) = (1)/(3 root(3)(8^2)) = (1)/(3 * (2^3)^((2)/(3))) = (1)/(3 * 2^2) = (1)/(12) = 0,08333

Resposta

Portanto, f^' ( 8 ) de f (x) = root(3)(x) será igual a 0,08333 .

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