Calcular f'(8) , se f(x) = raizCubica(x) . Passo 1Vamos 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 2Então, 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 RespostaPortanto, f^' ( 8 ) de f (x) = root(3)(x) será igual a 0,08333 . Compreendeu aí? Se tiver algo que não ficou bem claro, pode falar que a gente explica melhor.MOSTRAR SOLUÇÃO COMPLETA