Question

Cum se calculeaza derivata acestei functii?

Answer 1

$f(x) = \sqrt[3]{x^2-\ln^2(x+1)} = \Big[x^2-\ln^2(x+1)\Big]^\frac{1}{3} \\ \\ \text{Aplicam formula }u^n = n\cdot u^{n-1}\cdot u'\\ \\ \Big[\ln^2(x+1)\Big]' = 2\ln(x+1)\cdot \dfrac{1}{x+1} = \dfrac{2\ln(x+1)}{x+1} \\ \\ \\f'(x) = \dfrac{1}{3}\cdot \Big[x^2-\ln^2(x+1)\Big]^{\frac{1}{3}-1}\cdot \left[2x-\dfrac{2\ln(x+1)}{x+1}\right]\\ \\f'(x) = \dfrac{2x-\dfrac{2\ln(x+1)}{x+1}}{3\sqrt[3]{\Big[x^2-\ln^2 (x+1)\Big]^2}}$