Question

Решите номер 5 .Есть вложение. 25 б

Answer 1

$(ln(ln(3-2x^3))'=\frac{1}{ln(3-2x^3)}\cdot(ln(3-2x^3))'=$

$\frac{1}{ln(3-2x^3)}\cdot\frac{1}{3-2x^3}\cdot(3-2x^3)'=$

$\frac{1}{ln(3-2x^3)}\cdot\frac{1}{3-2x^3}\cdot(-6x^2)=$

$\frac{-6x^2}{(3-2x^3)ln(3-2x^3)}=\frac{6x^2}{(2x^3-3)ln(3-2x^3)}$