Решите номер 5 .Есть вложение. 25 б

Решите задачу:

$(\frac{1}{\sqrt3}ln(\sqrt3x+\sqrt{3x^2-2})'=\frac{1}{\sqrt3}\cdot(ln(\sqrt3x+\sqrt{3x^2-2})'=$

$\frac{1}{\sqrt3}\cdot\frac{1}{\sqrt3x+\sqrt{3x^2-2}}\cdot(\sqrt3x+\sqrt{3x^2-2})'=$

$\frac{1}{\sqrt3}\cdot\frac{1}{\sqrt3x+\sqrt{3x^2-2}}\cdot\left(\sqrt3+ \frac{1}{2\sqrt{3x^2-2}} \cdot(3x^2-2)'\right) =$

$\frac{1}{\sqrt3\cdot(\sqrt3x+\sqrt{3x^2-2})}\cdot\left(\sqrt3+ \frac{6x}{2\sqrt{3x^2-2}}\right) =$

$\frac{1}{\sqrt3\cdot(\sqrt3x+\sqrt{3x^2-2})}\cdot\frac{\sqrt3\cdot2\sqrt{3x^2-2}+6x}{2\sqrt{3x^2-2}} =$

$\frac{1}{\sqrt3\cdot(\sqrt3x+\sqrt{3x^2-2})}\cdot\frac{2\sqrt3(\sqrt{3x^2-2}+\sqrt3x)}{2\sqrt{3x^2-2}} =$

$\frac{1}{\sqrt{3x^2-2}}$