# How do you differentiate (sqrt(1+3x^2))lnx^2?

$y = {\left(1 + 3 {x}^{2}\right)}^{\frac{1}{2}} \ln \left({x}^{2}\right)$
$y ' = \frac{1}{2} {\left(1 + 3 {x}^{2}\right)}^{\frac{1}{2} - 1} \cdot \left(6 x\right) \ln \left({x}^{2}\right) + {\left(1 + 3 {x}^{2}\right)}^{\frac{1}{2}} \frac{1}{x} ^ 2 \cdot 2 x =$
$= 3 x \ln {x}^{2} / \left(\sqrt{1 + 3 {x}^{2}}\right) + 2 \frac{\sqrt{1 + 3 {x}^{2}}}{x} =$
$= \frac{3 {x}^{2} \ln {x}^{2} + 2 + 6 {x}^{2}}{x \sqrt{1 + 3 {x}^{2}}}$