# How do you use the product rule to differentiate sqrt(1+3x^2 )lnx^2?

Jun 15, 2016

Solution

#### Explanation:

Let us assume the following

$u = \sqrt{1 + 3 {x}^{2}}$
$v = \ln {x}^{2}$
$d \left(u v\right) = u . \mathrm{dv} + v . \mathrm{du}$
$d \left(u v\right) = \sqrt{1 + 3 {x}^{2}} \setminus \times \frac{1}{x} ^ 2 \setminus \times 2 x + \ln {x}^{2} \setminus \times \frac{1}{2 \sqrt{1 + 3 {x}^{2}}} \setminus \times 6 x$
$= \frac{2 \sqrt{1 + 3 {x}^{2}}}{x} + \frac{3 x \ln {x}^{2}}{\sqrt{1 + 3 {x}^{2}}}$

On further reduction will lead to
$\frac{d \left(u v\right)}{\mathrm{dx}} = \frac{2 + 6 {x}^{2} + 3 {x}^{2} \ln {x}^{2}}{x \sqrt{1 + 3 {x}^{2}}}$