# What is the derivative of ln((2x^2 - 3)/(2x^2 + 3))^(1/2) ?

Mar 9, 2018

$y ' = \frac{12 x}{4 {x}^{4} - 9}$

#### Explanation:

$y = \ln {\left(\frac{2 {x}^{2} - 3}{2 {x}^{2} + 3}\right)}^{\frac{1}{2}}$

With the logarithm properties

$y = \left(\frac{1}{2}\right) \ln \left(\frac{2 {x}^{2} - 3}{2 {x}^{2} + 3}\right)$

Then

$y ' = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{\frac{2 {x}^{2} - 3}{2 {x}^{2} + 3}}\right) \cdot \left(\frac{\left(4 x\right) \left(2 {x}^{2} + 3\right) - \left(2 {x}^{2} - 3\right) \left(4 x\right)}{2 {x}^{2} + 3} ^ 2\right)$

$y ' = \left(\frac{1}{2}\right) \cdot \left(\frac{2 {x}^{2} + 3}{2 {x}^{2} - 3}\right) \cdot \left(\frac{\cancel{8 {x}^{3}} + 12 x - \cancel{8 {x}^{3}} + 12 x}{2 {x}^{2} + 3} ^ 2\right)$

$y ' = \left(\frac{1}{2}\right) \cdot \left(\frac{\cancel{2 {x}^{2} + 3}}{2 {x}^{2} - 3}\right) \cdot \left(\frac{24 x}{\cancel{{\left(2 {x}^{2} + 3\right)}^{2}}}\right)$

$y ' = \left(\frac{1}{\cancel{2}}\right) \cdot \left(\frac{1}{2 {x}^{2} - 3}\right) \cdot \left(\frac{\cancel{24} x}{2 {x}^{2} + 3}\right)$

$y ' = \left(1\right) \cdot \left(\frac{1}{2 {x}^{2} - 3}\right) \cdot \left(\frac{12 x}{2 {x}^{2} + 3}\right)$

$y ' = \frac{12 x}{\left(2 {x}^{2} - 3\right) \left(2 {x}^{2} + 3\right)}$

$y ' = \frac{12 x}{4 {x}^{4} - 9}$