Answer
$y' = \frac{8x}{x^{4}-16}$
Work Step by Step
$y = \ln \sqrt {\frac{x^{2}+4}{x^{2}-4}}$
$y = \ln (\frac{x^{2}+4}{x^{2}-4})^{\frac{1}{2}}$
$y = \frac{1}{2}\ln (\frac{x^{2}+4}{x^{2}-4})$
$y = \frac{1}{2}\ln(x^{2}+4) - \frac{1}{2}\ln(x^{2}-4)$
$y' = [0\ln (x^{2}+4) + \frac{1}{2}(\frac{1}{x^{2}-4})(2x)]-[0\ln(x^{2}-4)+\frac{1}{2}(\frac{1}{x^{2}-4})(2x))$
$y' = [0 + \frac{1}{2}(\frac{2x}{x^{2}+4})]-[0+\frac{1}{2}(\frac{2x}{x^{2}-4})]$
$y' = \frac{2x}{2(x^{2}+4)}-\frac{2x}{2(x^{2}-4)}$
$y' = \frac{x}{(x^{2}+4)} - \frac{x}{(x^{2}-4)}$
$y' = \frac{x(x^{2}-4)-[x((x^{2}+4))}{(x^{2}+4)(x^{2}-4)}$
$y' = \frac{x^{3}-4x-x^{3}-4x}{(x^{2}+4)(x^{2}-4)}$
$y' = \frac{8x}{(x^{2}+4)(x^{2}-4)}$
$y' = \frac{8x}{x^{2}(x^{2}+4)-4(x^{2}+4)}$
$y' = \frac{8x}{x^{4}+4x^{2}-4x^{2}-16}$
$y' = \frac{8x}{x^{4}-16}$
