# What is the derivative of the function f (x)=ln (ln ((x+4)/ln (x^2+4) ?

Apr 27, 2018

$f ' \left(x\right) = \left(\frac{1}{\ln \left(\frac{x + 4}{\ln \left({x}^{2} + 4\right)}\right)}\right) \left(\frac{1}{\left(x + 4\right)}\right) . \left(\frac{\left({x}^{2} + 4\right) \left(\ln \left({x}^{2} + 4\right)\right) - \left(2 {x}^{2} + 4 x\right)}{\left({x}^{2} + 4\right) \left(\ln \left({x}^{2} + 4\right)\right)}\right)$

#### Explanation:

$f ' \left(x\right) = \left(\frac{1}{\ln \left(\frac{x + 4}{\ln \left({x}^{2} + 4\right)}\right)}\right) \left(\frac{1}{\frac{x + 4}{\ln \left({x}^{2} + 4\right)}}\right) . \left(\frac{\left(1\right) \left(\ln \left({x}^{2} + 4\right)\right) - \left(x + 4\right) \frac{1}{\left({x}^{2} + 4\right)} \left(2 x\right)}{\left(\ln \left({x}^{2} + 4\right)\right)} ^ 2\right)$

$f ' \left(x\right) = \left(\frac{1}{\ln \left(\frac{x + 4}{\ln \left({x}^{2} + 4\right)}\right)}\right) \left(\ln \frac{{x}^{2} + 4}{\left(x + 4\right)}\right) . \left(\frac{\ln \left({x}^{2} + 4\right) - \frac{2 {x}^{2} + 4 x}{\left({x}^{2} + 4\right)}}{\left(\ln \left({x}^{2} + 4\right)\right)} ^ 2\right)$

$f ' \left(x\right) = \left(\frac{1}{\ln \left(\frac{x + 4}{\ln \left({x}^{2} + 4\right)}\right)}\right) \left(\frac{\cancel{\ln \left({x}^{2} + 4\right)}}{\left(x + 4\right)}\right) . \left(\frac{\left({x}^{2} + 4\right) \left(\ln \left({x}^{2} + 4\right)\right) - \left(2 {x}^{2} + 4 x\right)}{\left({x}^{2} + 4\right) {\left(\ln \left({x}^{2} + 4\right)\right)}^{\cancel{2}}}\right)$

$f ' \left(x\right) = \left(\frac{1}{\ln \left(\frac{x + 4}{\ln \left({x}^{2} + 4\right)}\right)}\right) \left(\frac{1}{\left(x + 4\right)}\right) . \left(\frac{\left({x}^{2} + 4\right) \left(\ln \left({x}^{2} + 4\right)\right) - \left(2 {x}^{2} + 4 x\right)}{\left({x}^{2} + 4\right) \left(\ln \left({x}^{2} + 4\right)\right)}\right)$