# Question #6aa72

Apr 16, 2017

$\left[\frac{\ln \left(\ln x\right)}{1 + {x}^{2}} + \frac{{\tan}^{- 1} x}{x \ln x}\right] {\left(\ln x\right)}^{{\tan}^{- 1} x}$

#### Explanation:

By taking the natural log of both sides,

$\ln \left(y\right) = \ln {\left(\ln x\right)}^{{\tan}^{- 1} x}$

By the log property: $\ln {x}^{r} = r \ln x$,

$R i g h t a r r o w \ln \left(y\right) = {\tan}^{- 1} x \cdot \ln \left(\ln x\right)$

By differentiating using Product Rule,

$R i g h t a r r o w \frac{y '}{y} = \left({\tan}^{- 1} x\right) ' \cdot \ln \left(\ln x\right) + {\tan}^{- 1} x \cdot \left(\ln \left(\ln x\right)\right) '$

By $\left({\tan}^{- 1} x\right) ' = \frac{1}{1 + {x}^{2}}$ & $\left[\ln \left(g \left(x\right)\right)\right] ' = \frac{g ' \left(x\right)}{g \left(x\right)}$,

$R i g h t a r r o w \frac{y '}{y} = \frac{1}{1 + {x}^{2}} \cdot \ln \left(\ln x\right) + {\tan}^{- 1} x \cdot \frac{\frac{1}{x}}{\ln x}$

By multiplying both sides by $y$,

$y ' = \left[\frac{\ln \left(\ln x\right)}{1 + {x}^{2}} + \frac{{\tan}^{- 1} x}{x \ln x}\right] {\left(\ln x\right)}^{{\tan}^{- 1} x}$

I hope that this was clear.