Question #6aa72

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Answer 1 · 2017-04-16T13:10:58

By taking the natural log of both sides,

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

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

$\Rightarrow ln (y) = {tan}^{- 1} x \cdot ln (ln x)$ $Rightarrow ln(y)=tan^(-1)x cdot ln(ln x)$

By differentiating using Product Rule,

$Rightarrow (y')/y=(tan^(-1)x)'cdot ln(lnx)+tan^(-1)x cdot(ln(ln x))'$

By $(tan^(-1)x)'=1/(1+x^2)$ & $[ln(g(x))]'=(g'(x))/(g(x))$ ,

$Rightarrow (y')/y=1/(1+x^2) cdot ln(ln x)+tan^(-1)x cdot (1/x)/(ln x)$

By multiplying both sides by $y$ ,

I hope that this was clear.