Answer
$y'=(tanx)^{\frac{1}{x}}[\frac{sec^{2}x}{xtanx}-\frac{1}{x^{2}}ln(tanx)]$
Work Step by Step
Given: $y =(tanx)^{\frac{1}{x}}$
Taking logarithmic on both sides of the function
$y =(tanx)^{\frac{1}{x}}$
$lny=\frac{1}{x} ln(tanx)$
Take implicit differentiation with respect to $x$.
Apply product rule of differentiation.
$\frac{d}{dx}(lny)=\frac{d}{dx}[\frac{1}{x} ln(tanx)]$
$\frac{1}{y}\frac{d}{dx}(y)=\frac{1}{x}\frac{d}{dx}[ln(tanx)]+ln(tanx)\frac{d}{dx}(\frac{1}{x})$
$\frac{d}{dx}(y)=y[\frac{1}{x}\frac{d}{dx}[ln(tanx)]+ln(tanx)
(-\frac{1}{x^{2}})]$
To make the tedious calculations easier, we will solve term$\frac{d}{dx}[ln(tanx)]$ separately such as:
$\frac{d}{dx}[ln(tanx)]=\frac{1}{tanx}\frac{d}{dx}(tanx)=\frac{1}{tanx}\frac{d}{dx}(sec^{2}x)$
Thus, $y'=(tanx)^{\frac{1}{x}}[\frac{1}{x}\frac{1}{tanx}\frac{d}{dx}(sec^{2}x)-\frac{1}{x^{2}}ln(tanx)]$
Hence, $y'=(tanx)^{\frac{1}{x}}[\frac{sec^{2}x}{xtanx}-\frac{1}{x^{2}}ln(tanx)]$
