## Calculus 8th Edition

$y'=(tanx)^{\frac{1}{x}}[\frac{sec^{2}x}{xtanx}-\frac{1}{x^{2}}ln(tanx)]$
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, $y'=(tanx)^{\frac{1}{x}}[\frac{sec^{2}x}{xtanx}-\frac{1}{x^{2}}ln(tanx)]$