Answer
$f'(x)<0$ when $x>e$ and $n>0$
Work Step by Step
Given the function $f(x)=\frac{lnx^{n}}{x}$, using the chain rule, the derivative function is found to be:
$f'(x)=n(\frac{1}{x^{2}} - \frac{ln}{x^{2}})$
Since we are looking at an inequality where the value of the derivative is less than $0$, we start by equating the function to $0$.
Solving for $f'(x)=0$:
$f'(x)=n(\frac{1}{x^{2}} - \frac{ln}{x^{2}})=0$
$1-lnx=0$
$x=e$
Using the value $x=e$ as a reference point, we realize that any value of $x>e$ will yield a denominator with a negative value. Assuming $n>0$ then $f'(x)<0$.
