Answer
$f'(x)=\frac{-3\sin(x)+x\cos(x)}{x^4}$ .
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=\sin(x); u'(x)=\cos(x)$
$v(x)=x^3; v'(x)=3x^2$
$f'(x)=\frac{(\cos(x))(x^3)-(3x^2)(sin(x))}{(x^3)^2}$
$=\frac{(x^2)(x\cos(x)-3\sin(x))}{x^6}$
$=\frac{-3\sin(x)+x\cos(x)}{x^4}$