Answer
$y'=\dfrac{x\cos{x}-4\sin{x}}{x^5}.$
Work Step by Step
Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=\sin{x}; u'(x)=\cos{x}$
$v(x)=x^4; v'(x)=4x^3$
$y'=\dfrac{(\cos{x})(x^4)-(4x^3)(\sin{x})}{(x^4)^2}$
$=\dfrac{x^3(x\cos{x}-4\sin{x})}{x^8}$
$=\dfrac{x\cos{x}-4\sin{x}}{x^5}.$
