## Calculus 10th Edition

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