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