Answer
$U'(y)=\dfrac{10y(y^{4}+1)^{4}(y^{4}+2y^{2}-1)}{(y^{2}+1)^{6}}$
Work Step by Step
$U(y)=(\dfrac{y^{4}+1}{y^{2}+1})^{5}$
Differentiate using the chain rule:
$U'(y)=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}(\dfrac{y^{4}+1}{y^{2}+1})'=...$
Now, use the quotient rule to find $(\dfrac{y^{4}+1}{y^{2}+1})'$:
$...=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}[\dfrac{(y^{2}+1)(y^{4}+1)'-(y^{4}+1)(y^{2}+1)'}{(y^{2}+1)^{2}}]=...$
$...=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}[\dfrac{(y^{2}+1)(4y^{3})-(y^{4}+1)(2y)}{(y^{2}+1)^{2}}]=...$
Simplify:
$...=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}[\dfrac{4y^{5}+4y^{3}-2y^{5}-2y}{(y^{2}+1)^{2}}]=5(\dfrac{y^{4}+1}{y^{2}+1})^{4}\dfrac{2y^{5}+4y^{3}-2y}{(y^{2}+1)^{2}}=\dfrac{10y(y^{4}+1)^{4}(y^{4}+2y^{2}-1)}{(y^{2}+1)^{6}}$
