Answer
$$ y'=-180x^3(1+\cot^5(x^4+1))^8\cot^4(x^4+1)\csc^2(x^4+1).$$
Work Step by Step
Since $ y=(1+\cot^5(x^4+1))^9$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\cot x)'=-\csc^2 x $, the derivative $ y'$ is given by
$$ y'=9(1+\cot^5(x^4+1))^8(1+\cot^5(x^4+1))'
=9(1+\cot^5(x^4+1))^8(5\cot^4(x^4+1))(\cot^4(x^4+1))'
\\
\\=9(1+\cot^5(x^4+1))^8(5\cot^4(x^4+1)(-4x^3\csc^2(x^4+1)))\\ =-180x^3(1+\cot^5(x^4+1))^8\cot^4(x^4+1)\csc^2(x^4+1).$$
