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