## Calculus (3rd Edition)

$$h'(z)=-\frac{3}{2}(z+(z+1)^{1/2})^{-5/2}\left(1+\frac{1}{2}(z+1)^{-1/2}\right) .$$
Recall that $(x^n)'=nx^{n-1}$ Since $h(z)=(z+(z+1)^{1/2})^{-3/2}$, the derivative $h'(z)$, by using the chain rule, is given by $$h'(z)=-\frac{3}{2}(z+(z+1)^{1/2})^{-5/2}\left(1+\frac{1}{2}(z+1)^{-1/2}\right) .$$