Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 122: 25

$-\frac{3}{4}$

Using quotient rule, we have $\frac{d}{dx}(\frac{1}{x^{3}+1})=\frac{(x^{3}+1)\times\frac{d}{dx}(1)-1\times\frac{d}{dx}(x^{3}+1)}{(x^{3}+1)^{2}}$ $=\frac{-3x^{2}}{(x^{3}+1)^{2}}$ $\frac{dz}{dx}|_{x=1}=\frac{-3(1)^{2}}{((1)^{3}+1)^{2}}=-\frac{3}{4}$