Answer
-0.00125
Work Step by Step
$ f(x)=\frac{1}{x+1}$, $ a=3$, $\Delta x=0.02$ and
$ f'(x)=\frac{-1}{(x+1)^{2}}$ (Using the reciprocal rule for differentiation)
Recall: $ f(a+\Delta x)-f(a)\approx f'(a)\Delta x $
$\implies f(3+0.02)-f(3)\approx f'(3)\Delta x=\frac{-1}{(3+1)^{2}}\times0.02=-0.00125$
