Answer
$$\frac{1}{{3 + \Delta x}} \approx \frac{1}{3} - \frac{1}{9}\Delta x$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{2 + x}};{\text{ }}\frac{1}{{3 + \Delta x}} \approx \frac{1}{3} - \frac{1}{9}\Delta x,{\text{ }}{x_0} = 1,{\text{ where }}\Delta x = x - 1 \cr
& {\text{Calculate }}f'\left( x \right) \cr
& f\left( x \right) = {\left( {2 + x} \right)^{ - 1}} \cr
& f'\left( x \right) = - {\left( {2 + x} \right)^{ - 2}} \cr
& f'\left( x \right) = - \frac{1}{{{{\left( {2 + x} \right)}^2}}} \cr
& {\text{Evaluate }}f\left( {{x_0}} \right){\text{ and }}f'\left( {{x_0}} \right){\text{ }} \cr
& f\left( 1 \right) = \frac{1}{{2 + 1}} = \frac{1}{3} \cr
& f'\left( 1 \right) = - \frac{1}{{{{\left( {2 + 1} \right)}^2}}} = - \frac{1}{9} \cr
& {\text{Apply }}f\left( {{x_0} + \Delta x} \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\Delta x{\text{ and substitute the}} \cr
& {\text{ known values, then}} \cr
& {\text{ }}\frac{1}{{3 + \Delta x}} \approx \frac{1}{3} + \left( { - \frac{1}{9}} \right)\Delta x \cr
& {\text{ }}\frac{1}{{3 + \Delta x}} \approx \frac{1}{3} - \frac{1}{9}\Delta x \cr} $$
