Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {x + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{x + 1}} \cr
& {\text{Using the formula }}\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{1}{{x + \Delta x + 1}} - \frac{1}{{x + 1}}}}{{\Delta x}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{{x + 1 - x - \Delta x - 1}}{{\left( {x + \Delta x + 1} \right)\left( {x + 1} \right)}}}}{{\Delta x}} \cr
& \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{ - \Delta x}}{{\Delta x\left( {x + \Delta x + 1} \right)\left( {x + 1} \right)}} \cr
& \frac{{dy}}{{dx}} = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\left( {x + \Delta x + 1} \right)\left( {x + 1} \right)}} \cr
& \cr
& \Delta x \to 0 \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left( {x + 0 + 1} \right)\left( {x + 1} \right)}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {x + 1} \right)}^2}}} \cr} $$
