The problem asks us to evaluate the difference quotient for the given function $f(x) = \frac{1}{x+1}$. The difference quotient is the expression:

$\frac{f(x+h) - f(x)}{h}$

Step-by-step solution:

1. Substitute $f(x) = \frac{1}{x+1}$ and $f(x+h)$ into the difference quotient formula:

   $f(x+h) = \frac{1}{(x+h)+1} = \frac{1}{x+h+1}$

   So the difference quotient becomes:

   $\frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{x+h+1} - \frac{1}{x+1}}{h}$

2. Combine the fractions in the numerator:

   We need a common denominator to subtract these two fractions. The common denominator will be $(x+h+1)(x+1)$. So:

   $\frac{1}{x+h+1} - \frac{1}{x+1} = \frac{(x+1) - (x+h+1)}{(x+h+1)(x+1)}$

3. Simplify the numerator:

   The numerator simplifies as follows:

   $(x+1) - (x+h+1) = x+1 - x - h - 1 = -h$

   So we now have:

   $\frac{-h}{(x+h+1)(x+1)}$

4. Substitute back into the difference quotient:

   Now, the expression for the difference quotient becomes:

   $\frac{\frac{-h}{(x+h+1)(x+1)}}{h}$

5. Simplify the expression:

   Dividing by $h$ is the same as multiplying by $\frac{1}{h}$, so:

   $\frac{-h}{h(x+h+1)(x+1)} = \frac{-1}{(x+h+1)(x+1)}$

Final result:

$\frac{f(x+h) - f(x)}{h} = \frac{-1}{(x+h+1)(x+1)}$

This is the simplified form of the difference quotient for the function $f(x) = \frac{1}{x+1}$.

Would you like further details on any step or have any other questions?

Here are 5 related questions:

1. How does this difference quotient relate to the derivative of the function?
2. Can we simplify the difference quotient further when $h \to 0$?
3. What is the geometric interpretation of the difference quotient?
4. How do you compute the derivative of a rational function like $f(x) = \frac{1}{x+1}$?
5. How does this method apply to other functions of similar forms?

Tip: The difference quotient is a fundamental concept in calculus as it forms the basis for the definition of the derivative.