Find the difference quotient \( \frac{f(x+h)-f(x)}{h} \), where \( h \neq 0 \), for the function below. \[ f(x)=\frac{7}{x+1} \] Simplify your answer as much as possible.

To find the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x)=\frac{7}{x+1} \), we start by calculating \( f(x+h) \). 1. **Calculate \( f(x+h) \)**: \[ f(x+h) = \frac{7}{(x+h)+1} = \frac{7}{x+h+1} \] 2. **Now, calculate \( f(x+h) - f(x) \)**: \[ f(x+h) - f(x) = \frac{7}{x+h+1} - \frac{7}{x+1} \] To subtract these fractions, we need a common denominator, which is \((x+h+1)(x+1)\): \[ = \frac{7(x+1) - 7(x+h+1)}{(x+h+1)(x+1)} \] Now simplify the numerator: \[ 7(x+1) - 7(x+h+1) = 7x + 7 - (7x + 7h + 7) = 7 - 7h = 7(1-h) \] Thus, \[ f(x+h) - f(x) = \frac{7(1-h)}{(x+h+1)(x+1)} \] 3. **Now plug that back into the difference quotient**: \[ \frac{f(x+h) - f(x)}{h} = \frac{\frac{7(1-h)}{(x+h+1)(x+1)}}{h} \] This simplifies to: \[ = \frac{7(1-h)}{h(x+h+1)(x+1)} \] 4. **Finally, we can split the expression as \( h \neq 0 \)**: \[ = \frac{7(1-h)}{(x+h+1)(x+1)} \cdot \frac{1}{h} = \frac{7(1-h)}{h(x+h+1)(x+1)} \] Therefore, the simplified difference quotient is: \[ \frac{7(1-h)}{h(x+h+1)(x+1)} \]