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

To find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the function \(f(x)=\frac{x}{x+1}\), we first need to calculate \(f(x+h)\). 1. **Calculate \(f(x+h)\)**: \[ f(x+h) = \frac{x+h}{(x+h)+1} = \frac{x+h}{x+h+1} \] 2. **Calculate \(f(x+h) - f(x)\)**: \[ f(x+h) - f(x) = \frac{x+h}{x+h+1} - \frac{x}{x+1} \] To subtract these fractions, we need a common denominator, which is \((x+h+1)(x+1)\): \[ f(x+h) - f(x) = \frac{(x+h)(x+1) - x(x+h+1)}{(x+h+1)(x+1)} \] 3. **Expand the numerators**: - Expand \((x+h)(x+1)\): \[ (x+h)(x+1) = x^2 + x + hx + h = x^2 + (1+h)x + h \] - Expand \(x(x+h+1)\): \[ x(x+h+1) = x^2 + hx + x \] 4. **Combine the results**: \[ f(x+h) - f(x) = \frac{x^2 + (1+h)x + h - (x^2 + (h+1)x)}{(x+h+1)(x+1)} \] Simplifying the numerator: \[ = \frac{x^2 + (1+h)x + h - x^2 - hx - x}{(x+h+1)(x+1)} \] \[ = \frac{(1+h-1)x + h}{(x+h+1)(x+1)} = \frac{hx + h}{(x+h+1)(x+1)} \] 5. **Factor out \(h\)**: \[ = \frac{h(x + 1)}{(x+h+1)(x+1)} \] 6. **Now, divide by \(h\)** in the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{h(x+1)}{h(x+h+1)(x+1)} \] Cancel \(h\) (since \(h \neq 0\)): \[ = \frac{x+1}{(x+h+1)(x+1)} = \frac{1}{x+h+1} \] Thus, the simplified difference quotient is: \[ \frac{1}{x+h+1} \]