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. \( \frac{f(x+h)-f(x)}{h}= \) Ø.

To find the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x) = \frac{x}{x+1} \), we need to first calculate \( f(x+h) \): \[ f(x+h) = \frac{x+h}{(x+h)+1} = \frac{x+h}{x+h+1} \] Now, substituting \( f(x+h) \) and \( f(x) \) into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{\frac{x+h}{x+h+1} - \frac{x}{x+1}}{h} \] To combine the fractions in the numerator, we find a common denominator, which is \((x+h+1)(x+1)\): \[ \frac{(x+h)(x+1) - x(x+h+1)}{(x+h+1)(x+1)} \] Now we expand both parts in the numerator: \[ (x+h)(x+1) = x^2 + x + hx + h \] \[ x(x+h+1) = x^2 + xh + x \] Subtracting these results gives: \[ x^2 + x + hx + h - (x^2 + xh + x) = h \] So the numerator simplifies to \( h \). Now we can substitute this back into our difference quotient: \[ \frac{h}{h \cdot (x+h+1)(x+1)} \] Assuming \( h \neq 0 \), we can cancel \( h \) in the numerator and denominator: \[ \frac{1}{(x+h+1)(x+1)} \] Thus, the simplified difference quotient is: \[ \frac{f(x+h)-f(x)}{h} = \frac{1}{(x+h+1)(x+1)} \]