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Derivative via limit definition for a rational function

Consider the function f(x)=2x−1x+1f(x) = \dfrac{2x-1}{x+1}.

Use the limit definition of the derivative, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, to find f′(x)f'(x) for f(x)=2x−1x+1f(x) = \dfrac{2x-1}{x+1}. Show the combination of fractions and simplification that leads to a limit you can evaluate directly, and state the resulting formula for f′(x)f'(x).

Hint

Write f(x+h)−f(x)f(x+h) - f(x) as a single fraction with common denominator (x+h+1)(x+1)(x+h+1)(x+1), simplify the numerator, then divide by hh and cancel before letting h→0h \to 0.

Step-by-step solution

Start with f(x+h)−f(x)h=1h(2(x+h)−1x+h+1−2x−1x+1).\frac{f(x+h)-f(x)}{h} = \frac{1}{h}\left(\frac{2(x+h)-1}{x+h+1} - \frac{2x-1}{x+1}\right). Combine the two fractions over the common denominator (x+h+1)(x+1)(x+h+1)(x+1): the numerator becomes (2x+2h−1)(x+1)−(2x−1)(x+h+1)(2x+2h-1)(x+1) - (2x-1)(x+h+1).

Expanding, (2x+2h−1)(x+1)=2x2+2x+2hx+2h−x−1(2x+2h-1)(x+1) = 2x^2+2x+2hx+2h-x-1 and (2x−1)(x+h+1)=2x2+2hx+2x−x−h−1(2x-1)(x+h+1) = 2x^2+2hx+2x-x-h-1. Subtracting these gives a numerator of 3h3h, so the difference quotient becomes 3hh(x+h+1)(x+1)=3(x+h+1)(x+1)\dfrac{3h}{h(x+h+1)(x+1)} = \dfrac{3}{(x+h+1)(x+1)}.

Now let h→0h \to 0: the expression 3(x+h+1)(x+1)\dfrac{3}{(x+h+1)(x+1)} approaches 3(x+1)2\dfrac{3}{(x+1)^2}, so f′(x)=3(x+1)2.f'(x) = \frac{3}{(x+1)^2}.

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