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Derivative via limit definition

Let f(x)=1x+3f(x) = \dfrac{1}{x+3}.

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)=1x+3f(x) = \dfrac{1}{x+3}. Show the steps of combining the fractions in the numerator, simplifying, and evaluating the limit.

Hint

Write f(x+h)−f(x)f(x+h) - f(x) as a single fraction over a common denominator (x+h+3)(x+3)(x+h+3)(x+3), then combine that with the outer division by hh. Simplify the numerator before taking the limit as h→0h \to 0.

Step-by-step solution

Start with f(x+h)−f(x)h=1h(1x+h+3−1x+3)\frac{f(x+h)-f(x)}{h} = \frac{1}{h}\left(\frac{1}{x+h+3} - \frac{1}{x+3}\right). Combine the two fractions over the common denominator (x+h+3)(x+3)(x+h+3)(x+3): 1h⋅(x+3)−(x+h+3)(x+h+3)(x+3)\frac{1}{h}\cdot\frac{(x+3) - (x+h+3)}{(x+h+3)(x+3)}. The numerator simplifies to (x+3)−(x+h+3)=−h(x+3)-(x+h+3) = -h, giving 1h⋅−h(x+h+3)(x+3)\frac{1}{h}\cdot\frac{-h}{(x+h+3)(x+3)}.

Cancel the factor of hh (valid since h→0h \to 0 but h≠0h \neq 0) to get −1(x+h+3)(x+3)\frac{-1}{(x+h+3)(x+3)}. Now take the limit as h→0h \to 0: the denominator (x+h+3)(x+3)→(x+3)(x+3)=(x+3)2(x+h+3)(x+3) \to (x+3)(x+3) = (x+3)^2. Therefore f′(x)=−1(x+3)2f'(x) = \dfrac{-1}{(x+3)^2}.

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