The Definition of the Derivative — Question 1

Question 1 - Solution

We are given: $$f(x) = \frac{1}{x + 2}$$

Using the definition of the derivative: $$f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}$$

Compute: $$f(1 + h) = \frac{1}{(1 + h) + 2} = \frac{1}{h + 3}, \quad f(1) = \frac{1}{1 + 2} = \frac{1}{3}$$

So: $$f'(1) = \lim_{h \to 0} \frac{\frac{1}{h + 3} - \frac{1}{3}}{h}$$

Find a common denominator for the numerator: $$\frac{1}{h + 3} - \frac{1}{3} = \frac{3 - (h + 3)}{3(h + 3)} = \frac{-h}{3(h + 3)}$$

Now plug back into the limit: $$f'(1) = \lim_{h \to 0} \frac{-h}{3(h + 3)} \cdot \frac{1}{h} = \lim_{h \to 0} \frac{-1}{3(h + 3)}$$

Now evaluate the limit: $$f'(1) = \frac{-1}{3(0 + 3)} = \boxed{-\frac{1}{9}}$$