The Definition of the Derivative — Question 3

Question 3 - Solution

We are given: $$f(x) = \frac{1}{x + 3}$$

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

Compute: $$f(1 + h) = \frac{1}{1 + h + 3} = \frac{1}{h + 4}, \quad f(1) = \frac{1}{4}$$

So, $$f'(1) = \lim_{h \to 0} \frac{\frac{1}{h + 4} - \frac{1}{4}}{h}$$

Combine the terms in the numerator: $$= \lim_{h \to 0} \frac{\frac{4 - (h + 4)}{4(h + 4)}}{h} = \lim_{h \to 0} \frac{\frac{-h}{4(h + 4)}}{h}$$

Simplify: $$= \lim_{h \to 0} \frac{-1}{4(h + 4)}$$

Now take the limit: $$f'(1) = \frac{-1}{4(4)} = \boxed{-\frac{1}{16}}$$