The Definition of the Derivative — Question 2

Question 2 - Solution

We are given: $$f(x) = \sqrt{4x + 1}$$

Using the limit definition of the derivative: $$f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}$$

Compute: $$f(2 + h) = \sqrt{4(2 + h) + 1} = \sqrt{8 + 4h + 1} = \sqrt{4h + 9}$$ $$f(2) = \sqrt{4(2) + 1} = \sqrt{9} = 3$$

So, $$f'(2) = \lim_{h \to 0} \frac{\sqrt{4h + 9} - 3}{h}$$

To evaluate this limit, multiply numerator and denominator by the conjugate of the numerator: $$f'(2) = \lim_{h \to 0} \frac{\sqrt{4h + 9} - 3}{h} \cdot \frac{\sqrt{4h + 9} + 3}{\sqrt{4h + 9} + 3}$$

Apply difference of squares in the numerator: $$= \lim_{h \to 0} \frac{(4h + 9) - 9}{h(\sqrt{4h + 9} + 3)} = \lim_{h \to 0} \frac{4h}{h(\sqrt{4h + 9} + 3)}$$

Cancel $h$ in numerator and denominator: $$= \lim_{h \to 0} \frac{4}{\sqrt{4h + 9} + 3}$$

Now evaluate the limit: $$f'(2) = \frac{4}{\sqrt{9} + 3} = \frac{4}{3 + 3} = \boxed{\frac{2}{3}}$$