The Definition of the Derivative — Question 9

Question 9 - Solution

We are given: $$f(x) = \sqrt{x^2 + 4x + 5}$$

Apply the definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{(x+h)^2 + 4(x+h) + 5} - \sqrt{x^2 + 4x + 5}}{h}$$

Expand the expression inside the first square root: $$(x+h)^2 + 4(x+h) + 5 = x^2 + 2xh + h^2 + 4x + 4h + 5$$

So we have: $$f'(x) = \lim_{h \to 0} \frac{\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} - \sqrt{x^2 + 4x + 5}}{h}$$

Multiply numerator and denominator by the conjugate: $$= \lim_{h \to 0} \frac{\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} - \sqrt{x^2 + 4x + 5}}{h} \cdot \frac{\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} + \sqrt{x^2 + 4x + 5}}{\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} + \sqrt{x^2 + 4x + 5}}$$

Use the identity $(a-b)(a+b) = a^2 - b^2$: $$= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 + 4x + 4h + 5) - (x^2 + 4x + 5)} {h\left(\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} + \sqrt{x^2 + 4x + 5}\right)}$$

Simplify the numerator: $$= \lim_{h \to 0} \frac{2xh + h^2 + 4h} {h\left(\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} + \sqrt{x^2 + 4x + 5}\right)}$$

Factor $h$ from the numerator: $$= \lim_{h \to 0} \frac{h(2x + h + 4)} {h\left(\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} + \sqrt{x^2 + 4x + 5}\right)}$$

Cancel $h$: $$= \lim_{h \to 0} \frac{2x + h + 4} {\sqrt{x^2 + 2xh + h^2 + 4x + 4h + 5} + \sqrt{x^2 + 4x + 5}}$$

Now take the limit as $h \to 0$: $$f'(x) = \frac{2x + 4} {2\sqrt{x^2 + 4x + 5}}$$

Final Answer: $$f'(x) = \boxed{\frac{2x + 4}{2\sqrt{x^2 + 4x + 5}}}$$