The Definition of the Derivative — Question 10

Question 10

Let $$f(x) = \sqrt{x + 3}$$ Use the definition of the derivative to find $f'(x)$. Do not apply any derivative shortcuts — use first principles only.

Question 10 - Solution

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

Apply the definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h}$$

To evaluate this limit, multiply numerator and denominator by the conjugate of the numerator: $$= \lim_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h} \cdot \frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}}$$

Use the identity $(a - b)(a + b) = a^2 - b^2$: $$= \lim_{h \to 0} \frac{(x + h + 3) - (x + 3)}{h(\sqrt{x + h + 3} + \sqrt{x + 3})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h + 3} + \sqrt{x + 3})}$$

Cancel $h$ in numerator and denominator: $$= \lim_{h \to 0} \frac{1}{\sqrt{x + h + 3} + \sqrt{x + 3}}$$

Now take the limit as $h \to 0$: $$f'(x) = \frac{1}{2\sqrt{x + 3}}$$

Final Answer: $$f'(x) = \boxed{\frac{1}{2\sqrt{x + 3}}}$$