The Definition of the Derivative — Question 6

Question 6

Let $$f(x) = \sqrt{x + 1}$$ Use the definition of the derivative to find $f'(x)$. Show all steps clearly without using derivative shortcuts.

Question 6 - Solution

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

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

To simplify the numerator, multiply by the conjugate: $$\lim_{h \to 0} \frac{\sqrt{x + h + 1} - \sqrt{x + 1}}{h} \cdot \frac{\sqrt{x + h + 1} + \sqrt{x + 1}}{\sqrt{x + h + 1} + \sqrt{x + 1}}$$

This becomes: $$= \lim_{h \to 0} \frac{(x + h + 1) - (x + 1)}{h\left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)} = \lim_{h \to 0} \frac{h}{h\left(\sqrt{x + h + 1} + \sqrt{x + 1}\right)}$$

Cancel the $h$’s: $$= \lim_{h \to 0} \frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}}$$

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

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