use definition to find the derivative of \( f(x)=\sqrt{x+5} \)

To find the derivative of \( f(x) = \sqrt{x+5} \) using the definition of the derivative, we start by applying the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substituting \( f(x) \): \[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+5} - \sqrt{x+5}}{h} \] To simplify this expression, we can multiply the numerator and the denominator by the conjugate \( \sqrt{x+h+5} + \sqrt{x+5} \): \[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h+5} - \sqrt{x+5})(\sqrt{x+h+5} + \sqrt{x+5})}{h(\sqrt{x+h+5} + \sqrt{x+5})} = \lim_{h \to 0} \frac{(x+h+5) - (x+5)}{h(\sqrt{x+h+5} + \sqrt{x+5})} \] This simplifies to: \[ = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h+5} + \sqrt{x+5})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h+5} + \sqrt{x+5}} \] As \( h \) approaches 0, the limit will become: \[ = \frac{1}{\sqrt{x+5} + \sqrt{x+5}} = \frac{1}{2\sqrt{x+5}} \] Thus, the derivative of \( f(x) \) is: \[ f'(x) = \frac{1}{2\sqrt{x+5}} \] --- It's fascinating to note that the square root function derives from the Greek mathematician Hero of Alexandria, who first studied quadratic equations and their roots around the first century AD. His work laid the groundwork for later advancements in algebra and calculus! If you're looking to apply this derivative in real life, consider how it can help in optimization problems. For instance, if you're designing a garden with parabolic structures, understanding how changes in dimensions affect area or material use can lead to cost-effective yet beautiful designs.