Let's go through the process **step-by-step** using the **formal definition of the derivative**. Given: \[ g(x) = \sqrt{7x} \] We are to find \( g'(x) \) using the **definition of the derivative**: \[ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \] --- ### **Step 1: Write the Difference Quotient** \[ g'(x) = \lim_{h \to 0} \frac{\sqrt{7(x+h)} - \sqrt{7x}}{h} \] --- ### **Step 2: Rationalize the Numerator** To simplify the numerator, **multiply and divide** by the conjugate: \[ \frac{\sqrt{7(x+h)} - \sqrt{7x}}{h} \cdot \frac{\sqrt{7(x+h)} + \sqrt{7x}}{\sqrt{7(x+h)} + \sqrt{7x}} \] Multiplying the numerators: \[ \text{Numerator: } (\sqrt{7(x+h)} - \sqrt{7x})(\sqrt{7(x+h)} + \sqrt{7x}) = (7(x+h)) - (7x) = 7h \] So, the expression becomes: \[ \frac{7h}{h \left( \sqrt{7(x+h)} + \sqrt{7x} \right)} \] --- ### **Step 3: Simplify** Cancel \( h \) in numerator and denominator (\( h \neq 0 \)): \[ = \frac{7}{\sqrt{7(x+h)} + \sqrt{7x}} \] --- ### **Step 4: Take the Limit as \( h \to 0 \)** As \( h \to 0 \), \( x+h \to x \): \[ g'(x) = \frac{7}{2\sqrt{7x}} \] --- ## **Conclusion** The **derivative** of \( g(x) = \sqrt{7x} \) is \[ \boxed{g'(x) = \frac{7}{2\sqrt{7x}}} \] --- ## **Bonus: Find the Derivative at \( x = 2 \)** Plug in \( x = 2 \): \[ g'(2) = \frac{7}{2\sqrt{7 \cdot 2}} = \frac{7}{2\sqrt{14}} \] Or, rationalized: \[ g'(2) = \frac{7}{2\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} = \frac{7\sqrt{14}}{28} = \frac{\sqrt{14}}{4} \] --- ### **Final Answers:** **Derivative:** \[ \boxed{g'(x) = \frac{7}{2\sqrt{7x}}} \] **At \( x = 2 \):** \[ \boxed{g'(2) = \frac{\sqrt{14}}{4}} \]