Let's find the derivative of \( g(x) = \sqrt{7x} \) **using the formal definition**: ### 1. **Formal Definition of the Derivative** \[ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \] ### 2. **Plug in the Function** \[ g'(x) = \lim_{h \to 0} \frac{\sqrt{7(x+h)} - \sqrt{7x}}{h} \] ### 3. **Rationalize the Numerator** Multiply numerator and denominator by the conjugate: \[ = \lim_{h \to 0} \frac{\sqrt{7(x+h)} - \sqrt{7x}}{h} \cdot \frac{\sqrt{7(x+h)} + \sqrt{7x}}{\sqrt{7(x+h)} + \sqrt{7x}} \] \[ = \lim_{h \to 0} \frac{[7(x+h) - 7x]}{h\left(\sqrt{7(x+h)} + \sqrt{7x}\right)} \] \[ = \lim_{h \to 0} \frac{7h}{h\left(\sqrt{7(x+h)} + \sqrt{7x}\right)} \] ### 4. **Simplify** \[ = \lim_{h \to 0} \frac{7}{\sqrt{7(x+h)} + \sqrt{7x}} \] As \( h \to 0 \), \( \sqrt{7(x+h)} \to \sqrt{7x} \): \[ = \frac{7}{2\sqrt{7x}} \] --- ## **Final Answer** \[ \boxed{g'(x) = \frac{7}{2\sqrt{7x}}} \] --- **If you want the value at \( x = 2 \):** \[ g'(2) = \frac{7}{2\sqrt{7 \cdot 2}} = \frac{7}{2\sqrt{14}} \] Or, rationalized: \[ g'(2) = \frac{7}{2\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} = \frac{7\sqrt{14}}{28} = \frac{\sqrt{14}}{4} \] --- ### **Summary** - \( g'(x) = \dfrac{7}{2\sqrt{7x}} \) - At \( x=2 \), \( g'(2) = \dfrac{\sqrt{14}}{4} \)