1. Stating the problem: Find the derivatives of the functions $$V(t) = \sqrt{14 + 3t}$$, $$f(z) = z^2 + 3$$, and $$W(t) = \frac{1}{\sqrt{t}}$$ using the definition of the derivative. 2. Definition of the derivative: For a function $$g(x)$$, the derivative is defined as $$ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} $$ --- ### For $$V(t) = \sqrt{14 + 3t}$$: 3. Compute $$V'(t)$$ using the definition: $$ V'(t) = \lim_{h \to 0} \frac{\sqrt{14 + 3(t+h)} - \sqrt{14 + 3t}}{h} $$ 4. Simplify the numerator by multiplying top and bottom by the conjugate: $$ = \lim_{h \to 0} \frac{(\sqrt{14 + 3(t+h)} - \sqrt{14 + 3t})(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}{h (\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})} $$ $$ = \lim_{h \to 0} \frac{14 + 3(t+h) - (14 + 3t)}{h (\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})} = \lim_{h \to 0} \frac{3h}{h(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})} $$ 5. Cancel $$h$$ and take the limit as $$h \to 0$$: $$ V'(t) = \frac{3}{2\sqrt{14 + 3t}} $$ --- ### For $$f(z) = z^2 + 3$$: 6. Use the definition: $$ f'(z) = \lim_{h \to 0} \frac{(z+h)^2 + 3 - (z^2 + 3)}{h} = \lim_{h \to 0} \frac{z^2 + 2zh + h^2 + 3 - z^2 - 3}{h} = \lim_{h \to 0} \frac{2zh + h^2}{h} $$ 7. Simplify and take the limit: $$ f'(z) = \lim_{h \to 0} (2z + h) = 2z $$ --- ### For $$W(t) = \frac{1}{\sqrt{t}} = t^{-1/2}$$: 8. Use the definition: $$ W'(t) = \lim_{h \to 0} \frac{(t+h)^{-1/2} - t^{-1/2}}{h} $$ 9. Multiply numerator and denominator by the conjugate $$ (\sqrt{t+h} + \sqrt{t}) $$: $$ = \lim_{h \to 0} \frac{\frac{1}{\sqrt{t+h}} - \frac{1}{\sqrt{t}}}{h} = \lim_{h \to 0} \frac{\sqrt{t} - \sqrt{t+h}}{h \sqrt{t+h} \sqrt{t}} $$ 10. Multiply numerator and denominator by $$\sqrt{t} + \sqrt{t+h}$$: $$ = \lim_{h \to 0} \frac{(\sqrt{t} - \sqrt{t+h})(\sqrt{t} + \sqrt{t+h})}{h \sqrt{t+h} \sqrt{t}(\sqrt{t} + \sqrt{t+h})} = \lim_{h \to 0} \frac{t - (t+h)}{h \sqrt{t+h} \sqrt{t} (\sqrt{t} + \sqrt{t+h})} $$ 11. Simplify numerator: $$ = \lim_{h \to 0} \frac{-h}{h \sqrt{t+h} \sqrt{t} (\sqrt{t} + \sqrt{t+h})} = \lim_{h \to 0} \frac{-1}{\sqrt{t+h} \sqrt{t} (\sqrt{t} + \sqrt{t+h})} $$ 12. Take the limit as $$h \to 0$$: $$ W'(t) = \frac{-1}{\sqrt{t} \sqrt{t} (\sqrt{t} + \sqrt{t})} = \frac{-1}{t (2\sqrt{t})} = -\frac{1}{2 t^{3/2}} $$ --- Final answers: $$ V'(t) = \frac{3}{2\sqrt{14 + 3t}}, \quad f'(z) = 2z, \quad W'(t) = -\frac{1}{2 t^{3/2}} $$