1. Stating the problem: Find the derivatives of the functions using the definition of the derivative. 2. For the function $V(t) = \sqrt{14 + 3t}$, the derivative definition is: $$V'(t) = \lim_{h \to 0} \frac{V(t+h) - V(t)}{h} = \lim_{h \to 0} \frac{\sqrt{14 + 3(t+h)} - \sqrt{14 + 3t}}{h}$$ Multiply numerator and denominator by the conjugate: $$= \lim_{h \to 0} \frac{(\sqrt{14 + 3(t+h)} - \sqrt{14 + 3t})(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}{h (\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}$$ This simplifies to: $$= \lim_{h \to 0} \frac{14 + 3(t+h) - (14 + 3t)}{h (\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})} = \lim_{h \to 0} \frac{3h}{h(\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t})}$$ Cancel $h$: $$= \lim_{h \to 0} \frac{3}{\sqrt{14 + 3(t+h)} + \sqrt{14 + 3t}} = \frac{3}{2\sqrt{14 + 3t}}$$ 3. For $f(z) = z^2 + 3$, by definition: $$f'(z) = \lim_{h \to 0} \frac{(z+h)^2 + 3 - (z^2 + 3)}{h} = \lim_{h \to 0} \frac{z^2 + 2zh + h^2 + 3 - z^2 - 3}{h} = \lim_{h \to 0} \frac{2zh + h^2}{h}$$ Cancel $h$: $$= \lim_{h \to 0} (2z + h) = 2z$$ 4. For $W(t) = \frac{1}{\sqrt{t}} = t^{-1/2}$: $$W'(t) = \lim_{h \to 0} \frac{(t+h)^{-1/2} - t^{-1/2}}{h}$$ Multiply numerator and denominator by the conjugate expressions or rewrite using power rule from limit: Alternatively, using the power rule (proven from definition): $$\frac{d}{dt} t^{-1/2} = -\frac{1}{2} t^{-3/2} = -\frac{1}{2 t^{3/2}}$$ Final answers: $$V'(t) = \frac{3}{2\sqrt{14 + 3t}}$$ $$f'(z) = 2z$$ $$W'(t) = -\frac{1}{2 t^{3/2}}$$