1. **Problem Statement:** Given the function $f(x) = \frac{1}{\sqrt{3x^3}}$, we want to find: 2.1.1 $f(x+h)$ 2.1.2 $f(x+h) - f(x)$ 2.1.3 $\frac{f(x+h) - f(x)}{h}$ 2.1.4 $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ 2. **Step 1: Simplify $f(x)$** We have $$f(x) = \frac{1}{\sqrt{3x^3}} = \frac{1}{\sqrt{3} \sqrt{x^3}} = \frac{1}{\sqrt{3} x^{3/2}} = \frac{1}{\sqrt{3}} x^{-\frac{3}{2}}$$ 3. **Step 2: Find $f(x+h)$** Substitute $x+h$ in place of $x$: $$f(x+h) = \frac{1}{\sqrt{3}} (x+h)^{-\frac{3}{2}}$$ 4. **Step 3: Find $f(x+h) - f(x)$** $$f(x+h) - f(x) = \frac{1}{\sqrt{3}} (x+h)^{-\frac{3}{2}} - \frac{1}{\sqrt{3}} x^{-\frac{3}{2}} = \frac{1}{\sqrt{3}} \left[(x+h)^{-\frac{3}{2}} - x^{-\frac{3}{2}} \right]$$ 5. **Step 4: Find $\frac{f(x+h) - f(x)}{h}$** $$\frac{f(x+h) - f(x)}{h} = \frac{1}{\sqrt{3}} \cdot \frac{(x+h)^{-\frac{3}{2}} - x^{-\frac{3}{2}}}{h}$$ 6. **Step 5: Find the limit as $h \to 0$** This limit represents the derivative $f'(x)$: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{1}{\sqrt{3}} \lim_{h \to 0} \frac{(x+h)^{-\frac{3}{2}} - x^{-\frac{3}{2}}}{h}$$ Using the derivative formula for $x^n$: $$\frac{d}{dx} x^{-\frac{3}{2}} = -\frac{3}{2} x^{-\frac{5}{2}}$$ So, $$f'(x) = \frac{1}{\sqrt{3}} \left(-\frac{3}{2} x^{-\frac{5}{2}}\right) = -\frac{3}{2 \sqrt{3}} x^{-\frac{5}{2}}$$ **Final answers:** 2.1.1 $f(x+h) = \frac{1}{\sqrt{3}} (x+h)^{-\frac{3}{2}}$ 2.1.2 $f(x+h) - f(x) = \frac{1}{\sqrt{3}} \left[(x+h)^{-\frac{3}{2}} - x^{-\frac{3}{2}}\right]$ 2.1.3 $\frac{f(x+h) - f(x)}{h} = \frac{1}{\sqrt{3}} \cdot \frac{(x+h)^{-\frac{3}{2}} - x^{-\frac{3}{2}}}{h}$ 2.1.4 $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x) = -\frac{3}{2 \sqrt{3}} x^{-\frac{5}{2}}$