1. **State the problem:** Given the function $$f(x) = \frac{1}{\sqrt{3x^3}}$$, find expressions for: 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 $$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ (the derivative) 2. **Find $$f(x+h)$$:** $$f(x+h) = \frac{1}{\sqrt{3(x+h)^3}} = \frac{1}{\sqrt{3(x+h)^3}}$$ 3. **Calculate $$f(x+h) - f(x)$$:** $$f(x+h) - f(x) = \frac{1}{\sqrt{3(x+h)^3}} - \frac{1}{\sqrt{3x^3}}$$ 4. **Find the difference quotient $$\frac{f(x+h) - f(x)}{h}$$:** $$\frac{f(x+h) - f(x)}{h} = \frac{1}{h} \left(\frac{1}{\sqrt{3(x+h)^3}} - \frac{1}{\sqrt{3x^3}}\right)$$ 5. **Find the limit as $$h \to 0$$:** This is the derivative of $$f(x)$$. Rewrite $$f(x)$$ to a power form to differentiate easily: $$f(x) = (3x^3)^{-1/2} = 3^{-1/2} x^{-3/2}$$ Using the power rule: $$f'(x) = 3^{-1/2} \cdot \left(-\frac{3}{2}\right) x^{-5/2} = -\frac{3}{2 \sqrt{3}} x^{-5/2} = -\frac{3}{2 \sqrt{3} x^{5/2}}$$ **Final answers:** 1. $$f(x+h) = \frac{1}{\sqrt{3(x+h)^3}}$$ 2. $$f(x+h) - f(x) = \frac{1}{\sqrt{3(x+h)^3}} - \frac{1}{\sqrt{3x^3}}$$ 3. $$\frac{f(x+h) - f(x)}{h} = \frac{1}{h} \left(\frac{1}{\sqrt{3(x+h)^3}} - \frac{1}{\sqrt{3x^3}}\right)$$ 4. $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = -\frac{3}{2 \sqrt{3} x^{5/2}}$$