**Problem:** Find the derivative of the following functions. **2.3** $$f(x) = \left(\sqrt[3]{x} - \frac{2}{5\sqrt{x^6}}\right)^2$$ **2.4** $$f(x) = \left(3x^3 + 7\right)^2 (x + 5)$$ --- **Step 1: Simplify and identify parts (2.3).** Rewrite the terms inside the parentheses using exponents: $$\sqrt[3]{x} = x^{1/3}$$ $$\frac{2}{5\sqrt{x^6}} = \frac{2}{5x^{6/2}} = \frac{2}{5x^3} = \frac{2}{5} x^{-3}$$ So, $$f(x) = \left(x^{1/3} - \frac{2}{5}x^{-3}\right)^2$$ This is a composition of functions: a square of another function, so use chain rule. --- **Step 2: Differentiate (2.3) using chain rule.** Let $$g(x) = x^{1/3} - \frac{2}{5}x^{-3}$$ Then $$f(x) = (g(x))^2$$ By chain rule, $$f'(x) = 2g(x) \cdot g'(x)$$ Calculate $$g'(x) = \frac{d}{dx} x^{1/3} - \frac{2}{5} \frac{d}{dx} x^{-3} = \frac{1}{3} x^{-2/3} - \frac{2}{5}(-3)x^{-4} = \frac{1}{3} x^{-2/3} + \frac{6}{5} x^{-4}$$ Therefore, $$f'(x) = 2 \left(x^{1/3} - \frac{2}{5} x^{-3}\right) \left(\frac{1}{3} x^{-2/3} + \frac{6}{5} x^{-4}\right)$$ --- **Step 3: Differentiate (2.4) using product and chain rule.** We have $$f(x) = \left(3x^3 + 7\right)^2 (x + 5)$$ Let $$u = \left(3x^3 + 7\right)^2$$ $$v = x + 5$$ By product rule, $$f'(x) = u'v + uv'$$ Find derivatives: Use chain rule for $$u'$$: $$u' = 2(3x^3 + 7) \cdot (9x^2) = 18x^2 (3x^3 + 7)$$ Derivative of $$v$$: $$v' = 1$$ Therefore, $$f'(x) = 18x^2 (3x^3 + 7)(x + 5) + (3x^3 + 7)^2 \cdot 1$$ --- **Final answers:** $$\boxed{f'(x) = 2 \left(x^{1/3} - \frac{2}{5} x^{-3}\right) \left(\frac{1}{3} x^{-2/3} + \frac{6}{5} x^{-4}\right)}$$ $$\boxed{f'(x) = 18x^2 (3x^3 + 7)(x + 5) + (3x^3 + 7)^2}$$