Solution — 2025 Question 3 Differentiation & Applications| Integration & Applications

$$\begin{array}{cc}& {(3\sqrt{x}+\frac{1}{\sqrt{3x}})}^2\\ & =9x+2\left(3\sqrt{x}\right)\left(\frac{1}{\sqrt{3x}}\right)+\frac{1}{3x}\\ & =9x+\frac{6\sqrt{x}}{\sqrt{3x}}+\frac{1}{3x}\\ & =9x+2+\frac{1}{3x}\\ & =9x+2+\frac{1}{3}{x}^{-1}\end{array}$$

$$\begin{array}{cc}& \frac{\mathrm{d}}{\mathrm{d}x}(9x+2+\frac{1}{3}{x}^{-1})\\ & =9-\frac{1}{3}{x}^{-2}\\ & =9-\frac{1}{3x^2}∎\end{array}$$

$$\begin{array}{cc}& {\displaystyle \int (x-\frac{1}{3x}+\frac{2}{3x^3})\mathrm{d}x}\\ & ={\displaystyle \int (x-\frac{1}{3}{x}^{-1}+\frac{2}{3}{x}^{-3})\mathrm{d}x}\\ & =\frac{x^2}{2}-\frac{1}{3}\ln x-\frac{1}{3x^2}+C∎\end{array}$$