Solution — 2023 Question 2 Differentiation & Applications| Integration & Applications

$$\begin{array}{cc}y& ={\mathrm{e}}^{5-4x}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =(-4){\mathrm{e}}^{5-4x}\\ & =-4{\mathrm{e}}^{5-4x}∎\end{array}$$

$$\begin{array}{cc}y& =\frac{3x^2-2x}{4x^4}\\ & =\frac{3}{4x^2}-\frac{1}{2x^3}\\ & =\frac{3}{4}{x}^{-2}-\frac{1}{2}{x}^{-3}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =-\frac{3}{2}{x}^{-3}+\frac{3}{2}{x}^{-4}\\ & =-\frac{3}{2x^3}+\frac{3}{2x^4}∎\end{array}$$

$$\begin{array}{cc}& {\displaystyle \int {(\frac{1}{2x}-\sqrt{x})}^2\mathrm{d}x}\\ & ={\displaystyle \int (\frac{1}{4x^2}-\frac{\sqrt{x}}{x}+x)\mathrm{d}x}\\ & ={\displaystyle \int (\frac{1}{4}{x}^{-2}-x^{-\frac{1}{2}}+x)\mathrm{d}x}\\ & =-\frac{1}{4x}-2x^{\frac{1}{2}}+\frac{x^2}{2}+c\\ & =-\frac{1}{4x}-2\sqrt{x}+\frac{x^2}{2}+c∎\end{array}$$