Solution — 2019 Question 4 Differentiation & Applications| Integration & Applications

$$\begin{array}{cc}y& =\frac{1}{{(2-3x)}^4}\\ & ={(2-3x)}^{-4}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =-4{(2-3x)}^{-5}(-3)\\ & =12{(2-3x)}^{-5}\\ & =\frac{12}{{(2-3x)}^5}∎\end{array}$$

$$\begin{array}{cc}y& ={(2\sqrt{x}-\frac{3}{\sqrt{x}})}^2\\ & =4x-12+\frac{9}{x}\\ & =4x-12+9x^{-1}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =4-9x^{-2}\\ & =4-\frac{9}{x^2}∎\end{array}$$

$$\begin{array}{cc}& {\displaystyle {\int }_0^1(x^2+2-{\mathrm{e}}^{-2x})\mathrm{d}x}\\ & ={[\frac{x^3}{3}+2x+\frac{{\mathrm{e}}^{-2x}}{2}]}_0^1\\ & =\frac{1^3}{3}+2\left(1\right)+\frac{{\mathrm{e}}^{-2\left(1\right)}}{2}-(\frac{0^3}{3}+2\left(0\right)+\frac{{\mathrm{e}}^{-2\left(0\right)}}{2})\\ & =\frac{11}{6}+\frac{{\mathrm{e}}^{-2}}{2}∎\end{array}$$