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

(a)

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

(b)

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

(c)

\begin{matrix} \int_{0}^{1} (x^{2} + 2 - e^{- 2 x}) d x \\ = {[\frac{x^{3}}{3} + 2 x + \frac{e^{- 2 x}}{2}]}_{0}^{1} \\ = \frac{1^{3}}{3} + 2 (1) + \frac{e^{- 2 (1)}}{2} - (\frac{0^{3}}{3} + 2 (0) + \frac{e^{- 2 (0)}}{2}) \\ = \frac{11}{6} + \frac{e^{- 2}}{2} ∎ \end{matrix}