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

(a)

\begin{matrix} y & = e^{5 - 4 x} \\ \frac{d y}{d x} & = (- 4) e^{5 - 4 x} \\ = - 4 e^{5 - 4 x} ∎ \end{matrix}

(b)

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

(c)

\begin{matrix} \int {(\frac{1}{2 x} - \sqrt{x})}^{2} d x \\ = \int (\frac{1}{4 x^{2}} - \frac{\sqrt{x}}{x} + x) d x \\ = \int (\frac{1}{4} x^{- 2} - x^{- \frac{1}{2}} + x) d x \\ = - \frac{1}{4 x} - 2 x^{\frac{1}{2}} + \frac{x^{2}}{2} + c \\ = - \frac{1}{4 x} - 2 \sqrt{x} + \frac{x^{2}}{2} + c ∎ \end{matrix}