Solution — 2024 Question 2 Differentiation & Applications| Integration & Applications

(a)

(i)

\begin{matrix} y & = 4 \ln {(2 x + 7)}^{2} \\ \frac{d y}{d x} & = 4 (\frac{2 (2 x + 7) (2)}{{(2 x + 7)}^{2}}) \\ = \frac{16}{2 x + 7} ∎ \end{matrix}

(ii)

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

(b)

\begin{matrix} \int \frac{3}{\sqrt{5 x + 1}} d x \\ = \int 3 {(5 x + 1)}^{- \frac{1}{2}} d x \\ = 3 (\frac{{(5 x + 1)}^{\frac{1}{2}}}{\frac{1}{2} (5)}) + c \\ = \frac{6 {(5 x + 1)}^{\frac{1}{2}}}{5} + c ∎ \end{matrix}