Solution — 2016 Question 1 Differentiation & Applications

(a)

\begin{matrix} y & = 2 \ln (3 x^{2} + 4) \\ \frac{d y}{d x} & = 2 (\frac{6 x}{3 x^{2} + 4}) \\ = \frac{12 x}{3 x^{2} + 4} ∎ \end{matrix}

(b)

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