Solution — 2022 Question 2 Differentiation & Applications

(a)

$\begin{matrix} y & = 3 \ln (4 - 5 x^{2}) \\ \frac{d y}{d x} & = 3 (\frac{- 10 x}{4 - 5 x^{2}}) \\ = - \frac{30 x}{4 - 5 x^{2}} ∎ \end{matrix}$

(b)

(i)

When $x = 1,$

$\begin{matrix} y & = 4 e^{2 - 3 x} \\ = 4 e^{2 - 3 (1)} \\ = 4 e^{- 1} \end{matrix}$

Differentiating $y = 4 e^{2 - 3 x}$ with respect to $x,$

$\begin{matrix} y & = 4 e^{2 - 3 x} \\ \frac{d y}{d x} & = - 12 e^{2 - 3 x} \end{matrix}$

When $x = 1,$

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

Equation of tangent:

$\begin{matrix} y - 4 e^{- 1} & = - 12 e^{- 1} (x - 1) \end{matrix}$

$\begin{matrix} y & = - 12 e^{- 1} x + \frac{16}{e} \\ = - 12 e^{- 1} x + 16 e^{- 1} ∎ \end{matrix}$

(ii)

When $\frac{d y}{d x} = - 2,$

$\begin{matrix} - 12 e^{2 - 3 x} & = - 2 \\ e^{2 - 3 x} & = \frac{1}{6} \\ 2 - 3 x & = \ln \frac{1}{6} \\ 3 x & = - \ln \frac{1}{6} + 2 \\ x & = \frac{- \ln \frac{1}{6} + 2}{3} ∎ \\ = \frac{\ln 6 + 2}{3} ∎ \end{matrix}$