\begin{align*} y & = x^2 + {2 \over x} \\ y & = x^2 + 2 x^{-1} \\ \\ {dy \over dx} & = 2x + 2(-1)(x^{-2}) \\ & = 2x - {2 \over x^2} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = \left(2x - {2 \over x^2} \right) \times 2 \\ & = 4x - {4 \over x^2} \\ \\ \text{When } & x = 2, \\ {dy \over dt} & = 4(2) - {4 \over 2^2} \\ & = 7 \text{ units per second} \end{align*}

\begin{align*} z & = y^3 + 2y \\ \\ {dz \over dy} & = 3y^2 + 2 \\ \\ {dz \over dt} & = {dz \over dy} \times \underbrace{ {dy \over dt} }_\text{Part i} \\ & = (3y^2 + 2) \times 7 \\ & = 21y^2 + 14 \phantom{00000000} [\text{Need to find value of } y \text{ to sub in}] \\ \\ \text{Substitute } & x = 2 \text{ into } y = x^2 + {2 \over x}, \\ y & = (2)^2 + {2 \over 2} \\ y & = 5 \\ \\ \therefore {dz \over dt} & = 21(5)^2 + 14 \\ & = 539 \text{ units per second} \end{align*}

\begin{align*} y & = \ln (x^2 - 9) \\ \\ {dy \over dx} & = {2x \over x^2 - 9} \phantom{000000} \left[ {d \over dx} [ \ln f(x)] = {f'(x) \over f(x)} \right] \end{align*}

\begin{align*} {dx \over dt} & = 2t \\ \\ x & = \int 2t \phantom{.} dt \\ x & = 2 \left(t^2 \over 2\right) + c \\ x & = t^2 + c \\ \\ \text{When } & t = 0 \text{ and } x = 5, \phantom{000000} [\text{Initially } x = 5] \\ 5 & = 0^2 + c \\ 5 & = c \\ \\ \therefore x & = t^2 + 5 \end{align*}

\begin{align*} {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {2x \over x^2 - 9} \times 2t \\ & = {4tx \over x^2 - 9} \phantom{0000000} [\text{Need to sub in } t \text{ and } x ] \\ \\ \text{Substitute } & t = 1 \text{ into } x = t^2 + 5, \\ x & = 1^2 + 5 \\ x & = 6 \\ \\ \text{Substitute } & t = 1 \text{ and } x = 6 \text{ into } {dy \over dt}, \\ {dy \over dt} & = {4(1)(6) \over (6)^2 - 9} \\ & = {8 \over 9} \text{ units per second} \end{align*}

\begin{align*} T & = 18 - 40e^{-0.54t} \\ \\ \text{When } & t = 0, \\ T & = 18 - 40e^{-0.54(0)} \\ T & = -22 ^\circ \text{C} \end{align*}

\begin{align*} T & = 18 - 40e^{-0.54t} \\ \\ {dT \over dt} & = 0 - 40 (-0.54) e^{-0.54t} \phantom{000000} \left[ {d \over dx} [ e^{f(x)} ] = f'(x) e^{f(x)} \right]\\ & = 21.6 e^{-0.54t} \\ \\ \text{It represents } & \text{the rate of change of temperature } (T) \text{ with respect to time } (t) \end{align*}

\begin{align*} \text{Volume of sphere after 21 s} & = 288\pi - 21(12 \pi) \\ & = 36 \pi \\ \\ \text{Volume of sphere, } V & = {4 \over 3} \pi r^3 \\ 36 \pi & = {4 \over 3} \pi r^3 \\ {36 \pi \over {4 \over 3} \pi} & = r^3 \\ 27 & = r^3 \\ \sqrt[3]{27} & = r \\ 3 & = r \\ \\ \text{Radius} & = 3 \text{ cm} \end{align*}

\begin{align*} V & = {4 \over 3} \pi r^3 \\ \\ {dV \over dr} & = {4 \over 3} \pi (3) (r^2) \\ & = 4\pi r^2 \\ \\ {dr \over dt} & = {dr \over dV} \times {dV \over dt} \\ & = {1 \over 4\pi r^2} \times -12 \pi \\ & = - {12 \pi \over 4\pi r^2} \\ & = - {3 \over r^2} \\ \\ \text{When } & r = 3, \phantom{000000} [\text{from (i)}] \\ {dr \over dt} & = -{3 \over 3^2} \\ & = -{1 \over 3} \\ \\ \text{Rate of decrease} & = {1 \over 3} \text{ cm per second} \end{align*}

\begin{align*} \text{Curved surface area, } A & = 4 \pi r^2 \\ \\ {dA \over dr} & = 4 \pi (2) (r) \\ & = 8 \pi r \\ \\ {dA \over dt} & = {dA \over dr} \times {dr \over dt} \\ & = 8 \pi r \times \underbrace{ -{1 \over 3} }_\text{from (ii)} \\ & = -{8 \over 3} \pi r \\ \\ \text{When } & r = 3, \\ {dA \over dt} & = -{8 \over 3} \pi (3) \\ & = -8 \pi \\ \\ \text{Rate of decrease} & = 8 \pi \text{ cm}^2 \text{ per second} \end{align*}