(a)

\begin{align} y & = 5x^3 + \sqrt{x} - {1 \over 2x} \\ y & = 5x^3 + x^{1 \over 2} - {1 \over 2} x^{-1} \\ \\ {dy \over dx} & = 5(3)x^2 + {1 \over 2}x^{-{1 \over 2}} - {1 \over 2} (-1) x^{-2} \\ & = 15x^2 + {1 \over 2} \left(1 \over \sqrt{x}\right) + {1 \over 2} \left(1 \over x^2\right) \\ & = 15x^2 + {1 \over 2 \sqrt{x}} + {1 \over 2x^2} \\ \\ \\ {d^2 y \over dx^2} & = 15(2)x + {1 \over 2} \left(-{1 \over 2}\right) x^{-{3 \over 2}} + {1 \over 2} (-2) x^{-3} \\ & = 30x - {1 \over 4} \left(1 \over \sqrt{x^3}\right) - x^{-3} \\ & = 30x - {1 \over 4 \sqrt{x^3}} - {1 \over x^3} \end{align}

(b)

\begin{align} u & = 6x^2 &&& v & = 8x - 3 \\ {du \over dx} & = 12x &&& {dv \over dx} & = 8 \end{align} \begin{align} {dy \over dx} & = {(8x - 3)(12x) - (6x^2)(8) \over (8x - 3)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {96x^2 - 36x - 48x^2 \over (8x - 3)^2 } \\ & = { 48x^2 - 36x \over (8x - 3)^2 } \end{align}
\begin{align} u & = 48x^2 - 36x &&& v & = (8x - 3)^2 \\ {du \over dx} & = 96x - 36 &&& {dv \over dx} & = \underbrace{ 2(8x - 3).(8)}_\text{Chain rule} \\ & = 12(8x - 3) &&& & = 16(8x - 3) \end{align} \begin{align} {d^2 y \over dx^2} & = { (8x - 3)^2 [12(8x - 3)] - (48x^2 - 36x)[16(8x - 3)] \over [(8x - 3)^2]^2} \\ & = { 12(8x - 3)^3 - 12x(4x - 3)[16(8x - 3)] \over (8x - 3)^4 } \\ & = { 12(8x - 3)^3 - 192x(4x - 3)(8x - 3) \over (8x - 3)^4 } \\ & = { 12(8x - 3) [ (8x - 3)^2 - 16x(4x - 3) ] \over (8x - 3)^4 } \\ & = { 12 [ (8x)^2 - 2(8x)(3) + 3^2 - 64x^2 + 48x ] \over (8x - 3)^3 } \\ & = { 12 ( 64x^2 - 48x + 9 - 64x^2 + 48x ) \over (8x - 3)^3 } \\ & = { 12 (9) \over (8x - 3)^3 } \\ & = { 108 \over (8x - 3)^3 } \end{align}

\begin{align} f'(x) & = 8 \left({1 \over 4}x^2 - 3\right)^7 . \left({1 \over 2}x\right) \phantom{000000} [\text{Chain rule}] \\ & = 4x \left({1 \over 4}x^2 - 3\right)^7 \end{align}
\begin{align} u & = 4x &&& v & = \left({1 \over 4}x^2 - 3\right)^7 \\ {du \over dx} & = 4 &&& {dv \over dx} & = 7 \left({1 \over 4}x^2 - 3\right)^6 . \left({1 \over 2}x\right) \\ & &&& & = {7 \over 2} x \left({1 \over 4}x^2 - 3\right)^6 \end{align} \begin{align} f''(x) & = (4x) \left[ {7 \over 2} x \left({1 \over 4}x^2 - 3\right)^6 \right] + \left({1 \over 4}x^2 - 3\right)^7 (4) \phantom{000000} [ \text{Product rule}] \\ & = 14x^2 \left({1 \over 4}x^2 - 3\right)^6 + 4 \left({1 \over 4}x^2 - 3\right)^7 \\ & = \left({1 \over 4}x^2 - 3\right)^6 \left[ 14x^2 + 4 \left({1 \over 4}x^2 - 3\right) \right] \\ & = \left({1 \over 4}x^2 - 3\right)^6 (14x^2 + x^2 - 12) \\ & = \left({1 \over 4}x^2 - 3\right)^6 (15x^2 - 12) \\ & = 3 \left({1 \over 4}x^2 - 3\right)^6 (5x^2 - 4) \end{align}

(i)

\begin{align} x & = \sqrt{t^3} + 1 \\ x & = t^{3 \over 2} + 1 \\ \\ {dx \over dt} & = {3 \over 2} t^{1 \over 2} \\ & = {3 \over 2} \sqrt{t} \\ \\ {d^2 x \over dt^2} & = {3 \over 2} \left(1 \over 2\right) t^{-{1 \over 2}} \\ & = {3 \over 4} \left(1 \over \sqrt{t}\right) \\ & = {3 \over 4 \sqrt{t} } \end{align}

(ii)

\begin{align} \left({dx \over dt}\right)^2 & = \left({3 \over 2} \sqrt{t}\right)^2 \\ & = \left(3 \over 2\right)^2 (\sqrt{t})^2 \\ & = {9 \over 4} t \\ \\ \therefore {d^2 x \over dt^2} & \ne \left({dx \over dt}\right)^2 \end{align}

(iii)

\begin{align} {d^2 x \over dt^2} & = {3 \over 4 \sqrt{t} } \\ \\ x & = \sqrt{t^3} + 1 \\ x - 1 & = \sqrt{t^3} \\ x - 1 & = (\sqrt{t})^3 \\ \sqrt[3]{x - 1} & = \sqrt{t} \\ \\ \therefore {d^2 x \over dt^2} & = {3 \over 4 \sqrt[3]{x - 1}} \end{align}

\begin{align} y & = ax^2 + bx + c \\ \\ {dy \over dx} & = 2ax + b \\ \\ {d^2 y \over dx^2} & = 2a \\ \\ x^2 \left({d^2 y \over dx^2} - a\right) - x {dy \over dx} + y & = x^2 \left(2a - a \right) - x (2ax + b) + ax^2 + bx + c \\ & = x^2 (a) - 2a x^2 - bx + ax^2 + bx + c \\ & = ax^2 - 2a x^2 - bx + ax^2 + bx + c \\ & = c \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} \theta & = -0.002t^3 - 0.01t^2 + 1.2t + 2 \\ \\ \text{Let } & t = 0, \\ \theta & = -0.002(0)^3 - 0.01(0)^2 + 1.2(0) + 2 \\ & = 2 ^\circ \text{C} \end{align}

(ii)

\begin{align} \theta & = -0.002t^3 - 0.01t^2 + 1.2t + 2 \\ \\ {d \theta \over dt} & = -0.002(3)t^2 - 0.01 (2)t + 1.2 \\ & = -0.006t^2 - 0.02t + 1.2 \\ \\ {d^2 \theta \over dt^2} & = -0.006(2)t - 0.02 \\ & = -0.012t - 0.02 \end{align}

(iii)

\begin{align} {d \theta \over dt} & = -0.006t^2 - 0.02t + 1.2 \\ \\ \text{Let } & {d \theta \over dt} = 0, \\ 0 & = -0.006t^2 - 0.02t + 1.2 \\ \\ t & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-0.02) \pm \sqrt{ (-0.02)^2 - 4(-0.006)(1.2)} \over 2(-0.006)} \\ & = {0.02 \pm \sqrt{0.0292} \over -0.012} \\ & = -15.906 \text{ (NA) } \text{ or } 12.573 \\ \\ \therefore \text{Average } & \text{daily temperature stops increasing on 13th January} \end{align}

\begin{align} f''(x) & = 3x^2 - x \\ \\ f'(x) & = x^3 - {1 \over 2}x^2 + c, \text{ where } c \text{ is a constant} \\ \\ f(x) & = {1 \over 4}x^4 - {1 \over 2} \left(1 \over 3\right) x^3 + cx + d, \text{ where } d \text{ is a constant} \\ & = {1 \over 4}x^4 - {1 \over 6}x^3 + cx + d \end{align}
\begin{align} \text{Let } & c = 0, d = 5, &&& \text{Let } & c = -7, d = 0, \\ f(x) & = {1 \over 4}x^4 - {1 \over 6}x^3 + 5 &&& f(x) & = {1 \over 4}x^4 - {1 \over 6}x^3 - 7x \end{align}