(a)

\begin{align} {d \over dx} \left({3 \over 5} x^2 - 1 \right)^{10} & = 10 \left({3 \over 5}x^2 -1\right)^9 . \left( {6 \over 5}x \right) \phantom{000000} [\text{Chain rule}] \\ & = 12x \left({3 \over 5}x^2 - 1\right)^9 \end{align}
\begin{align} u & = 12x &&& v & = \left({3 \over 5}x^2 - 1 \right)^9 \\ {du \over dx} & = 12 &&& {dv \over dx} & = 9 \left({3 \over 5}x^2 - 1\right)^8 . \left({6 \over 5}x \right) \\ & &&& & = {54 \over 5}x \left({3 \over 5}x^2 - 1\right)^8 \end{align} \begin{align} {d \over dx} \left[ 12x \left({3 \over 5}x^2 - 1 \right)^9 \right] & = (12x) \left[ {54 \over 5}x \left({3 \over 5}x^2 - 1\right)^8 \right] + \left({3 \over 5}x^2 - 1 \right)^9 (12) \phantom{000000} [\text{Product rule}] \\ & = {648 \over 5} x^2 \left({3 \over 5}x^2 - 1\right)^8 + 12 \left({3 \over 5}x^2 - 1\right)^9 \\ & = {648 \over 5} x^2 \left({3 \over 5}x^2 - 1\right)^8 + {60 \over 5} \left({3 \over 5}x^2 - 1\right)^9 \\ & = {12 \over 5} \left({3 \over 5}x^2 - 1\right)^8 \left[ 54 x^2 + 5 \left({3 \over 5}x^2 - 1\right) \right] \\ & = {12 \over 5} \left({3 \over 5}x^2 - 1\right)^8 ( 54x^2 + 3x^2 - 5 ) \\ & = {12 \over 5} \left({3 \over 5}x^2 - 1\right)^8 ( 57x^2 - 5) \end{align}

(b)

\begin{align} u & = x &&& v & = \sqrt{x + 6} \\ & &&& & = (x + 6)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = \underbrace{ {1 \over 2} (x + 6)^{-{1 \over 2}}. (1) }_\text{Chain rule} \\ & &&& & = {1 \over 2} \left(1 \over \sqrt{x + 6}\right) \\ & &&& & = {1 \over 2 \sqrt{x + 6} } \end{align} \begin{align} {d \over dx} (x \sqrt{x + 6}) & = (x) \left(1 \over 2 \sqrt{x + 6}\right) + (\sqrt{x + 6})(1) \phantom{000000} [\text{Product rule}] \\ & = \left(x \over 1\right) \left(1 \over 2 \sqrt{x + 6}\right) + {\sqrt{x + 6} \over 1} \\ & = {x \over 2\sqrt{x + 6}} + {\sqrt{x + 6} (2 \sqrt{x + 6}) \over 2 \sqrt{x + 6} } \\ & = {x \over 2 \sqrt{x + 6}} + {2(x + 6) \over 2 \sqrt{x + 6}} \\ & = {x + 2(x + 6) \over 2 \sqrt{x + 6}} \\ & = {x + 2x + 12 \over 2 \sqrt{x + 6}} \\ & = {3x + 12 \over 2 \sqrt{x + 6}} \end{align}
\begin{align} u & = 3x + 12 &&& v & = 2 \sqrt{x + 6} \\ & &&& & = 2 (x + 6)^{1 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = \underbrace{ 2 \left(1 \over 2\right) (x + 6)^{-{1 \over 2}}. (1)}_\text{Chain rule} \\ & &&& & = (x + 6)^{-{1 \over 2}} \end{align} \begin{align} {d \over dx} \left( {3x + 12 \over 2 \sqrt{x + 6}} \right) & = { (2 (x + 6)^{1 \over 2})(3) - (3x + 12)(x + 6)^{-{1 \over 2}} \over [2 (x + 6)^{1 \over 2}]^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 6 (x + 6)^{1 \over 2} - (3x + 12) (x + 6)^{-{1 \over 2}} \over 2^2 [(x + 6)^{1 \over 2}]^2 } \\ & = { 6 (x + 6)^{1 \over 2} - (3x + 12) (x + 6)^{-{1 \over 2}} \over 4 (x + 6) } \times { (x + 6)^{1 \over 2} \over (x + 6)^{1 \over 2} }\\ & = { 6(x + 6) - (3x + 12)(x + 6)^0 \over 4 (x + 6)^{3 \over 2} } \\ & = { 6x + 36 - (3x + 12)(1) \over 4 (x + 6)^{3 \over 2} } \\ & = { 6x + 36 - 3x - 12 \over 4 \sqrt{(x + 6)^3} } \\ & = { 3x + 24 \over 4 \sqrt{(x + 6)^3} } \end{align}

\begin{align} u & = 9x^3 - 1 &&& v & = x^2 + 4 \\ {du \over dx} & = 27x^2 &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = { (x^2 + 4)(27x^2) - (9x^3 - 1)(2x) \over (x^2 + 4)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 27x^2 (x^2 + 4) - 2x (9x^3 - 1) \over (x^2 + 4)^2 } \\ & = { 27x^4 + 108x^2 - 18x^4 + 2x \over (x^2 + 4)^2 } \\ & = { 9x^4 + 108x^2 + 2x \over (x^2 + 4)^2 } \\ \\ \text{Let } & x = 0, \\ {dy \over dx} & = { 9(0)^4 + 108(0)^2 + 2(0) \over (0^2 +4)^2 } \\ & = 0 \end{align}

\begin{align} {d \over dx} \left(1 - 2x + {x^2 \over 2!} - {x^3 \over 3!} + {x^4 \over 4!} \right) & = {d \over dx} \left( 1 - 2x + {x^2 \over 2} - {x^3 \over 6} + {x^4 \over 24} \right) \\ & = -2 + {2x \over 2} - {3x^2 \over 6} + {4x^3 \over 24} \\ & = -2 + x - {x^2 \over 2} + {x^3 \over 6} \\ & = -2 + x - {1 \over 2}x^2 + {1 \over 6}x^3 \end{align}

\begin{align} u & = 3x &&& v & = \sqrt{2 + x} \\ & &&& & = (2 + x)^{1 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = \underbrace{{1 \over 2}(2 + x)^{-{1 \over 2}}. (1)}_\text{Chain rule} \\ & &&& & = {1 \over 2} (2 + x)^{-{1 \over 2}} \end{align} \begin{align} {dy \over dx} & = { (2 + x)^{1 \over 2} (3) - (3x)\left[{1 \over 2} (2 + x)^{-{1 \over 2}}\right] \over [(2 + x)^{1 \over 2}]^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 3(2 + x)^{1 \over 2} - {3 \over 2}x (2 + x)^{-{1 \over 2}} \over 2 + x } \\ \\ \text{When } & x = 7, \\ {dy \over dx} & = { 3(2 + 7)^{1 \over 2} - {3 \over 2}(7)(2 + 7)^{-{1 \over 2}} \over 2 + 7} \\ & = {11 \over 18} \end{align}

(i)

\begin{align} f(x) & = ax^2 + bx + c \\ \\ f'(x) & = 2ax + b \\ \\ \text{For increasing} & \text{ function, } f'(x) > 0 \\ \\ 2ax + b & > 0 \\ 2ax & > - b \\ x & > -{b \over 2a} \end{align}

(ii)

\begin{align} \text{For decreasing} & \text{ function, } f'(x) < 0 \\ \\ 2ax + b & < 0 \\ 2ax & < - b \\ x & < -{b \over 2a} \end{align}

(i)

\begin{align} y & = 16 - (3 - x)^4 \\ \\ {dy \over dx} & = - 4(3 - x)^3 . (-1) \phantom{000000} [\text{Chain rule}] \\ & = 4 (3 - x)^3 \\ \\ y = 27 \text{ is a } & \text{horizontal line with gradient} = 0 \\ \\ \text{When } & x = p \text{ and } {dy \over dx} = 0, \phantom{000000} [\text{Gradient at } (p, q) = 0] \\ 0 & = 4(3 - p)^3 \\ 0 & = (3 - p)^3 \\ \sqrt[3]{0} & = 3 - p \\ 0 & = 3 - p \\ p & = 3 \\ \\ \text{For } x < 3, & \phantom{0} (3 - x)^3 > 0 \\ \\ \implies {dy \over dx} > 0 & \text{ and } y \text{ is increasing for } x < 3 \end{align}

(ii)

\begin{align} {dy \over dx} & = 4 (3 - x)^3 \\ \\ \text{For } x > 3, & \phantom{0} (3 - x)^3 < 0 \\ \\ \implies {dy \over dx} < 0 & \text{ and } y \text{ is decreasing for } x > 3 \end{align}

\begin{align} u & = x - 1 &&& v & = \sqrt{x + 4} \\ & &&& & = (x + 4)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = \underbrace{ {1 \over 2} (x + 4)^{-{1 \over 2}}. (1)}_\text{Chain rule} \\ & &&& & = {1 \over 2}(x +4)^{-{1 \over 2}} \end{align} \begin{align} {dy \over dx} & = { (x + 4)^{1 \over 2} (1) - (x - 1)\left[{1 \over 2}(x + 4)^{-{1 \over 2}}\right] \over [(x + 4)^{1 \over 2}]^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { (x + 4)^{1 \over 2} - {1 \over 2}(x - 1)(x + 4)^{-{1 \over 2}} \over x + 4 } \times {(x + 4)^{1 \over 2 } \over (x + 4)^{1 \over 2}} \\ & = { x + 4 - {1 \over 2}(x - 1) (x + 4)^0 \over (x + 4)^{3 \over 2} } \\ & = { x + 4 - {1 \over 2}(x - 1)(1) \over (x + 4)^{3 \over 2} } \\ & = { x + 4 - {1 \over 2}x + {1 \over 2} \over (x + 4)^{3 \over 2}} \\ & = { {1 \over 2}x + {9 \over 2} \over (x + 4)^{3 \over 2}} \end{align}
\begin{align} y + 2 (x + 4){dy \over dx} & = {x - 1 \over \sqrt{x + 4}} + 2(x + 4) \left[ { {1 \over 2}x + {9 \over 2} \over (x + 4)^{3 \over 2}} \right] \\ & = {x - 1 \over (x + 4)^{1 \over 2}} + {2(x + 4) \over 1} \left[ { {1 \over 2}x + {9 \over 2} \over (x + 4)^{3 \over 2}} \right] \\ & = {x - 1 \over (x + 4)^{1 \over 2}} + {(x + 4)(x + 9) \over (x + 4)^{3 \over 2}} \\ & = {x - 1 \over (x + 4)^{1 \over 2}} + {x + 9 \over (x + 4)^{1 \over 2}} \\ & = {2x + 8 \over (x + 4)^{1 \over 2}} \\ & = {2 (x + 4) \over (x + 4)^{1 \over 2}} \\ & = 2 (x + 4)^{1 \over 2} \\ & = 2 \sqrt{x + 4} \phantom{0} \text{ (Shown)} \end{align}

\begin{align} \text{Let } y = {x^3 \over (1 - \sqrt{x})^6 } & = x^3 (1 - \sqrt{x})^{-6} \end{align} \begin{align} u & = x^3 &&& v & = (1 - \sqrt{x})^{-6} \\ & &&& & = (1 - x^{1 \over 2})^{-6} \\ {du \over dx} & = 3x^2 &&& {dv \over dx} & = (-6) (1 - x^{1 \over 2})^{-7} . \left(-{1 \over 2} x^{-{1 \over 2}}\right) \\ & &&& & = 3 x^{-{1 \over 2}} (1 - x^{1 \over 2})^{-7} \end{align} \begin{align} {dy \over dx} & = (x^3) \left[ 3 x^{-{1 \over 2}} (1 - x^{1 \over 2})^{-7} \right] + (1 - \sqrt{x})^{-6} (3x^2) \\ & = 3 x^{5 \over 2} (1 - x^{1 \over 2})^{-7} + 3x^2 (1 - \sqrt{x})^{-6} \\ & = {3x^{5 \over 2} \over (1 - \sqrt{x})^7} + {3x^2 \over (1 - \sqrt{x})^6} \times {1 - \sqrt{x} \over 1 - \sqrt{x}} \\ & = {3x^{5 \over 2} \over (1 - \sqrt{x})^7} + {3x^2 - 3x^{5 \over 2} \over (1 - \sqrt{x})^7} \\ & = {3x^2 \over (1 - \sqrt{x})^7} \end{align}
\begin{align} u & = x^3 &&& v & = (1 - \sqrt{x})^6 \\ & &&& & = (1 - x^{1 \over 2})^6 \\ {du \over dx} & = 3x^2 &&& {dv \over dx} & = 6(1 - x^{1 \over 2})^5 . \left(-{1 \over 2}x^{-{1 \over 2}}\right) \\ & &&& & = - 3 x^{-{1 \over 2}} (1 - x^{1 \over 2})^5 \end{align} \begin{align} {dy \over dx} & = { (1 - \sqrt{x})^6 (3x^2) - (x^3) [ - 3 x^{-{1 \over 2}} (1 - x^{1 \over 2})^5 ] \over [(1 - \sqrt{x})^6]^2 } \\ & = { 3x^2 (1 - \sqrt{x})^6 + 3 x^{5 \over 2} (1 - \sqrt{x})^5 \over (1 - \sqrt{x})^{12} } \\ & = { 3x^2 (1 - \sqrt{x})^5 \left( 1 - \sqrt{x} + x^{1 \over 2} \right) \over (1 - \sqrt{x})^{12} } \\ & = { 3x^2 (1 - \sqrt{x})^5 (1) \over (1 - \sqrt{x})^{12} } \\ & = { 3x^2 (1 - \sqrt{x})^5 \over (1 - \sqrt{x})^{12} } \\ & = { 3x^2 \over (1 - \sqrt{x})^7} \end{align}