(a) By chain rule

\begin{align} {d \over dx}(\sqrt{1 - 3x^2}) & = {d \over dx}(1 - 3x^2)^{1 \over 2} \\ & = {1 \over 2}(1 - 3x^2)^{-{1 \over 2}}. (-6x) \\ & = -3x (1 - 3x^2)^{-{1 \over 2}} \\ & = {-3x \over \sqrt{1 - 3x^2}} \end{align}

(b) By product rule

\begin{align} u & = x &&& v & = (1 + 2x)^4 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 4(1 + 2x)^3 . (2) \phantom{00000} [\text{Chain rule}] \\ & &&& & = 8(1 + 2x)^3 \end{align} \begin{align} {d \over dx} \left[ x(1 + 2x)^4 \right] & = (x)[8(1 + 2x)^3] + (1 + 2x)^4 (1) \\ & = 8x(1 + 2x)^3 + (1 + 2x)^4 \\ & = (1 + 2x)^3[8x + (1 + 2x)] \\ & = (1 + 2x)^3 (8x + 1 + 2x) \\ & = (1 + 2x)^3(10x + 1) \end{align}

(c) Method 1: By quotient rule

\begin{align} u & = 1 + 2x &&& v & = \sqrt{x} \\ & &&& & = x^{1 \over 2} \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over 2} x^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x}} \end{align} \begin{align} {d \over dx}\left({1 + 2x \over \sqrt{x}}\right) & = {(\sqrt{x})(2) - (1 + 2x)\left(1 \over 2\sqrt{x}\right) \over (\sqrt{x})^2} \\ & = {2\sqrt{x} - {1 + 2x \over 2\sqrt{x}} \over x } \\ & = {2\sqrt{x} - {1 + 2x \over 2\sqrt{x}} \over x } \times {2\sqrt{x} \over 2\sqrt{x}} \\ & = {2\sqrt{x}(2\sqrt{x}) - (1 + 2x) \over x(2\sqrt{x}) } \\ & = {4x - 1 - 2x \over 2x\sqrt{x}} \\ & = {2x - 1 \over 2x\sqrt{x}} \end{align}

Method 2: By simplifying

\begin{align} {1 + 2x \over \sqrt{x}} & = {1 + 2x \over x^{1 \over 2}} \\ & = {1 \over x^{1 \over 2}} + {2x \over x^{1 \over 2}} \\ & = x^{-{1 \over 2}} + 2x^{1 \over 2} \\ \\ {d \over dx} \left( x^{-{1 \over 2}} + 2x^{1 \over 2} \right) & = -{1 \over 2} x^{-{3 \over 2}} + 2 \left(1 \over 2\right) x^{-{1 \over 2}} \\ & = -{1 \over 2} \left(1 \over x^{3 \over 2}\right) + {1 \over \sqrt{x}} \\ & = -{1 \over 2} \left[ 1 \over x (x^{1 \over 2}) \right] + {1 \over \sqrt{x}} \\ & = -{1 \over 2} \left( 1 \over x \sqrt{x} \right) + {1 \over \sqrt{x}} \\ & = -{1 \over 2x\sqrt{x}} + {1 \over \sqrt{x}} \times {2x \over 2x} \\ & = -{1 \over 2x\sqrt{x}} + {2x \over 2x\sqrt{x}} \\ & = {-1 + 2x \over 2x\sqrt{x}} \end{align}

Use product rule for the differentiation

\begin{align} u & = 2x + 3 &&& v & = \sqrt{5x - 9} \\ & &&& & = (5x - 9)^{1 \over 2} \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over 2} (5x - 9)^{-{1 \over 2}} . (5) \phantom{00000} [\text{Chain rule}] \\ & &&& & = {5 \over 2} (5x - 9)^{-{1 \over 2}} \\ & &&& & = {5 \over 2\sqrt{5x - 9}} \end{align} \begin{align} {dy \over dx} & = (2x + 3)\left(5 \over 2\sqrt{5x - 9}\right) + (\sqrt{5x - 9})(2) \\ & = {5(2x + 3) \over 2\sqrt{5x - 9}} + 2\sqrt{5x - 9} \\ & = {5(2x + 3) \over 2\sqrt{5x - 9}} + {2\sqrt{5x - 9} \over 1} \times {2\sqrt{5x - 9} \over 2\sqrt{5x - 9}} \\ & = {10x + 15 \over 2\sqrt{5x - 9}} + {2(2)\sqrt{5x - 9}\sqrt{5x - 9} \over \sqrt{5x - 9}} \\ & = {10x + 15 \over 2\sqrt{5x - 9}} + {4(5x - 9) \over \sqrt{5x - 9}} \\ & = {10x + 15 + 4(5x - 9) \over \sqrt{5x - 9}} \\ & = {10x + 15 + 20x - 36 \over 2\sqrt{5x - 9}} \\ & = {30x - 21 \over 2\sqrt{5x - 9}} \\ \\ \therefore k & = 30 \end{align}

I used quotient rule for this question (product rule is possible as well)

\begin{align} u & = x + 1 &&& v & = \sqrt{1 - 2x} \\ & &&& & = (1 - 2x)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (1 - 2x)^{-{1 \over 2}} . (-2) \phantom{00000} [\text{Chain rule}] \\ & &&& & = - (1 - 2x)^{-{1 \over 2}} \\ & &&& & = -{1 \over \sqrt{1 - 2x}} \end{align} \begin{align} {dy \over dx} & = {(\sqrt{1 - 2x})(1) - (x + 1) \left(-{1 \over \sqrt{1 - 2x}}\right) \over (\sqrt{1 - 2x})^2} \\ & = {\sqrt{1 - 2x} + {x + 1 \over \sqrt{1 - 2x}} \over 1 - 2x } \\ & = {\sqrt{1 - 2x} + {x + 1 \over \sqrt{1 - 2x}} \over 1 - 2x } \times {\sqrt{1 - 2x} \over \sqrt{1 - 2x}} \\ & = {\sqrt{1 - 2x}(\sqrt{1 - 2x}) + (x + 1) \over (1 - 2x) \sqrt{1 - 2x}} \\ & = {1 - 2x + x + 1 \over (1 - 2x) (1 - 2x)^{1 \over 2}} \\ & = {2 - x \over (1 - 2x)^{3 \over 2}} \\ & = {- x + 2 \over \sqrt{(1 - 2x)^3}} \\ \\ \therefore n & = 0 \end{align}

\begin{align} y & = {9 \over (2x - 5)^2} \\ & = 9(2x - 5)^{-2} \\ \\ {dy \over dx} & = (9)(-2)(2x - 5)^{-3} . (2) \phantom{000000} [\text{Chain rule}] \\ & = -36(2x - 5)^{-3} \\ & = {-36 \over (2x - 5)^3} \\ \\ \text{When } & x = -1, \\ {dy \over dx} & = {-36 \over [2(-1) - 5]^{3}} \\ & = {36 \over 343} \end{align}

The x-axis is a horizontal line, which means it's gradient is equals to 0. Since the tangent is parallel to the x-axis, the gradient of the tangent is 0.

\begin{align} u & = x^2 - 1 &&& v & = x^2 + 1 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2 + 1)(2x) - (x^2 - 1)(2x) \over (x^2 + 1)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {2x(x^2 + 1) - 2x(x^2 - 1) \over (x^2 + 1)^2} \\ & = {2x^3 + 2x - 2x^3 + 2x \over (x^2 + 1)^2} \\ & = {4x \over (x^2 + 1)^2} \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = { 4x \over (x^2 + 1)^2 } \\ 0 & = 4x \\ 0 & = x \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = {0^2 - 1 \over 0^2 + 1} \\ & = -1 \\ \\ \therefore \text{Coordi} & \text{nate is } (0, -1) \end{align}

(a)

\begin{align} {1 \over 9\left( {8 \over x} - 7 \right)^2} & = {1 \over 9}\left( {1 \over \left({8 \over x} - 7 \right)^2} \right) \\ & = {1 \over 9}\left( {8 \over x} - 7 \right)^{-2} \\ & = {1 \over 9}\left( 8x^{-1} - 7 \right)^{-2} \end{align} \begin{align} {d \over dx}\left[ {1 \over 9}\left( 8 x^{-1} - 7 \right)^{-2} \right] & = {1 \over 9} (-2)\left( 8 x^{-1} - 7 \right)^{-3}. (-8x^{-2}) \phantom{00000} [\text{Chain rule}] \\ & = -{2 \over 9} \left( 8 x^{-1} - 7 \right)^{-3} \left(-8 \over x^2\right) \\ & = {16 \over 9x^2} \left( 8 x^{-1} - 7 \right)^{-3} \\ & = {16 \over 9x^2} \left( {8 \over x} - 7 \right)^{-3} \\ & = {16 \over 9x^2 \left( {8 \over x} - 7 \right)^{3}} \end{align}

(b)

\begin{align} {d \over dx}\left[ 11(4\sqrt{x} - 1)^{-{3 \over 5}} \right] & = {d \over dx} \left[ 11(4x^{1 \over 2} - 1)^{-{3 \over 5}} \right] \\ & = 11 \left(-{3 \over 5}\right) (4x^{1 \over 2} - 1)^{-{8 \over 5}} . \left[ \left({1 \over 2}\right) (4)x^{-{1 \over 2}} \right] \phantom{00000} [\text{Chain rule}] \\ & = -{33 \over 5} (4x^{1 \over 2} - 1)^{-{8 \over 5}} \left(2x^{-{1 \over 2}}\right) \\ & = -{66 \over 5}x^{-{1 \over 2}} (4 \sqrt{x} - 1)^{-{8 \over 5}} \end{align}

\begin{align} 4y & = 2x + c \\ y & = {2 \over 4}x + {1 \over 4}c \\ y & = {1 \over 2}x + {1 \over 4}c \\ \\ \text{Comparing with } & y = mx + c, \\ m & = {1 \over 2} \\ \\ \therefore \text{Gradient of tangent} & = {1 \over 2} \\ \\ y & = 4x + {8 \over x} \\ & = 4x + 8x^{-1} \\ \\ {dy \over dx} & = 4 + 8(-1)x^{-2} \\ & = 4 - 8x^{-2} \\ & = 4 - {8 \over x^2} \\ \\ \text{When } & {dy \over dx} = {1 \over 2}, \\ {1 \over 2} & = 4 - {8 \over x^2} \\ {8 \over x^2} & = 4 - {1 \over 2} \\ {8 \over x^2} & = {7 \over 2} \\ 2(8) & = 7x^2 \\ 16 & = 7x^2 \\ x^2 & = {16 \over 7} \\ x & = \pm\sqrt{16 \over 7} \\ & = \pm {\sqrt{16} \over \sqrt{7}} \\ & = \pm{4 \over \sqrt{7}} \end{align} \begin{align} \text{Substitute } & x = {4 \over \sqrt{7}} \text{ into eqn of curve,} &&& \text{Substitute } & x = -{4 \over \sqrt{7}} \text{ into eqn of curve,} \\ y & = 4 \left(4 \over \sqrt{7}\right) + {8 \over {4 \over \sqrt{7}}} &&& y & = 4 \left(-{4 \over \sqrt{7}}\right) + {8 \over -{4 \over \sqrt{7}}} \\ & = {16 \over \sqrt{7}} + {8\sqrt{7} \over 4} &&& & = -{16 \over \sqrt{7}} - {8\sqrt{7} \over 4} \\ & = {16\sqrt{7} \over 7} + 2\sqrt{7} &&& & = -{16\sqrt{7} \over 7} - 2\sqrt{7} \\ & = {16 \over 7}\sqrt{7} + 2\sqrt{7} &&& & = -{16 \over 7} \sqrt{7} - 2\sqrt{7} \\ & = {30 \over 7} \sqrt{7} &&& & = -{30 \over 7} \sqrt{7} \\ \\ \therefore & \phantom{.} \left({4 \over \sqrt{7}}, {30 \over 7}\sqrt{7}\right) &&& \therefore & \phantom{.} \left(-{4 \over \sqrt{7}}, -{30 \over 7}\sqrt{7}\right) \\ \\ \text{Substitute } & \text{into eqn of tangent,} &&& \text{Substitute } & \text{into eqn of tangent,} \\ 4\left({30 \over 7}\sqrt{7}\right) & = 2\left(4 \over \sqrt{7}\right) + c &&& 4\left(-{30 \over 7}\sqrt{7}\right) & = 2\left(-{4 \over \sqrt{7}}\right) + c \\ {120 \over 7}\sqrt{7} & = {8 \over \sqrt{7}} + c &&& -{120 \over 7}\sqrt{7} & = -{8 \over \sqrt{7}} + c \\ {120 \over 7}\sqrt{7} & = {8 \over 7}\sqrt{7} + c &&& -{120 \over 7}\sqrt{7} & = -{8 \over 7}\sqrt{7} + c \\ {120 \over 7}\sqrt{7} - {8 \over 7} \sqrt{7} & = c &&& -{120 \over 7}\sqrt{7} + {8 \over 7}\sqrt{7} & = c \\ 16\sqrt{7} & = c &&& -16 \sqrt{7} & = c \end{align}

The x-axis is a horizontal line, which means it's gradient equals to 0.

Also, the y-coordinate of the point(s) where the x-axis meets the curve is equals to 0.

\begin{align} {dy \over dx} & = 3x^2 - 6x - 9 \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 6x - 9 \\ 0 & = x^2 - 2x - 3 \\ 0 & = (x - 3)(x + 1) \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x + 1 & = 0 \\ x & = 3 &&& x & = -1 \\ \\ \therefore & \phantom{.} (3, 0) &&& \therefore & \phantom{.} (-1, 0) \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ 0 & = (3)^3 - 3(3)^2 - 9(3) + k &&& 0 & = (-1)^3 - 3(-1)^2 - 9(-1) + k \\ 0 & = 27 - 27 - 27 + k &&& 0 & = -1 - 3 + 9 + k \\ 0 & = -27 + k &&& 0 & = 5 + k \\ 27 & = k &&& -5 & = k \end{align}

Recall that parallel lines have the same gradient.

\begin{align} y & = 2x + 1 \\ \\ \text{Comparing with } & y = mx + c, \\ m & = 2 \\ \\ \therefore \text{Gradient of tangent} & = 2 \phantom{00000} [\text{Parallel lines}] \end{align}
\begin{align} u & = x &&& v & = \sqrt{a - 3x^2} \\ & &&& & = (a - 3x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(a - 3x^2)^{-{1 \over 2}} . (-6x) \phantom{00000} [\text{Chain rule}] \\ & &&& & = -3x(a - 3x^2)^{-{1 \over 2}} \\ & &&& & = {-3x \over \sqrt{a - 3x^2}} \end{align} \begin{align} {dy \over dx} & = (x)\left(-3x \over \sqrt{a - 3x^2}\right) + (\sqrt{a - 3x^2})(1) \phantom{00000} [\text{Product rule}] \\ & = {-3x^2 \over \sqrt{a - 3x^2}} + \sqrt{a - 3x^2} \\ & = {-3x^2 \over \sqrt{a - 3x^2}} + {\sqrt{a - 3x^2} \over 1} \\ & = {-3x^2 \over \sqrt{a - 3x^2}} + {a - 3x^2 \over \sqrt{a - 3x^2}} \\ & = {-3x^2 + a - 3x^2 \over \sqrt{a - 3x^2}} \\ & = {-6x^2 + a \over \sqrt{a - 3x^2}} \\ \\ \text{When } x = 1 & \text{ and } {dy \over dx} = 2, \\ 2 & = {-6(1)^2 + a \over \sqrt{a - 3(1)^2}} \\ 2 & = {-6 + a \over \sqrt{a - 3}} \\ 2\sqrt{a - 3} & = -6 + a \\ (2 \sqrt{a - 3})^2 & = (-6 + a)^2 \\ 4(a - 3) & = (-6)^2 + 2(-6)(a) + a^2 \\ 4a - 12 & = 36 - 12a + a^2 \\ 0 & = a^2 - 16a + 48 \\ 0 & = (a - 12)(a - 4) \end{align} \begin{align} a - 12 & = 0 && \text{ or } & a - 4 & = 0 \\ a & = 12 &&& a & = 4 \phantom{0} \left(\text{Reject, since } {dy \over dx} \ne 2 \right)\\ \\ y & = x\sqrt{12 - 3x^2} \\ \\ \text{Using } & (1, b), \\ b & = (1)\sqrt{12 - 3(1)^2} \\ & = 3 \\ \\ \therefore a & = 12, b = 3 \end{align}

(i) Since the tangent to the curve is horizontal, the gradient of the tangent is equals to 0.

\begin{align} y & = {ax^2 \over x - b} \\ \\ \text{Using } & \left({1 \over 3}, 6\right), \\ 6 & = {a\left({1 \over 3}\right)^2 \over {1 \over 3} - b} \\ 6 & = { {a \over 9} \over {1 \over 3} - b} \\ 6\left( {1 \over 3} - b \right) & = {a \over 9} \\ 2 - 6b & = {a \over 9} \\ 18 - 54b & = a \phantom{00}\text{--- (1)} \\ \end{align}
\begin{align} u & = ax^2 &&& v & = x - b \\ {du \over dx} & = 2ax &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = { (x - b)(2ax) - (ax^2)(1) \over (x - b)^2 } \\ & = {2ax(x - b) - ax^2 \over (x - b)^2} \\ & = { 2ax^2 - 2abx - ax^2 \over (x - b)^2 } \\ & = { ax^2 - 2abx \over (x - b)^2 } \\ \\ \text{When } {dy \over dx} = 0 & \text{ and } x = {1 \over 3}, \\ 0 & = { a\left( {1 \over 3} \right)^2 - 2ab\left( {1 \over 3} \right) \over ({1 \over 3} - b)^2 } \\ \therefore 0 & = {1 \over 9}a - {2 \over 3}ab \\ 0 & = a - 6b\phantom{00}\text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 0 & = (18 - 54b) - 6(18 - 54b)(b) \\ 0 & = 18 - 54b - 108b + 324b^2 \\ 0 & = 324b^2 - 162b + 18 \\ 0 & = 18b^2 - 9b + 1 \\ 0 & = (6b - 1)(3b - 1) \end{align} \begin{align} 6b - 1 & = 0 && \text{ or } & 3b - 1 & = 0 \\ 6b & = 1 &&& 3b & = 1 \\ b & = {1 \over 6} &&& b & = {1 \over 3} \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ a & = 18 - 54\left(1 \over 6\right) &&& a & = 18 - 54\left(1 \over 3\right) \\ & = 9 &&& & = 0 \text{ (Reject, since } a \ne 0) \\ \\ \therefore a & = 9, b = {1 \over 6} \end{align}

(ii)

\begin{align} {dy \over dx} & = {ax^2 - 2ab x \over (x - b)^2} \\ & = {9x^2 - 2(9)({1 \over 6})x \over (x - {1 \over 6})^2} \\ & = {9x^2 - 3x \over (x - {1 \over 6})^2} \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = {9x^2 - 3x \over (x - {1 \over 6})^2} \\ \therefore 0 & = 9x^2 - 3x \\ 0 & = 3x(3x - 1) \end{align} \begin{align} 3x & = 0 && \text{ or } & 3x - 1 & = 0 \\ x & = 0 &&& 3x & = 1 \\ & &&& x & = {1 \over 3} \phantom{000} \left[ \text{The point } \left({1 \over 3}, 6\right) \right] \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = {9(0)^2 \over 0 - {1 \over 6}} \\ & = 0 \\ \\ \therefore & \phantom{.} (0, 0) \end{align}