(a)

\begin{align} u & = x &&& v & = (3x + 2)^2 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 2(3x + 2) . (3) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 6(3x + 2) \end{align} \begin{align} {d \over dx} \left[ x(3x + 2)^2 \right] & = x[6(3x + 2)] + (3x + 2)^2 (1) \\ & = 6x(3x + 2) + (3x + 2)^2 \\ & = (3x + 2) \left[ 6x + (3x + 2) \right] \\ & = (3x + 2) (6x + 3x + 2) \\ & = (3x + 2) (9x + 2) \end{align}

(b)

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

(c)

\begin{align} u & = 2 - 3x &&& v & = (4x + 1)^4 \\ {du \over dx} & = -3 &&& {dv \over dx} & = 4(4x + 1)^3 . (4) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 16(4x + 1)^3 \end{align} \begin{align} {d \over dx} \left[ (2 - 3x)(4x + 1)^4 \right] & = (2 - 3x)\left[ 16(4x + 1)^3 \right] + (4x + 1)^4 (-3) \\ & = 16(2 - 3x)(4x + 1)^3 - 3(4x + 1)^4 \\ & = (4x + 1)^3 [16(2 - 3x) - 3(4x + 1) ] \\ & = (4x + 1)^3 (32 - 48x - 12x - 3) \\ & = (4x + 1)^3 (29 - 60x) \end{align}

(d)

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

(a)

\begin{align} u & = \sqrt{x} &&& v & = (1 + x)^2 \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} &&& {dv \over dx} & = 2(1 + x) .(1) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2\sqrt{x}} &&& & = 2(1 + x) \end{align} \begin{align} {d \over dx} [\sqrt{x} (1 + x)^2] & = (\sqrt{x}) [2(1 + x)] + (1 + x)^2 \left(1 \over 2\sqrt{x}\right) \\ & = 2 \sqrt{x} (1 + x) + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {2 \sqrt{x} (1 + x) \over 1} + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {2 \sqrt{x} (1 + x)(2\sqrt{x}) \over 2\sqrt{x}} + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {4x(1 + x) \over 2\sqrt{x}} + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {4x(1 + x) + (1 + x)^2 \over 2\sqrt{x}} \\ & = {(1 + x) [4x + (1 + x)] \over 2\sqrt{x}} \\ & = {(1 + x) (4x + 1 + x) \over 2\sqrt{x}} \\ & = {(1 + x)(5x + 1) \over 2\sqrt{x}} \end{align}

(b)

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

(c)

\begin{align} u & = x^2 &&& v & = \sqrt{1 + 2x^2} \\ &&&& & = (1 + 2x^2)^{1 \over 2} \\ {du \over dx} & = 2x &&& {dv \over dx} & = {1 \over 2}(1 + 2x^2)^{-{1 \over 2}} . (4x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 2x (1 + 2x^2)^{-{1 \over 2}} \\ &&&& & = {2x \over \sqrt{1 + 2x^2}} \end{align} \begin{align} {d \over dx} \left[ x^2 \sqrt{1 + 2x^2} \right] & = (x^2) \left({2x \over \sqrt{1 + 2x^2}} \right) + \left(\sqrt{1 + 2x^2}\right)(2x) \\ & = {2x^3 \over \sqrt{1 + 2x^2} } + 2x \sqrt{1 + 2x^2} \\ & = {2x^3 \over \sqrt{1 + 2x^2}} + {2x \sqrt{1 + 2x^2} \over 1} \\ & = {2x^3 \over \sqrt{1 + 2x^2}} + {2x \sqrt{1 + 2x^2} \left(\sqrt{1 + 2x^2}\right) \over \sqrt{1 + 2x^2}} \\ & = {2x^3 \over \sqrt{1 + 2x^2}} + {2x(1 + 2x^2) \over \sqrt{1 + 2x^2}} \\ & = {2x^3 + 2x(1 + 2x^2) \over \sqrt{1 + 2x^2}} \\ & = {2x^3 + 2x + 4x^3 \over \sqrt{1 + 2x^2}} \\ & = {6x^3 + 2x \over \sqrt{1 + 2x^2}} \\ & = {2x(3x^2 + 1) \over \sqrt{1 + 2x^2}} \end{align}

(d)

\begin{align} u & = 2x - 1 &&& v & = \sqrt{5x^2 - 1} \\ &&&& & = (5x^2 - 1)^{1 \over 2} \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over 2}(5x^2 - 1)^{-{1 \over 2}} . (10x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 5x(5x^2 - 1)^{-{1 \over 2}} \\ &&&& & = {5x \over \sqrt{5x^2 - 1}} \end{align} \begin{align} {d \over dx} \left[ (2x - 1) \sqrt{5x^2 - 1} \right] & = (2x - 1) \left(5x \over \sqrt{5x^2 - 1}\right) + \left( \sqrt{5x^2 - 1} \right) (2) \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + 2 \sqrt{5x^2 - 1} \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + {2\sqrt{5x^2 - 1} \over 1} \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + {2\sqrt{5x^2 - 1} \left(\sqrt{5x^2 - 1}\right) \over \sqrt{5x^2 - 1}} \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + {2(5x^2 - 1) \over \sqrt{5x^2 - 1}} \\ & = {5x(2x - 1) + 2(5x^2 - 1) \over \sqrt{5x^2 - 1}} \\ & = {10x^2 - 5x + 10x^2 - 2 \over \sqrt{5x^2 - 1}} \\ & = {20x^2 - 5x - 2 \over \sqrt{5x^2 - 1}} \end{align}

(e)

\begin{align} u & = \sqrt{x + 2} + 1 &&& v & = x\sqrt{x} + 1 \\ & = (x + 2)^{1 \over 2} &&& & = x(x)^{1 \over 2} + 1 \\ & &&& & = x^{3 \over 2} + 1 \\ {du \over dx} & = {1 \over 2}(x + 2)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] &&& {dv \over dx} & = {3 \over 2}x^{1 \over 2} \\ & = {1 \over 2} (x + 2)^{-{1 \over 2}} &&& & = {3 \over 2} \sqrt{x} \\ & = {1 \over 2\sqrt{x + 2}} &&& & = {3 \sqrt{x} \over 2} \end{align} \begin{align} {d \over dx} \left[ \sqrt{x + 2} \left(x \sqrt{x} + 1\right) \right] & = \sqrt{x + 2} \left( {3 \sqrt{x} \over 2} \right) + \left(x \sqrt{x} + 1\right) \left( {1 \over 2\sqrt{x + 2}} \right) \\ & = {3\sqrt{x} \sqrt{x + 2} \over 2} + {x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3\sqrt{x} \sqrt{x + 2} \left(\sqrt{x + 2}\right) \over 2\sqrt{x + 2}} + {x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3\sqrt{x} (x + 2) \over 2\sqrt{x + 2}} + {x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3\sqrt{x} (x + 2) + x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3x \sqrt{x} + 6\sqrt{x} + x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {4x \sqrt{x} + 6\sqrt{x} + 1 \over 2\sqrt{x + 2}} \end{align}

(f)

\begin{align} u & = 3\sqrt{x} + 3 &&& v & = \sqrt{2 - 2x} \\ & = 3x^{1 \over 2} + 3 &&& & = (2 - 2x)^{1 \over 2} \\ {du \over dx} & = 3\left({1 \over 2}\right)x^{-{1 \over 2}} &&& {dv \over dx} & = {1 \over 2}(2 - 2x)^{-{1 \over 2}} . (-2) \phantom{00000} [\text{Chain rule}] \\ & = {3 \over 2} x^{-{1 \over 2}} &&& & = -(2 - 2x)^{-{1 \over 2}} \\ & = {3 \over 2\sqrt{x}} &&& & = -{1 \over \sqrt{2 - 2x}} \end{align} \begin{align} {d \over dx} \left[ \left( 3\sqrt{x} + 3 \right) \sqrt{2 - 2x} \right] & = \left( 3\sqrt{x} + 3 \right) \left( -{1 \over \sqrt{2 - 2x}} \right) + \sqrt{2 - 2x} \left( {3 \over 2\sqrt{x}} \right) \\ & = {-(3\sqrt{x} + 3) \over \sqrt{2 - 2x}} + {3\sqrt{2 - 2x} \over 2\sqrt{x}} \\ & = {-(3\sqrt{x} + 3)(2\sqrt{x}) \over 2\sqrt{x} \sqrt{2 - 2x}} + { 3\sqrt{2 - 2x} \left(\sqrt{2 - 2x}\right) \over 2\sqrt{x} \sqrt{2 - 2x}} \\ & = {-2\sqrt{x} (3 \sqrt{x} + 3) \over \sqrt{2 - 2x}} + {3(2 - 2x) \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {-2\sqrt{x} (3 \sqrt{x} + 3) + 3(2 - 2x) \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {-6x - 6\sqrt{x} + 6 - 6x \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {-12x - 6\sqrt{x} + 6 \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {6 \left( -2x - \sqrt{x} + 1 \right) \over 2 \sqrt{x(2 - 2x)}} \\ & = { 3(-2x - \sqrt{x} + 1) \over \sqrt{x(2 - 2x)} } \\ & = { 3(1 - \sqrt{x} - 2x) \over \sqrt{2x(1 - x)}} \end{align}

(a)

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

(b)

\begin{align} u & = x^2 (x - 1) &&& v & = \sqrt{6x + 5} \\ & = x^3 - x^2 &&& & = (6x + 5)^{1 \over 2} \\ {du \over dx} & = 3x^2 - 2x &&& {dv \over dx} & = {1 \over 2}(6x + 5)^{-{1 \over 2}} . (6) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 3(6x + 5)^{-{1 \over 2}} \\ &&&& & = {3 \over \sqrt{6x + 5}} \end{align} \begin{align} {d \over dx} \left[ (x^3 - x^2)\sqrt{6x + 5} \right] & = (x^3 - x^2) \left( {3 \over \sqrt{6x + 5}} \right) + \left( \sqrt{6x + 5} \right)(3x^2 - 2x) \\ & = {3(x^3 - x^2) \over \sqrt{6x + 5}} + (3x^2 - 2x)\sqrt{6x + 5} \\ & = {3(x^3 - x^2) \over \sqrt{6x + 5}} + { (3x^2 - 2x)\sqrt{6x + 5} \over 1} \\ & = {3(x^3 - x^2) \over \sqrt{6x + 5}} + {(3x^2 - 2x)\sqrt{6x + 5}(\sqrt{6x + 5}) \over \sqrt{6x + 5}} \\ & = {3x^3 - 3x^2 \over \sqrt{6x + 5}} + {(3x^2 - 2x)(6x + 5) \over \sqrt{6x + 5}} \\ & = {3x^3 - 3x^2 + (3x^2 - 2x)(6x + 5) \over \sqrt{6x + 5}} \\ & = {3x^3 - 3x^2 + 18x^3 + 15x^2 - 12x^2 - 10x \over \sqrt{6x + 5}} \\ & = {21x^3 - 10x \over \sqrt{6x + 5}} \\ & = {x(21x^2 - 10) \over \sqrt{6x + 5}} \end{align}

\begin{align} u & = x &&& v & = \sqrt{5 - x^2} \\ &&&& & = (5 - x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(5 - x^2)^{-{1 \over 2}} . (-2x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = -x (5 - x^2)^{-{1 \over 2}} \\ &&&& & = {-x \over \sqrt{5 - x^2}} \end{align} \begin{align} {dy \over dx} & = (x)\left({-x \over \sqrt{5 - x^2}}\right) + \left(\sqrt{5 - x^2} \right) (1) \\ & = {-x^2 \over \sqrt{5 - x^2}} + \sqrt{5 - x^2} \\ & = {-x^2 \over \sqrt{5 - x^2}} + {\sqrt{5 - x^2} \over 1} \\ & = {-x^2 \over \sqrt{5 - x^2}} + {\sqrt{5 - x^2}\sqrt{5 - x^2} \over \sqrt{5 - x^2}} \\ & = {-x^2 \over \sqrt{5 - x^2}} + {5 - x^2 \over \sqrt{5 - x^2}} \\ & = {-x^2 + 5 - x^2 \over \sqrt{5 - x^2}} \\ & = {5 - 2x^2 \over \sqrt{5 - x^2}} \phantom{0} \text{ (Shown)} \end{align}

(i)

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

(ii)

\begin{align} \text{Substitute } & x = 5 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {3(5) - 2 \over 2\sqrt{(5) - 1}} \\ & = 3{1 \over 4} \end{align}

The points where the curve the x-axis are the x-intercepts, where y = 0.

\begin{align} u & = (x - 1)^3 &&& v & = x + 1 \\ {du \over dx} & = 3(x - 1)^2 . (1) &&& {dv \over dx} & = 1 \\ & = 3(x - 1)^2 \end{align} \begin{align} {dy \over dx} & = (x - 1)^3 (1) + (x + 1)[3(x - 1)^2] \\ & = (x - 1)^3 + 3(x + 1)(x - 1)^2 \\ & = (x - 1)^2 [(x - 1) + 3(x + 1)] \\ & = (x - 1)^2 (x - 1 + 3x + 3) \\ & = (x - 1)^2 (4x + 2) \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = (x - 1)^3 (x + 1) \end{align} \begin{align} (x - 1)^3 & = 0 && \text{ or } & x + 1 & =0 \\ x - 1 & = 0 &&& x & = -1 \\ x & = 1 \\ \\ \text{Substitute} & \text{ into } {dy \over dx}, &&& \text{Substitute} & \text{ into } {dy \over dx}, \\ {dy \over dx} & = (1 - 1)^2 [4(1) + 2] &&& {dy \over dx} & = (-1 - 1)^2[4(-1) + 2] \\ & = 0 &&& & = -8 \end{align}

(i)

\begin{align} u & = x - 3 &&& v & = \sqrt{x - 1} \\ & &&& & = (x - 1)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(x - 1)^{-{1 \over 2}} . (1) \\ & &&& & = {1 \over 2} (x - 1)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x - 1}} \end{align} \begin{align} {dy \over dx} & = (x - 3) \left({1 \over 2\sqrt{x - 1}}\right) + \left( \sqrt{x - 1} \right) (1) \\ & = {x - 3 \over 2\sqrt{x - 1}} + \sqrt{x - 1} \\ & = {x - 3 \over 2\sqrt{x - 1}} + {\sqrt{x - 1} \over 1} \\ & = {x - 3 \over 2\sqrt{x - 1}} + {\sqrt{x - 1}\left(2 \sqrt{x - 1} \right) \over 2\sqrt{x - 1}} \\ & = {x - 3 \over 2\sqrt{x - 1}} + {2(x - 1) \over 2\sqrt{x - 1}} \\ & = {x - 3 + 2(x - 1) \over 2\sqrt{x - 1}} \\ & = {x - 3 + 2x - 2 \over 2\sqrt{x - 1}} \\ & = {3x - 5 \over 2\sqrt{x - 1}} \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {3x - 5 \over 2\sqrt{x - 1} } \\ \therefore 0 & = 3x - 5 \\ 5 & = 3x \\ {5 \over 3} & = x \\ \\ \text{Substitute } & x = {5 \over 3} \text{ into eqn of curve,} \\ y & = \left( {5 \over 3} - 3 \right) \sqrt{ {5 \over 3} - 1} \\ & = \left(-{4 \over 3}\right) \sqrt{ 2 \over 3} \\ & = \left(-{4 \over 3}\right) {\sqrt{2} \over \sqrt{3}} \\ & = -{4 \sqrt{2} \over 3\sqrt{3}} \\ & = -{4 \sqrt{2} \over 3\sqrt{3}} \times {\sqrt{3} \over \sqrt{3}} \phantom{00000} [\text{Rationalise denominator}] \\ & = -{4 \sqrt{6} \over 3(3)} \\ & = -{4 \sqrt{6} \over 9} \\ \\ \therefore & \phantom{.} \left(1{2 \over 3}, -{4\sqrt{6} \over 9}\right) \end{align}

(i)

\begin{align} u & = \sqrt{x} &&& v & = (x - 3)^4 \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} &&& {dv \over dx} & = 4(x - 3)^3 . (1) \\ & = {1 \over 2\sqrt{x}} &&& & = 4(x - 3)^3 \end{align} \begin{align} {dy \over dx} & = (\sqrt{x}) [4(x - 3)^3] + (x - 3)^4 \left(1 \over 2\sqrt{x}\right) \\ & = 4\sqrt{x} (x - 3)^3 + {1 \over 2\sqrt{x}}(x - 3)^4 \\ & = (x - 3)^3 \left[ 4\sqrt{x} + {1 \over 2\sqrt{x}}(x - 3) \right] \\ & = (x - 3)^3 \left[ {4\sqrt{x} \over 1} + {x - 3 \over 2\sqrt{x}} \right] \\ & = (x - 3)^3 \left[ {4\sqrt{x}(2\sqrt{x}) \over 2\sqrt{x}} + {x - 3 \over 2\sqrt{x}} \right] \\ & = (x - 3)^3 \left( {8x + x - 3 \over 2\sqrt{x}} \right) \\ & = (x - 3)^3 \left( {9x - 3 \over 2\sqrt{x}} \right) \\ & = {(x - 3)^3 (9x - 3) \over 2\sqrt{x}} \\ & = {3(x - 3)^3 (3x - 1) \over 2\sqrt{x}} \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {3(x - 3)^3 (3x - 1) \over 2\sqrt{x}} \\ \therefore 0 & = 3(x - 3)^3 (3x - 1) \\ \end{align} \begin{align} (x - 3)^3 & = 0 && \text{ or } & 3x - 1 & = 0 \\ x - 3 & = 0 &&& 3x & = 1 \\ x & = 3 &&& x & = {1 \over 3} \end{align}

\begin{align} y & = {x^2 - 1 \over x^2 + 1} \\ & = (x^2 - 1)\left(1 \over x^2 + 1\right) \\ & = (x^2 - 1)(x^2 + 1)^{-1} \end{align}
\begin{align} u & = x^2 - 1 &&& v & = (x^2 + 1)^{-1} \\ {du \over dx} & = 2x &&& {dv \over dx} & = (-1)(x^2 + 1)^{-2} . (2x) \\ & &&& & = -2x (x^2 + 1)^{-2} \\ & &&& & = {-2x \over (x^2 + 1)^2} \end{align} \begin{align} {dy \over dx} & = (x^2 - 1) \left[ {-2x \over (x^2 + 1)^2} \right] + (x^2 + 1)^{-1} (2x) \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + \left(1 \over x^2 + 1\right)(2x) \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + {2x \over x^2 + 1} \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + {2x (x^2 + 1) \over (x^2 + 1)(x^2 + 1)} \\ & = {-2x^3 + 2x \over (x^2 + 1)^2} + {2x^3 + 2x \over (x^2 + 1)^2} \\ & = {-2x^3 + 2x + 2x^3 + 2x \over (x^2 + 1)^2} \\ & = {4x \over (x^2 + 1)^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {4x \over (x^2 + 1)^2} \\ \therefore 0 & = 4x \\ 0 & = x \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = {0^2 - 1 \over 0^2 + 1} \\ & = -1 \end{align}

\begin{align} u & = x &&& v & = \sqrt{9 - x^2} \\ & &&& & = (9 - x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (9 - x^2)^{-{1 \over 2}} . (-2x) \\ & &&& & = -x (9 - x^2)^{-{1 \over 2}} \\ & &&& & = {-x \over \sqrt{9 - x^2}} \end{align} \begin{align} {dy \over dx} & = (x) \left( {-x \over \sqrt{9 - x^2}} \right) + \left( \sqrt{9 - x^2}\right) (1) \\ & = { -x^2 \over \sqrt{9 - x^2}} + \sqrt{9 - x^2} \\ \\ \\ y & = x\phantom{00}\text{--- (1)} \\ \\ y & = x\sqrt{9 - x^2}\phantom{00}\text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x & = x\sqrt{9 - x^2} \\ x - x\sqrt{9 - x^2} & = 0 \\ (x)\left(1 - \sqrt{9 - x^2}\right) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & 1 - \sqrt{9 - x^2} & = 0 \\ & &&& 1 & = \sqrt{9 - x^2} \\ & &&& 1^2 & = 9 - x^2 \\ & &&& 1 & = 9 - x^2 \\ & &&& x^2 & = 9 - 1 \\ & &&& x^2 & = 8 \\ & &&& x & = \pm \sqrt{8} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = { -(0)^2 \over \sqrt{9 - (0)^2} } + \sqrt{9 - (0)^2} &&& {dy \over dx} & = { -(\pm \sqrt{8})^2 \over \sqrt{9 - (\pm \sqrt{8})^2} } + \sqrt{9 - (\pm \sqrt{8})^2} \\ & = 3 &&& & = -7 \end{align}

Recall that parallel lines have the same gradient

\begin{align} 5x - y & = 0 \\ y & = 5x \\ y & = 5x + 0 \\ \\ \text{Comparing with } & y = mx + c, \\ m & = 5 \\ \\ \therefore \text{Gradient of tangent} & = 5 \end{align}
\begin{align} u & = 3x - 1 &&& v & = x - 2 \\ {du \over dx} & = 3 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = (3x - 1)(1) + (x - 2)(3) \\ & = 3x - 1 + 3x - 6 \\ & = 6x - 7 \\ \\ \text{When } & {dy \over dx} = 5, \\ 5 & = 6x - 7 \\ 12 & = 6x \\ {12 \over 6} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = [3(2) - 1](2 - 2) \\ & = 0 \\ \\ \therefore & \phantom{.} (2, 0) \end{align}

Since the curve meets the x-axis at point A, the y-coordinate of A is 0.

\begin{align} u & = x - a &&& v & = \sqrt{x - b} \\ & &&& & = (x - b)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (x - b)^{-{1 \over 2}} . (1) \\ & &&& & = {1 \over 2} (x - b)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x - b}} \end{align} \begin{align} {dy \over dx} & = (x - a) \left(1 \over 2\sqrt{x - b}\right) + \left( \sqrt{x - b} \right) (1) \\ & = {x - a \over 2\sqrt{x - b}} + \sqrt{x - b} \\ & = {x - a \over 2\sqrt{x - b}} + {\sqrt{x - b} \over 1} \\ & = {x - a \over 2\sqrt{x - b}} + {\sqrt{x - b} (2\sqrt{x - b}) \over 2\sqrt{x - b}} \\ & = {x - a \over 2\sqrt{x - b}} + {2(x - b) \over 2\sqrt{x - b}} \\ & = {x - a + 2(x - b) \over 2\sqrt{x - b}} \\ & = {x - a + 2x - 2b \over 2\sqrt{x - b}} \\ & = {3x - a - 2b \over 2\sqrt{x - b}} \\ \\ \\ \text{When } & x = b + 1, \\ {dy \over dx} & = {3(b + 1) - a - 2b \over 2\sqrt{b + 1 - b} } \\ & = {3b + 3 - a - 2b \over 2\sqrt{1}} \\ & = {b + 3 - a \over 2} \\ \\ \text{Substitute } & A(b + 1, 0) \text{ into eqn of curve,} \\ 0 & = (b + 1 - a)\sqrt{b + 1 - b} \\ 0 & = (b + 1 - a) \sqrt{1} \\ 0 & = (b + 1 - a)(1) \\ 0 & = b + 1 - a \\ a & = b + 1 \\ \\ \therefore {dy \over dx} & = {b + 3 - (b + 1) \over 2} \\ & = {b + 3 - b - 1 \over 2} \\ & = {2 \over 2} \\ & = 1 \phantom{0} \text{ (Shown)} \end{align}

(i)

$$ x - a $$

(ii)

\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ f(x) & = (x - a) Q(x) + 0 \\ f(x) & = (x - a) Q(x), \text{ where } Q(x) \text{ denotes a polynomial} \end{align} \begin{align} u & = x - a &&& v & = Q(x) \\ {du \over dx} & = 1 &&& {dv \over dx} & = Q'(x) \end{align} \begin{align} f'(x) & = (x - a)Q'(x) + Q(x) . (1) \\ & = (x - a)Q'(x) + Q(x) \\ \\ f'(a) & = (a - a)Q'(a) + Q(a) \\ & = 0[Q'(a)] + Q(a) \\ & = 0 + Q(a) \\ & = Q(a) \\ \\ \text{If } & f'(a) = 0, \\ Q(a) & = 0 \\ \\ \implies (x - a) & \text{ is a factor of } Q(x) \\ \therefore (x - a) & \text{ is a repeated factor of } f(x) \end{align}