(a)

\begin{align} {d \over dx} \left(4 \over 7x + 1\right) & = {d \over dx} [4 (7x + 1)^{-1} ] \\ & = 4 (-1)(7x + 1)^{-2} . (7) \phantom{000000} [\text{Chain rule}] \\ & = -28 (7x + 1)^{-2} \\ & = -{28 \over (7x + 1)^2} \end{align}

(b)

\begin{align} {d \over dx} \left(2 \over 6 - x^3\right) & = {d \over dx} [2 (6 - x^3)^{-1} ] \\ & = 2 (-1) (6 - x^3)^{-2} . (-3x^2) \phantom{000000} [\text{Chain rule}] \\ & = 6x^2 (6 - x^3)^{-2} \\ & = {6x^2 \over (6 - x^3)^2} \end{align}

(c)

\begin{align} {d \over dx} \left(11 \over 8x^3 + 1\right) & = {d \over dx} [11 (8x^3 + 1)^{-1} ] \\ & = 11 (-1) (8x^3 + 1)^{-2} . (24x^2) \phantom{000000} [\text{Chain rule}] \\ & = - 264 x^2 (8x^3 + 1)^{-2} \\ & = - {264 x^2 \over (8x^3 + 1)^2 } \end{align}

(d)

\begin{align} u & = x + 9 &&& v & = x - 9 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 1 \end{align} \begin{align} {d \over dx} \left(x + 9 \over x - 9\right) & = { (x - 9)(1) - (x + 9)(1) \over (x - 9)^2} \\ & = { x - 9 - x - 9 \over (x - 9)^2 } \\ & = { - 18 \over (x - 9)^2 } \end{align}

(e)

\begin{align} u & = 6x^2 &&& v & = 2 - 5x \\ {du \over dx} & = 12x &&& {dv \over dx} & = -5 \end{align} \begin{align} {d \over dx} \left(6x^2 \over 2 - 5x\right) & = { (2 - 5x)(12x) - (6x^2)(-5) \over (2 - 5x)^2} \\ & = { 24x - 60x^2 + 30x^2 \over (2 - 5x)^2 } \\ & = { 24x - 30x^2 \over (2 - 5x)^2 } \end{align}

(f)

\begin{align} u & = 10x - 1 &&& v & = 3x^2 + 1 \\ {du \over dx} & = 10 &&& {dv \over dx} & = 6x \end{align} \begin{align} {d \over dx} \left(10x - 1 \over 3x^2 + 1\right) & = { (3x^2 + 1)(10) - (10x - 1)(6x) \over (3x^2 + 1)^2} \\ & = { 30x^2 + 10 - 60x^2 + 6x \over (3x^2 + 1)^2 } \\ & = { 10 + 6x - 30x^2 \over (3x^2 + 1)^2 } \end{align}

(g)

\begin{align} u & = 7x &&& v & = \sqrt{4x - 9} \\ & &&& & = (4x - 9)^{1 \over 2} \\ {du \over dx} & = 7 &&& {dv \over dx} & = \underbrace{ {1 \over 2}(4x - 9)^{-{1 \over 2}}. (4) }_\text{Chain rule} \\ & &&& & = 2(4x - 9)^{-{1 \over 2}} \end{align} \begin{align} {d \over dx} \left(7x \over \sqrt{4x - 9} \right) & = { (\sqrt{4x - 9}) (7) - (7x)[2(4x - 9)^{-{1 \over 2}}] \over (\sqrt{4x - 9})^2 } \\ & = { 7 (4x - 9)^{1 \over 2} - 14x (4x - 9)^{-{1 \over 2}} \over 4x - 9 } \times { (4x - 9)^{1 \over 2} \over (4x - 9)^{1 \over 2} } \\ & = { 7 (4x - 9) - 14x (4x - 9)^0 \over (4x - 9)^{3 \over 2} } \\ & = { 28x - 63 - 14x(1) \over \sqrt{ (4x - 9)^3 } } \\ & = { 28x - 63 - 14x \over \sqrt{ (4x - 9)^3 } } \\ & = { 14x - 63 \over \sqrt{ (4x - 9)^3 } } \end{align}

(h)

\begin{align} u & = 8x^2 &&& v & = \sqrt{x} - 4 \\ & &&& & = x^{1 \over 2} - 4 \\ {du \over dx} & = 16x &&& {dv \over dx} & = {1 \over 2}x^{-{1 \over 2}} \end{align} \begin{align} {d \over dx} \left( 8x^2 \over \sqrt{x} - 4 \right) & = { (\sqrt{x} - 4)(16x) - (8x^2)\left({1 \over 2} x^{-{1 \over 2}} \right) \over (\sqrt{x} - 4)^2 } \\ & = { 16 x \sqrt{x} - 64x - 4 x^{3 \over 2} \over (\sqrt{x} - 4)^2 } \\ & = { 16 x \sqrt{x} - 64x - 4 x (x^{1 \over 2}) \over (\sqrt{x} - 4)^2 } \\ & = { 16 x \sqrt{x} - 64x - 4 x \sqrt{x} \over (\sqrt{x} - 4)^2 } \\ & = { 12 x \sqrt{x} - 64x \over (\sqrt{x} - 4)^2 } \end{align}

(i)

\begin{align} u & = \sqrt{6 - x} &&& v & = \sqrt{x^2 + 36} \\ & = (6 - x)^{1 \over 2} &&& & = (x^2 + 36)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(6 - x)^{-{1 \over 2}}. (-1) &&& {dv \over dx} & = {1 \over 2} (x^2 + 36)^{-{1 \over 2}}. (2x) \phantom{000000} [\text{Chain rule}] \\ & = -{1 \over 2} (6 - x)^{-{1 \over 2}} &&& & = x (x^2 + 36)^{-{1 \over 2}} \end{align} \begin{align} {d \over dx} \left( \sqrt{6 - x \over x^2 + 36} \right) & = {d \over dx} \left( \sqrt{6 - x} \over \sqrt{x^2 + 36} \right) \\ & = { (\sqrt{x^2 + 36}) \left[ -{1 \over 2} (6 - x)^{-{1 \over 2}} \right] - (\sqrt{6 - x}) \left[x (x^2 + 36)^{-{1 \over 2}}\right] \over (\sqrt{x^2 + 36})^2 } \\ & = { - {1 \over 2} (x^2 + 36)^{1 \over 2} (6 - x)^{-{1 \over 2}} - x (6 - x)^{1 \over 2} (x^2 + 36)^{-{1 \over 2}} \over x^2 + 36 } \times { (x^2 + 36)^{1 \over 2} (6 - x)^{1 \over 2} \over (x^2 + 36)^{1 \over 2} (6 - x)^{1 \over 2}} \\ & = { - {1 \over 2} (x^2 + 36) - x (6 - x) \over (x^2 + 36)^{3 \over 2} (6 - x)^{1 \over 2} } \\ & = { - {1 \over 2}x^2 - 18 - 6x + x^2 \over \sqrt{ (x^2 + 36)^3 } \sqrt{6 - x} } \\ & = { {1 \over 2}x^2 - 6x - 18 \over \sqrt{ (x^2 + 36)^3 (6 - x) } } \times {2 \over 2} \\ & = { x^2 - 12x - 36 \over 2 \sqrt{ (x^2 + 36)^3 (6 - x)} } \end{align}

(j)

\begin{align} u & = 5 \sqrt{x} &&& v & = \sqrt{9 - x^2} \\ & = 5 x^{1 \over 2} &&& & = (9 - x^2)^{1 \over 2} \\ {du \over dx} & = 5 \left(1 \over 2\right) x^{-{1 \over 2}} &&& {dv \over dx} & = \underbrace{ {1 \over 2} (9 - x^2)^{-{1 \over 2}} . (-2x)}_\text{Chain rule} \\ & = {5 \over 2} x^{-{1 \over 2}} &&& & = - x (9 - x^2)^{-{1 \over 2}} \end{align} \begin{align} {d \over dx} \left( 5 \sqrt{x} \over \sqrt{9 - x^2} \right) & = { (\sqrt{9 - x^2}) \left( {5 \over 2} x^{-{1 \over 2}} \right) - (5 \sqrt{x}) \left[ - x (9 - x^2)^{-{1 \over 2}} \right] \over ( \sqrt{9 - x^2})^2 } \\ & = { {5 \over 2} x^{-{1 \over 2}} (9 - x^2)^{1 \over 2} + 5 x^{3 \over 2} (9 - x^2)^{-{1 \over 2}} \over 9 - x^2 } \times { x^{1 \over 2} (9 - x^2)^{1 \over 2} \over x^{1 \over 2} (9 - x^2)^{1 \over 2} } \\ & = { {5 \over 2} (9 - x^2) + 5x^2 \over x^{1 \over 2} (9 - x^2)^{3 \over 2} } \\ & = { {45 \over 2} - {5 \over 2}x^2 + 5x^2 \over \sqrt{x} \sqrt{ (9 - x^2)^3 } } \\ & = { {45 \over 2} + {5 \over 2}x^2 \over \sqrt{ x(9 - x^2)^3} } \times {2 \over 2} \\ & = { 45 + 5x^2 \over 2 \sqrt{ x(9 - x^2)^3} } \end{align}

(i)

\begin{align} u & = 5x^2 &&& v & = 4x - 1 \\ {du \over dx} & = 10x &&& {dv \over dx} & = 4 \end{align} \begin{align} {d \over dx} \left( 5x^2 \over 4x - 1 \right) & = { (4x - 1)(10x) - (5x^2)(4) \over (4x - 1)^2 } \\ & = { 40x^2 - 10x - 20x^2 \over (4x - 1)^2 } \\ & = { 20x^2 - 10x \over (4x - 1)^2 } \\ & = { 10x (2x - 1) \over (4x - 1)^2 } \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} u & = 10x(2x - 1) &&& v & = (4x - 1)^2 \\ & = 20x^2 - 10x \\ {du \over dx} & = 40x - 10 &&& {dv \over dx} & = \underbrace{2(4x - 1).(4)}_\text{Chain rule} \\ & &&& & = 8(4x - 1) \end{align} \begin{align} {dz \over dx} & = { (4x - 1)^2 (40x - 10) - 10x(2x - 1)[ 8(4x - 1)] \over [ (4x - 1)^2 ]^2 } \\ & = { (4x - 1)^2 (40x - 10) - 80x (2x - 1) (4x - 1) \over (4x - 1)^4 } \\ & = { 10(4x - 1)^2 (4x - 1) - 80x (2x - 1)(4x - 1) \over (4x - 1)^4 } \\ & = { 10 (4x - 1) [ (4x - 1)^2 - 8x(2x - 1)] \over (4x - 1)^4 } \\ & = { 10 (4x - 1) [ (4x)^2 - 2(4x)(1) + 1^2 - 16x^2 + 8x ] \over (4x - 1)^4 } \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2 ] \\ & = { 10 (4x - 1) ( 16x^2 - 8x + 1 - 16x^2 + 8x) \over (4x - 1)^4 } \\ & = { 10 (4x - 1) (1) \over (4x - 1)^4 } \\ & = { 10 \over (4x - 1)^3} \end{align}

(i)

\begin{align} u & = 5x - 1 &&& v & = x^3 \\ {du \over dx} & = 5 &&& {dv \over dx} & = 3x^2 \end{align} \begin{align} {d \over dx} \left( 5x - 1 \over x^3 \right) & = { (x^3)(5) - (5x - 1)(3x^2) \over (x^3)^2 } \\ & = { 5x^3 - 15x^3 + 3x^2 \over x^6 } \\ & = { 3x^2 - 10x^3 \over x^6} \\ & = { x^2 ( 3 - 10x ) \over x^6 } \\ & = { 3 - 10x \over x^4 } \end{align}

(ii)

\begin{align} {d \over dx} \left( 5x - 1 \over x^3 \right) & = {d \over dx} \left( {5x \over x^3} - {1 \over x^3} \right) \\ & = {d \over dx} \left( {5 \over x^2} - x^{-3} \right) \\ & = {d \over dx} ( 5x^{-2} - x^{-3} ) \\ & = 5(-2)x^{-3} - (-3) x^{-4} \\ & = -10 x^{-3} + 3 x^{-4} \\ & = {-10 \over x^3} + {3 \over x^4} \\ & = {-10x \over x^4} + {3 \over x^4} \\ & = { -10x + 3 \over x^4} \\ \\ \text{Prefer method } & \text{used in (ii), since it's shorter} \end{align}

(i)

\begin{align} u & = x - 8 &&& v & = 10 - x \\ {du \over dx} & = 1 &&& {dv \over dx} & = -1 \end{align} \begin{align} {dy \over dx} & = { (10 - x)(1) - (x - 8)(-1) \over (10 - x)^2 } \\ & = { 10 - x + x - 8 \over (10 - x)^2 } \\ & = { 2 \over (10 - x)^2 } \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = {x - 8 \over 10 - x} \\ 0 & = x - 8 \\ 8 & = x \\ \\ \text{Substitute } & x = 8 \text{ into eqn of curve,} \\ {dy \over dx} & = {2 \over (10 - 8)^2 } \\ & = {1 \over 2} \end{align}

(ii)

\begin{align} \text{Substitute } & x = 0 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {2 \over (10 - 0)^2} \\ & = {1 \over 50} \end{align}

(i)

\begin{align} u & = x &&& v & = \sqrt{6x - 7} \\ & &&& & = (6x - 7)^{1 \over 2} \\ \\ {du \over dx} & = 1 &&& {dv \over dx} & = \underbrace{ {1 \over 2} (6x - 7)^{-{1 \over 2}} . (6) }_\text{Chain rule} \\ & &&& & = 3 (6x - 7)^{-{1 \over 2}} \end{align} \begin{align} {dy \over dx} & = { (\sqrt{6x - 7})(1) - (x)[3 (6x - 7)^{-{1 \over 2}}] \over (\sqrt{6x - 7})^2 } \\ & = { (6x - 7)^{1 \over 2} - 3x (6x - 7)^{-{1 \over 2}} \over 6x - 7} \times { (6x - 7)^{1 \over 2} \over (6x - 7)^{1 \over 2} } \\ & = { 6x - 7 - 3x (6x - 7)^0 \over (6x - 7)^{3 \over 2} } \\ & = { 6x - 7 - 3x (1) \over \sqrt{ (6x - 7)^3 } } \\ & = { 3x - 7 \over \sqrt{ (6x - 7)^3 } } \\ \\ \therefore a & = 3, b = -7, n = 3 \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {3x - 7 \over \sqrt{ (6x - 7)^3 } } \\ 0 & = 3x - 7 \\ 7 & = 3x \\ {7 \over 3} & = x \\ \\ \text{Substitute } & x = {7 \over 3} \text{ into eqn of curve, } \\ y & = {{7 \over 3} \over \sqrt{ 6 \left(7 \over 3\right) - 7 } } \\ & = { {7 \over 3} \over \sqrt{7 } } \\ & = {7 \over 3} \div \sqrt{7} \\ & = {7 \over 3} \times {1 \over \sqrt{7}} \\ & = {7 \over 3 \sqrt{7}} \times {\sqrt{7} \over \sqrt{7}} \\ & = {7 \sqrt{7} \over 3(7) } \\ & = {\sqrt{7} \over 3} \\ \\ \therefore & \phantom{.} \left( {7 \over 3}, {\sqrt{7} \over 3} \right) \end{align}

\begin{align} y & = \sqrt[3] {x - a \over x + b} \\ & = \left(x - a \over x + b\right)^{1 \over 3} \\ & = { (x - a)^{1 \over 3} \over (x + b)^{1 \over 3} } \end{align}
\begin{align} u & = (x - a)^{1 \over 3} &&& v & = (x + b)^{1 \over 3} \\ {du \over dx} & = {1 \over 3} (x - a)^{-{2 \over 3}} . (1) &&& {dv \over dx} & = {1 \over 3} (x + b)^{-{2 \over 3}}. (1) \phantom{000000} [\text{Chain rule}] \\ & = {1 \over 3} (x - a)^{-{2 \over 3}} &&& & = {1 \over 3} (x + b)^{-{2 \over 3}} \end{align} \begin{align} {dy \over dx} & = { (x + b)^{1 \over 3} \left[ {1 \over 3} (x - a)^{-{2 \over 3}} \right] - (x - a)^{1 \over 3} \left[ {1 \over 3} (x + b)^{-{2 \over 3}} \right] \over [(x + b)^{1 \over 3}]^2 } \\ & = { {1 \over 3} (x + b)^{1 \over 3} (x - a)^{-{2 \over 3}} - {1 \over 3} (x - a)^{1 \over 3} (x + b)^{-{2 \over 3}} \over (x + b)^{2 \over 3} } \times {3 \over 3} \\ & = { (x + b)^{1 \over 3} (x - a)^{-{2 \over 3}} - (x - a)^{1 \over 3} (x + b)^{-{2 \over 3}} \over 3(x + b)^{2 \over 3} } \end{align}
\begin{align} \require{cancel} 3 y^2 (x + b)^2 {dy \over dx} - a - b & = 3 \left[ (x - a)^{1 \over 3} \over (x + b)^{1 \over 3} \right]^2 (x + b)^2 \left[ { (x + b)^{1 \over 3} (x - a)^{-{2 \over 3}} - (x - a)^{1 \over 3} (x + b)^{-{2 \over 3}} \over 3(x + b)^{2 \over 3} } \right] - a - b \\ & = {3 \over 1} \left[ (x - a)^{2 \over 3} \over (x + b)^{2 \over 3} \right] \left[ (x + b)^2 \over 1\right] \left[ { (x + b)^{1 \over 3} (x - a)^{-{2 \over 3}} - (x - a)^{1 \over 3} (x + b)^{-{2 \over 3}} \over 3(x + b)^{2 \over 3} } \right] - a - b \\ & = { 3 (x - a)^{2 \over 3} \over (x + b)^{2 \over 3} } \left[ (x + b)^2 \over 1\right] \left[ { (x + b)^{1 \over 3} (x - a)^{-{2 \over 3}} - (x - a)^{1 \over 3} (x + b)^{-{2 \over 3}} \over 3(x + b)^{2 \over 3} } \right] - a - b \\ & = { 3(x - a)^{2 \over 3} (x + b)^2 \over (x + b)^{2 \over 3} } \left[ { (x + b)^{1 \over 3} (x - a)^{-{2 \over 3}} - (x - a)^{1 \over 3} (x + b)^{-{2 \over 3}} \over 3(x + b)^{2 \over 3} } \right] - a - b \\ & = { 3 (x - a)^0 (x + b)^{7 \over 3} - 3(x - a)(x + b)^{4 \over 3} \over 3 (x + b)^{4 \over 3}} - a - b \\ & = { 3 (x + b)^{7 \over 3} - 3(x - a)(x + b)^{4 \over 3} \over 3 (x + b)^{4 \over 3} } - a - b \\ & = { \cancel{3 (x + b)^{4 \over 3}} [ (x + b) - (x - a) ] \over \cancel{3 (x + b)^{4 \over 3} }} - a - b \\ & = [ (x + b) - (x - a) ] - a - b \\ & = x + b - x + a - a - b \\ & = 0 \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} u & = ax &&& v & = x^2 + b \\ {du \over dx} & = a &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = { (x^2 + b)(a) - (ax)(2x) \over (x^2 + b)^2 } \\ & = { ax^2 + ab - 2a x^2 \over (x^2 + b)^2} \\ & = { ab - a x^2 \over (x^2 + b)^2 } \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {ab - a x^2 \over (x^2 + b)^2 } \\ 0 & = ab - ax^2 \\ \\ \text{Perfect squares} & = 1^2, 2^2, 3^2, 4^2, ... \\ & = 1, 4, 9, 16, ... \\ \\ \therefore a & = 1, b = 4 \end{align}

(ii)

\begin{align} 0 & = ab - ax^2 \\ 0 & = (1)(4) - (1)x^2 \\ 0 & = 4 - x^2 \\ x^2 & = 4 \\ x & = \pm \sqrt{4} \\ x & = 2 \text{ (Reject, since } x < 0 \text{ or } - 2 \\ \\ \text{Substitute } & x = -2 \text{ into eqn of curve,} \\ y & = {a(-2) \over (-2)^2 + b} \\ & = {-2a \over 4 + b} \\ & = {-2(1) \over 4 + 4} \\ & = -{1 \over 4} \\ \\ \therefore & \phantom{.} \left(-2, -{1 \over 4}\right) \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 - 21x + 20 \\ \\ f(1) & = (1)^3 - 21(1) + 20 \\ & = 0 \\ \\ \therefore x - 1 & \text{ is a factor of } f(x) \end{align} $$ \require{enclose} \begin{array}{rll} x^2 + x - 20 \phantom{00000000}\\ x - 1 \enclose{longdiv}{ x^3 + 0x^2 - 21x + 20 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - x^2){\phantom{00000000000}}} \\ x^2 - 21x + 20 \phantom{0} \\ -\underline{( x^2 - x ){\phantom{000000.}}} \\ -20x + 20 \phantom{0} \\ -\underline{( -20x + 20){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} f(x) & = (x - 1)(x^2 + x - 20) \\ & = (x - 1)(x - 4)(x + 5) \end{align}

(ii)

$$ \require{enclose} \begin{array}{rll} 1 \phantom{00000000000000000.}\\ x^3 + 0x^2 - 21x + 20 \enclose{longdiv}{ x^3 - 12x^2 + 12x - 55 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 + 0x^2 - 21x + 20){\phantom{0}}} \\ - 12x^2 + 33x - 75 \phantom{0} \end{array} $$
\begin{align} {x^3 - 12x^2 + 12x - 55 \over x^3 - 21x + 20} & = 1 + {-12x^2 + 33x - 75 \over x^3 - 21x + 20} \\ & = 1 + {-12x^2 + 33x - 75 \over \underbrace{ (x - 1)(x - 4)(x + 5)}_\text{From (i)} } \\ \\ {-12x^2 + 33x - 75 \over (x - 1)(x - 4)(x + 5)} & = {A \over x - 1} + {B \over x - 4} + {C \over x + 5} \\ & = {A(x - 4)(x + 5) \over (x - 1)(x - 4)(x + 5)} + {B(x - 1)(x + 5) \over (x - 1)(x - 4)(x + 5)} + {C(x - 1)(x - 4) \over (x - 1)(x - 4)(x + 5)} \\ & = {A(x - 4)(x + 5) + B(x - 1)(x + 5) + C(x - 1)(x - 4) \over (x - 1)(x - 4)(x + 5)} \\ \\ -12x^2 + 33x - 75 & = A(x - 4)(x + 5) + B(x - 1)(x + 5) + C(x - 1)(x - 4) \\ \\ \text{Let } & x = 4, \\ -12(4)^2 + 33(4) - 75 & = A(0)(9) + B(3)(9) + C(3)(0) \\ -135 & = 27B \\ {-135 \over 27} & = B \\ -5 & = B \\ \\ -12x^2 + 33x - 75 & = A(x - 4)(x + 5) -5(x - 1)(x + 5) + C(x - 1)(x - 4) \\ \\ \text{Let } & x = -5, \\ -12(-5)^2 + 33(-5) - 75 & = A(-9)(0) - 5(-6)(0) + C(-6)(-9) \\ -540 & = 54C \\ {-540 \over 54} & = C \\ -10 & = C \\ \\ -12x^2 + 33x - 75 & = A(x - 4)(x + 5) -5(x - 1)(x + 5) - 10(x - 1)(x - 4) \\ \\ \text{Let } & x = 0, \\ 0 + 0 - 75 & = A(-4)(5) - 5(-1)(5) - 10(-1)(-4) \\ -75 & = -20A + 25 - 40 \\ 20A & = 25 - 40 + 75 \\ 20A & = 60 \\ A & = {60 \over 20} \\ A & = 3 \\ \\ {-12x^2 + 33x - 75 \over (x - 1)(x - 4)(x + 5)} & = {3 \over x - 1} + {-5 \over x - 4} + {-10 \over x + 5} \\ \\ \therefore {x^3 - 12x^2 + 12x - 55 \over x^3 - 21x + 20} & = 1 + {3 \over x - 1} - {5 \over x - 4} - {10 \over x + 5} \end{align}

(iii)

\begin{align} {d \over dx} \left( {x^3 - 12x^2 + 12x - 55 \over x^3 - 21x + 20} \right) & = {d \over dx} \left( 1 + {3 \over x - 1} - {5 \over x - 4} - {10 \over x + 5} \right) \\ & = {d \over dx} \left[ 1 + 3(x - 1)^{-1} - 5(x - 4)^{-1} - 10 (x + 5)^{-1} \right] \\ & = 0 + 3(-1)(x - 1)^{-2}.(1) - 5(-1)(x - 4)^{-2}.(1) - 10(-1)(x + 5)^{-2}.(1) \phantom{00000} [\text{Chain rule}] \\ & = - 3(x - 1)^{-2} + 5 (x - 4)^{-2} + 10(x + 5)^{-2} \\ & = - {3 \over (x - 1)^2} + {5 \over (x - 4)^2} + {10 \over (x + 5)^2} \end{align}

(i)

\begin{align} y & = 8x^3 - 36x^2 + 22x + 21 \\ \\ \text{Let } & x = 0, \\ y & = 8(0)^3 - 36(0)^2 + 22(0) + 21 \\ y & = 21 \\ \\ \text{Let } & y = 0, \\ 0 & = 8x^3 - 36x^2 + 22x + 21 \\ \\ \text{Let } f(x) & = 8x^3 - 36x^2 + 22x + 21 \\ \\ f(-0.5) & = 8(-0.5)^3 - 36(-0.5)^2 + 22(-0.5) + 21 \\ & = 0 \\ \\ \therefore 2x + 1 & \text{ is a factor of } f(x) \end{align} $$ \require{enclose} \begin{array}{rll} 4x^2 - 20x + 21 \phantom{0000000}\\ 2x + 1 \enclose{longdiv}{ 8x^3 - 36x^2 + 22x + 21 \phantom{0}}\kern-.2ex \\ -\underline{( 8x^3 + 4x^2){\phantom{00000000000}}} \\ -40x^2 + 22x + 21 \phantom{0} \\ -\underline{( -40x^2 - 20x){\phantom{0000.}}} \\ 42x + 21 \phantom{0} \\ -\underline{( 42x + 21){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} f(x) & = (2x + 1)(4x^2 - 20x + 21) \\ & = (2x + 1)(2x - 3)(2x - 7) \\ \\ 0 & = (2x + 1)(2x - 3)(2x - 7) \end{align} \begin{align} 2x + 1 & = 0 && \text{ or } & 2x - 3 & = 0 && \text{ or } & 2x - 7 & = 0 \\ 2x & = -1 &&& 2x & = 3 &&& 2x & = 7 \\ x & = -0.5 &&& x & = 1.5 &&& x & = 3.5 \end{align}

(ii)

\begin{align} y & = 8x^3 - 36x^2 + 22x + 21 \\ \\ {dy \over dx} & = 8(3)x^2 - 36(2)x + 22 \\ & = 24x^2 - 72x + 22 \\ & = 24(x^2 - 3x) + 22 \\ & = 24 \left[ x^2 - 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 \right] + 22 \phantom{000000} [\text{Complete the square}] \\ & = 24 \left[ (x - 1.5)^2 - 2.25 \right] + 22 \\ & = 24(x - 1.5)^2 - 54 + 22 \\ & = 24(x - 1.5)^2 - 32 \\ \\ \text{For } &\text{all real values of } x, \\ (x - 1.5)^2 & \ge 0 \\ 24(x - 1.5)^2 & \ge 0 \\ 24(x - 1.5)^2 - 32 & \ge - 32 \\ \\ \therefore \text{Gradient of } & \text{curve, } {dy \over dx}, \text{ cannot be less than } -32 \end{align}