(a)

\begin{align} {d \over dx} (5x - 7)^8 & = 8(5x - 7)^7 (5) \\ & = 40(5x - 7)^7 \end{align}

(b)

\begin{align} {d \over dx} (10 + 3x)^6 & = 6(10 + 3x)^5 (3) \\ & = 18 (10 + 3x)^5 \end{align}

(c)

\begin{align} {d \over dx} \left(10 - {1 \over 2}x\right)^4 & = 4 \left(10 - {1 \over 2}x\right)^3 \left(-{1 \over 2}\right) \\ & = -2 \left(10 - {1 \over 2}x\right)^3 \end{align}

(d)

\begin{align} {d \over dx} [ 8(2x + 9)^6] & = 8(6)(2x + 9)^5 (2) \\ & = 96(2x + 9)^5 \end{align}

(e)

\begin{align} {d \over dx} \left({1 \over 4x - 11}\right) & = {d \over dx} (4x - 11)^{-1} \\ & = (-1)(4x - 11)^{-2} (4) \\ & = -4 (4x - 11)^{-2} \\ & = - \left(4 \over 1\right)\left[1 \over (4x - 11)^2\right] \\ & = - {4 \over (4x - 11)^2} \end{align}

(f)

\begin{align} {d \over dx} \left[ 8 \over (6x^2 - 1)^3 \right] & = {d \over dx} [ 8 (6x^2 - 1)^{-3} ] \\ & = 8(-3)(6x^2 - 1)^{-4} (12x) \\ & = -288x (6x^2 - 1)^{-4} \\ & = - \left(288x \over 1\right) \left[ 1 \over (6x^2 - 1)^4 \right] \\ & = - {288x \over (6x^2 - 1)^4} \end{align}

(g)

\begin{align} {d \over dx} \sqrt{5x^3 - 6x + 2} & = {d \over dx} (5x^3 - 6x + 2)^{1 \over 2} \\ & = {1 \over 2} (5x^3 - 6x + 2)^{-{1 \over 2}} . (15x^2 - 6) \\ & = {1 \over 2} \left(1 \over \sqrt{5x^3 - 6x + 2}\right) \left(15x^2 - 6 \over 1\right) \\ & = {15x^2 - 6 \over 2 \sqrt{5x^3 - 6x + 2} } \end{align}

(h)

\begin{align} {d \over dx} \left(7 \over \sqrt[3]{x + 9} \right) & = {d \over dx} \left[ 7 (x + 9)^{-{1 \over 3}} \right] \\ & = 7 \left(-{1 \over 3}\right) (x + 9)^{-{4 \over 3}} (1) \\ & = -{7 \over 3} (x + 9)^{-{4 \over 3}} \\ & = -{7 \over 3} \left(1 \over \sqrt[3]{(x + 9)^4}\right) \\ & = - {7 \over 3 \sqrt[3]{(x + 9)^4}} \end{align}

(i)

\begin{align} {d \over dx} \left( {4 \over x} - \sqrt[3]{x^2} \right)^6 & = {d \over dx} \left( 4x^{-1} - x^{2 \over 3} \right)^6 \\ & = 6 \left( 4x^{-1} - x^{2 \over 3} \right)^5 \left( -4 x^{-2} - {2 \over 3} x^{-{1 \over 3}} \right) \\ & = -6 \left( 4x^{-1} - x^{2 \over 3} \right)^5 \left( 4 x^{-2} + {2 \over 3} x^{-{1 \over 3}} \right) \\ & = -6 \left( {4 \over x} - \sqrt[3]{x^2} \right)^5 \left( {4 \over x^2} + {2 \over 3 \sqrt[3]{x}} \right) \end{align}

(j)

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

(a)

\begin{align} {dy \over dx} & = 6(x^2 + 4x - 7)^5 (2x + 4) \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 6[1^2 + 4(1) - 7]^5 [2(1) + 4] \\ & = -1152 \end{align}

(b)

\begin{align} y & = {5 \over \sqrt{4 - 2x}} \\ y & = 5 (4 - 2x)^{-{1 \over 2}} \\ \\ {dy \over dx} & = 5 \left(-{1 \over 2}\right) (4 - 2x)^{-{3 \over 2}} . (-2) \\ & = -{5 \over 2}(-2)(4- 2x)^{-{3 \over 2}} \\ & = 5 (4 - 2x)^{-{3 \over 2}} \\ & = {5 \over \sqrt{(4 - 2x)^3}} \phantom{000000} [\text{Need to find value of } x \text{ to sub in}] \\ \\ \text{Substitute } & y = 2.5 \text{ into eqn of curve,} \\ 2.5 & = {5 \over \sqrt{4 - 2x}} \\ {5 \over 2} & = {5 \over \sqrt{4 - 2x}} \\ 5 \sqrt{4 - 2x} & = 2(5) \\ \sqrt{4 - 2x} & = 2 \\ 4 - 2x & = 2^2 \\ 4 - 2x & = 4 \\ -2x & = 0 \\ x & = 0 \\ \\ \text{Substitute } & x = 0 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {5 \over \sqrt{[4 - 2(0)]^3}} \\ & = {5 \over 8} \end{align}

(c)

\begin{align} y & = 3x - 1 - {1 \over 3x - 1} \\ y & = 3x - 1 - (3x - 1)^{-1} \\ \\ {dy \over dx} & = 3 - 0 - (-1)(3x - 1)^{-2} . (3) \\ & = 3 + 3(3x - 1)^{-2} \\ & = 3 + {3 \over (3x - 1)^2} \\ \\ \text{When } & x = 0, \phantom{000000} [y \text{-intercept}] \\ {dy \over dx} & = 3 + {3 \over [3(0) - 1]^2} \\ & = 6 \end{align}

(d)

\begin{align} y & = \sqrt[3]{8x^3 + 1} \\ y & = (8x^3 + 1)^{1 \over 3} \\ \\ {dy \over dx} & = {1 \over 3}(8x^3 + 1)^{-{2 \over 3}} . (24x^2) \\ & = 8x^2 (8x^3 + 1)^{-{2 \over 3}} \\ & = {8x^2 \over \sqrt[3]{(8x^3 + 1)^2}} \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = \sqrt[3]{8x^3 + 1} \\ 0^3 & = 8x^3 + 1 \\ 0 & = 8x^3 + 1 \\ -8x^3 & = 1 \\ x^3 & = -{1 \over 8} \\ x & = \sqrt[3]{-{1 \over 8}} \\ x & = -{1 \over 2} \\ \\ \text{Substitute } & x = -{1 \over 2} \text{ into } {dy \over dx}, \\ {dy \over dx} & = {8 \left(-{1 \over 2}\right)^2 \over \sqrt[3]{ \left[ 8 \left(-{1 \over 2}\right)^3 + 1 \right]^2} } \\ & = {2 \over \sqrt[3]{ 0 } } \\ & = {2 \over 0} \\ \\ \therefore \text{Gradient } & \text{of tangent is undefined} \end{align}

\begin{align} a^3 + b^3 & = (a + b)(a^2 - ab + b^2) \phantom{000000} [\text{from chapter 4.3}] \\ \\ y & = {x + 1 \over x^3 + 1} \\ y & = {x + 1 \over x^3 + 1^3} \\ y & = {x + 1 \over (x + 1)[x^2 - (x)(1) + 1^2]} \\ y & = {x + 1 \over (x + 1)(x^2 - x + 1)} \\ y & = {1 \over x^2 - x + 1} \\ y & = (x^2 - x + 1)^{-1} \\ \\ {dy \over dx} & = (-1)(x^2 - x + 1)^{-2} . (2x - 1) \\ & = -(2x - 1)(x^2 - x + 1)^{-2} \\ & = - {2x -1 \over (x^2 - x + 1)^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = - {2(1) - 1 \over [ (1)^2 - 1 + 1]^2 } \\ & = -1 \end{align}

\begin{align} f(x) & = (x^p + q)^n \\ \\ f'(x) & = n (x^p + q)^{n - 1} . (p x^{p - 1}) \\ & = n p x^{p - 1} (x^p + q)^{n - 1} \end{align}

\begin{align} y & = {2 \over 3} (ax - 1)^n \\ \\ \text{Using } & \left({1 \over 3}, {2 \over 3}\right), \\ {2 \over 3} & = {2 \over 3} \left( {1 \over 3}a - 1 \right)^n \\ { {2 \over 3} \over {2 \over 3} } & = \left( {1 \over 3}a - 1 \right)^n \\ 1 & = \left( {1 \over 3}a - 1 \right)^n \\ \sqrt[n]{1} & = {1 \over 3}a - 1 \\ 1 & = {1 \over 3}a - 1 \\ -{1 \over 3}a & = -2 \\ {1 \over 3}a & = 2 \\ a & = 6 \\ \\ {dy \over dx} & = {2 \over 3} n (ax - 1)^{n - 1} . (a) \\ & = {2 \over 3} an (ax - 1)^{n - 1} \\ & = {2 \over 3} (6)n (6x - 1)^{n - 1} \\ & = 4n (6x - 1)^{n - 1} \\ \\ \text{When } & x = {1 \over 3} \text{ and } {dy \over dx} = 20, \\ 20 & = 4n \left[ 6 \left(1 \over 3\right) - 1 \right]^{n - 1} \\ 20 & = 4n (1)^{n - 1} \\ 20 & = 4n (1) \\ 20 & = 4n \\ {20 \over 4} & = n \\ 5 & = n \\ \\ \therefore a & = 6, n = 5 \end{align}

\begin{align} y & = \sqrt[3]{4x^3 + 9x^2 - 12x} \\ y & = (4x^3 + 9x^2 - 12x)^{1 \over 3} \\ \\ {dy \over dx} & = {1 \over 3} (4x^3 + 9x^2 - 12x)^{-{2 \over 3}}. (12x^2 + 18x - 12) \\ & = {1 \over 3} (12x^2 + 18x - 12) (4x^3 + 9x^2 - 12x)^{-{2 \over 3}} \\ & = (4x^2 + 6x - 4)(4x^3 + 9x^2 - 12x)^{-{2 \over 3}} \\ & = {4x^2 + 6x - 4 \over \sqrt[3]{(4x^3 + 9x^2 - 12x)^2}} \\ \\ x = 7 \text{ is} & \text{ a vertical line} \implies \text{tangents are horizontal and gradient} = 0 \\ \\ \text{When } {dy \over dx} & = 0, \\ 0 & = {4x^2 + 6x - 4 \over \sqrt[3]{(4x^3 + 9x^2 - 12x)^2}} \\ 0 & = 4x^2 + 6x - 4 \\ 0 & = 2x^3 + 3x - 2 \\ 0 & = (2x - 1)(x + 2) \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & x + 2 & =0 \\ 2x & = 1 &&& x & = -2 \\ x & = {1 \over 2} \end{align}

(a)

\begin{align} y & = - (2 \sqrt{x} + 1)^4 \\ y & = - (2 x^{1 \over 2} + 1)^4 \\ \\ {dy \over dx} & = - (4) (2 x^{1 \over 2} + 1)^3 . \left[ 2 \left(1 \over 2\right) x^{-{1 \over 2}} \right] \\ & = - 4 (2 x^{1 \over 2} + 1)^3 (x^{-{1 \over 2}}) \\ & = - {4 (2 \sqrt{x} + 1)^3 \over \sqrt{x} } \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = - {4 [ 2 \sqrt{1} + 1 ]^3 \over \sqrt{1}} \\ & = -108 \\ \\ \therefore m & = -108 \\ \\ y & = mx + c \\ y & = -108x + c \\ \\ \text{Using } & (0, 3), \\ 3 & = -108(0) + c \\ 3 & = c \\ \\ \text{Possible eqn: } & y = -108x + 3 \end{align}

(b)

\begin{align} \text{Let } y & = ax^2 + b, \text{ where } a \text{ and } b \text{ are constants} \\ \\ {dy \over dx} & = 2ax \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = -108, \\ -108 & = 2a(1) \\ -108 & = 2a \\ {-108 \over 2} & = a \\ -54 & = a \\ \\ y & = -54x^2 + b \\ \\ \text{Let } & b = -1, \\ y & = -54x^2 - 1 \end{align}

(i)

\begin{align} \text{Let } f(x) & = 2x^3 + 21x^2 + 60x + 25 \\ \\ f(-5) & = 2(-5)^3 + 21(-5)^2 + 60(-5) + 25 \\ & = 0 \\ \\ \therefore x + 5 & \text{ is a factor of } f(x) \end{align}
$$ \require{enclose} \begin{array}{rll} 2x^2 + 11x + 5 \phantom{00000000}\\ x + 5 \enclose{longdiv}{ 2x^3 + 21x^2 + 60x + 25 \phantom{0}}\kern-.2ex \\ -\underline{(2x^3 + 10x^2){\phantom{0000000000}}} \\ 11x^2 + 60x + 25 \phantom{0} \\ -\underline{( 11x^2 + 55x){\phantom{0000.}}} \\ 5x + 25 \phantom{0} \\ -\underline{(5x + 25){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} 2x^3 + 21x^2 + 60x + 25 & = (x + 5)(2x^2 + 11x + 5) \\ & = (x + 5)(x + 5)(2x + 1) \\ & = (x + 5)^2 (2x + 1) \end{align}

(ii)

\begin{align} {10x^2 + 62x + 69 \over 2x^3 + 21x^2 + 60x + 25} & = {10x^2 + 62x + 69 \over (x + 5)^2 (2x + 1)} \\ & = {A \over x + 5} + {B \over (x + 5)^2} + {C \over 2x + 1} \\ & = {A(x + 5)(2x + 1) \over (x + 5)^2 (2x + 1)} + {B(2x + 1) \over (x + 5)^2 (2x + 1)} + { C(x + 5)^2 \over (x + 5)^2 (2x + 1)} \\ & = {A(x + 5)(2x + 1) + B(2x + 1) + C(x + 5)^2 \over (x + 5)^2 (2x + 1)} \\ \\ 10x^2 + 62x + 69 & = A(x + 5)(2x + 1) + B(2x + 1) + C(x + 5)^2 \\ \\ \text{Let } & x = -5, \\ 10(-5)^2 + 62(-5) + 69 & = 0 + B[2(-5) + 1] + 0 \\ 9 & = B(-9) \\ 9 & = -9B \\ {9 \over -9} & = B \\ -1 & = B \\ \\ 10x^2 + 62x + 69 & = A(x + 5)(2x + 1) - (2x + 1) + C(x + 5)^2 \\ \\ \text{Let } & x = -0.5, \\ 10(-0.5)^2 + 62(-0.5) + 69 & = 0 - 0 + C(-0.5 + 5)^2 \\ 40.5 & = C(20.25) \\ 40.5 & = 20.25 C \\ {40.5 \over 20.25} & = C \\ 2 & = C \\ \\ 10x^2 + 62x + 69 & = A(x + 5)(2x + 1) - (2x + 1) + 2(x + 5)^2 \\ \\ \text{Let } & x = 0, \\ 0 + 0 + 69 & = A(5)(1) - 1 + 2(5)^2 \\ 69 & = 5A - 1 + 50 \\ 69 & = 5A + 49 \\ 20 & = 5A \\ {20 \over 5} & = A \\ 4 & = A \\ \\ {10x^2 + 62x + 69 \over 2x^3 + 21x^2 + 60x + 25} & = {4 \over x + 5} + {-1 \over (x + 5)^2} + {2 \over 2x + 1} \\ & = {4 \over x + 5} - {1 \over (x + 5)^2} + {2 \over 2x + 1} \end{align}

(iii)

\begin{align} {d \over dx} \left( {10x^2 + 62x + 69 \over 2x^3 + 21x^2 + 60x + 25} \right) & = {d \over dx} \left({4 \over x + 5} - {1 \over (x + 5)^2} + {2 \over 2x + 1}\right) \phantom{000000} [\text{Use ans from (ii)}] \\ & = {d \over dx} [ 4(x + 5)^{-1} - (x + 5)^{-2} + 2(2x + 1)^{-1} ] \\ & = 4(-1)(x + 5)^{-2} (1) - (-2)(x + 5)^{-3} (1) + 2(-1)(2x + 1)^{-2}(2) \\ & = - 4(x + 5)^{-2} + 2 (x + 5)^{-3} - 4 (2x + 1)^{-2} \\ & = - {4 \over (x + 5)^2} + {2 \over (x + 5)^3} - {4 \over (2x + 1)^2} \end{align}

\begin{align} 243 & = 3^5 \\ \\ 32t^5 & = (2t)^5 \\ \\ \therefore 243 + 810t + 1080t^2 + 720t^3 + 240t^4 + 32t^5 & = 3^5 + {5 \choose 1} (3^4)(2t) + {5 \choose 2} (3^3)(2t)^2 + {5 \choose 3} (3^2)(2t)^3 + {5 \choose 4}(3)(2t)^4 + (2t)^5 \\ & = (3 + 2t)^5 \\ \\ {d \over dt} \sqrt[5]{(3 + 2t)^5} & = {d \over dt} (3 + 2t) \\ & = 2 \\ \\ \therefore k & = 2 \end{align}