(a)

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

(b)

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

(c)

\begin{align} f'(x) & = 0 \end{align}

(d)

\begin{align} {d \over dx} \left(1 \over x^4\right) & = {d \over dx} (x^{-4}) \\ & = (-4)(x^{-5}) \\ & = (-4) \left(1 \over x^5\right) \\ & = - {4 \over x^5} \end{align}

(e)

\begin{align} {dy \over dx} & = -{3 \over 2}(x^{-{5 \over 2}}) \\ & = - {3 \over 2} \left(1 \over x^{5 \over 2}\right) \\ & = - {3 \over 2} \left(1 \over \sqrt{x^5}\right) \\ & = -{3 \over 2 \sqrt{x^5}} \end{align}

(f)

\begin{align} f'(x) & = 0 \end{align}

(a)

\begin{align} \text{Let } y & = {3 \over x} + \sqrt[3]{x} + \sqrt{2} \\ y & = 3x^{-1} + x^{1 \over 3} + \sqrt{2} \\ \\ {dy \over dx} & = 3(-1)(x^{-2}) + {1 \over 3}(x^{-{2 \over 3}}) + 0 \\ & = -3 \left(1 \over x^2\right) + {1 \over 3} \left(1 \over x^{2 \over 3}\right) \\ & = - {3 \over x^2} + {1 \over 3} \left(1 \over \sqrt[3]{x^2}\right) \\ & = - {3 \over x^2} + {1 \over 3 \sqrt[3]{x^2}} \end{align}

(b)

\begin{align} \text{Let } y & = {4 \over 9x^3} + ax \\ y & = {4 \over 9} \left(1 \over x^3\right) + ax \\ y & = {4 \over 9} x^{-3} + ax \\ \\ {dy \over dx} & = {4 \over 9}(-3)(x^{-4}) + a \\ & = -{4 \over 3} \left(1 \over x^4\right) + a \\ & = - {4 \over 3 x^4} + a \end{align}

\begin{align} f(x) & = {8 + 3x \over \sqrt{x}} \\ f(x) & = {8 \over \sqrt{x}} + {3x \over \sqrt{x}} \\ f(x) & = {8 \over x^{1 \over 2}} + {3x \over x^{1 \over 2}} \\ f(x) & = 8x^{-{1 \over 2}} + 3x^{1 \over 2} \\ \\ f'(x) & = 8 \left(-{1 \over 2}\right) x^{-{3 \over 2}} + 3 \left(1 \over 2\right) x^{-{1 \over 2}} \\ & = - 4 x^{-{3 \over 2}} + {3 \over 2} x^{-{1 \over 2}} \\ & = - {4 \over x^{3 \over 2}} + {3 \over 2} \left(1 \over x^{1 \over 2}\right) \\ & = - {4 \over \sqrt{x^3}} + {3 \over 2} \left(1 \over \sqrt{x}\right) \\ & = - {4 \over \sqrt{x^3}} + {3 \over 2 \sqrt{x}} \end{align}

\begin{align} y & = (3x - 2)(x + 4) \\ y & = 3x^2 + 12x - 2x - 8 \\ y & = 3x^2 + 10x - 8 \\ \\ {dy \over dx} & = 3(2)x + 10 \\ & = 6x + 10 \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = 6(3) + 10 \\ & = 28 \end{align}

The point where the curve crosses the x-axis is the x-intercept (when y = 0)

\begin{align} y & = {3x - 9 \over 2x^2} \\ y & = {3x \over 2x^2} - {9 \over 2x^2} \\ y & = {3 \over 2x} - {9 \over 2x^2} \\ y & = {3 \over 2}\left(1 \over x\right) - {9 \over 2}\left(1 \over x^2\right) \\ y & = {3 \over 2}x^{-1} - {9 \over 2} x^{-2} \\ \\ {dy \over dx} & = {3 \over 2}(-1)(x^{-2}) - {9 \over 2}(-2)(x^{-3}) \\ & = -{3 \over 2} \left(1 \over x^2\right) + 9 \left(1 \over x^3\right) \\ & = - {3 \over 2x^2} + {9 \over x^3} \\ \\ \text{Substitute } & y = 0 \text{ into } y = {3x - 9 \over 2x^2}, \\ 0 & = {3x - 9 \over 2x^2} \\ 0 & = 3x - 9 \\ 9 & = 3x \\ {9 \over 3} & = x \\ 3 & = x \phantom{000000} [x \text{-intercept}] \\ \\ \text{Substitute } & x = 3 \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{3 \over 2(3)^2} + {9 \over (3)^3} \\ & = {1 \over 6} \end{align}

\begin{align} y & = ax + {b \over x^2} \\ \\ \text{Using } & (1, 11), \\ 11 & = a(1) + {b \over (1)^2} \\ 11 & = a + {b \over 1} \\ 11 & = a + b \\ 11 - b & = a \phantom{00} \text{--- (1)} \\ \\ \\ y & = ax + bx^{-2} \\ \\ {dy \over dx} & = a + b(-2)(x^{-3}) \\ & = a - 2b \left(1 \over x^3\right) \\ & = a - {2b \over x^3} \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = 2, \\ 2 & = a -{2b \over (1)^3} \\ 2 & = a - {2b \over 1} \\ 2 & = a - 2b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2 & = 11 - b - 2b \\ 2 & = 11 - 3b \\ 3b & = 11 - 2 \\ 3b & = 9 \\ b & = {9 \over 3} \\ b & = 3 \\ \\ \text{Substitute } & b = 3 \text{ into (1),} \\ a & = 11- 3 \\ a & = 8 \\ \\ \\ \therefore a & = 8, b = 3 \end{align}

\begin{align} y & = ax^2 + bx \\ \\ {dy \over dx} & = 2ax + b \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = 4, \\ 4 & = 2a(1) + b \\ 4 & = 2a + b \\ 4 - 2a & = b \phantom{00} \text{--- (1)} \\ \\ \text{When } & x = 2 \text{ and } {dy \over dx} = 14, \\ 14 & = 2a(2) + b \\ 14 & = 4a + b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 14 & = 4a + 4 - 2a \\ 14 -4 & = 4a - 2a \\ 10 & = 2a \\ {10 \over 2} & = a \\ 5 & = a \\ \\ \text{Substitute } & a = 5 \text{ into (1),} \\ b & = 4 - 2(5) \\ b & = -6 \\ \\ \\ \therefore a & = 5, b = -6 \end{align}

(a)

\begin{align} {dy \over dx} & = (7)(4x^3 - 1)^6 . (12x^2) \\ & = 84x^2 (4x^3 - 1)^6 \end{align}

(b)

\begin{align} \text{Let } y & = 8 \sqrt{x^2 + 3x - 4} \\ y & = 8 (x^2 + 3x - 4)^{1 \over 2} \\ \\ {dy \over dx} & = 8 \left(1 \over 2\right) (x^2 + 3x - 4)^{-{1 \over 2}} . (2x + 3) \\ & = 4(2x + 3) (x^2 + 3x - 4)^{-{1 \over 2}} \\ & = 4(2x + 3) \left( 1 \over \sqrt{x^2 + 3x - 4}\right) \\ & = {4(2x + 3) \over \sqrt{x^2 + 3x - 4} } \end{align}

(c)

\begin{align} f(x) & = {6 \over (2 - 5x^4)^3} \\ & = 6 (2 - 5x^4)^{-3} \\ \\ f'(x) & = 6(-3) (2 - 5x^4)^{-4} . (-20x^3) \\ & = (360x^3) (2 - 5x^4)^{-4} \\ & = {360x^3 \over (2 - 5x^4)^4} \end{align}

\begin{align} {dy \over dx} & = 6(3x - 8)^5 . (3) \\ & = 18(3x - 8)^5 \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = 18 [3(3) - 8]^5 \\ & = 18 \end{align}

(a)

\begin{align} u & = 4x - 3 \phantom{0000000} {du \over dx} = 4 \\ v & = (2x + 7)^6 \phantom{00000} {dv \over dx} = 6(2x + 7)^5 . (2) = 12(2x + 7)^5 \\ \\ {dy \over dx} & = (4x - 3)[12(2x + 7)^5] + (2x + 7)^6 (4) \\ & = 12(4x - 3)(2x + 7)^5 + 4(2x + 7)^6 \\ & = 4(2x + 7)^5 [3(4x - 3) + 2x + 7] \\ & = 4(2x + 7)^5 (12x - 9 + 2x + 7) \\ & = 4(2x + 7)^5 (14x - 2) \\ & = 4(2)(2x + 7)^5 (7x - 1) \\ & = 8(2x + 7)^5 (7x - 1) \end{align}

(b)

\begin{align} u & = 3 - x^2 \phantom{0000000} {du \over dx} = -2x \\ v & = \sqrt{5x + 4} \\ & = (5x + 4)^{1 \over 2} \phantom{00000} {dv \over dx} = {1 \over 2}(5x + 4)^{-{1 \over 2}}. (5) = {5 \over 2\sqrt{5x + 4}} \\ \\ f'(x) & = (3 - x^2) \left(5 \over 2\sqrt{5x + 4}\right) + \left(\sqrt{5x + 4}\right)(-2x) \\ & = {5(3 - x^2) \over 2 \sqrt{5x + 4}} - 2x \sqrt{5x + 4} \\ & = {5(3 - x^2) \over 2 \sqrt{5x + 4}} - {2x \sqrt{5x + 4} \left(2 \sqrt{5x + 4}\right) \over 2 \sqrt{5x + 4}} \\ & = {5(3 - x^2) \over 2 \sqrt{5x + 4}} - {4x (5x + 4) \over 2 \sqrt{5x + 4}} \\ & = {5(3 - x^2) - 4x(5x + 4) \over 2 \sqrt{5x + 4}} \\ & = {15 - 5x^2 - 20x^2 - 16x \over 2 \sqrt{5x + 4}} \\ & = {15 - 16x - 25x^2 \over 2 \sqrt{5x + 4}} \end{align}

(a)

\begin{align} {3x \over a - bx^2} & = 3x (a - bx^2)^{-1} \\ \\ u & = 3x \phantom{00000000000} {du \over dx} = 3 \\ v & = (a - bx^2)^{-1} \phantom{000.} {dv \over dx} = (-1)(a - bx^2)^{-2} . (-2bx) = 2bx (a - bx^2)^{-2} \\ \\ {d \over dx} [3x (a - bx^2)^{-1}] & = (3x)[2bx (a - bx^2)^{-2}] + (a - bx^2)^{-1} (3) \\ & = 6b x^2 (a - bx^2)^{-2} + 3 (a - bx^2)^{-1} \\ & = {6bx^2 \over (a - bx^2)^2} + {3 \over a - bx^2} \times {a - bx^2 \over a - bx^2} \\ & = {6bx^2 \over (a - bx^2)^2} + {3a - 3bx^2 \over (a - bx^2)^2} \\ & = {6bx^2 + 3a - 3bx^2 \over (a - bx^2)^2} \\ & = {3bx^2 + 3a \over (a - bx^2)^2} \end{align}

(b)

\begin{align} u & = a^2 - x^2 \phantom{00000000} {du \over dx} = -2x \\ v & = (5ax - 3x^3)^4 \phantom{0000} {dv \over dx} = 4(5ax - 3x^3)^3 (5a - 9x^2) = 4(5a - 9x^2)(5ax - 3x^3)^3 \\ \\ {d \over dx} [ (a^2 - x^2) (5ax - 3x^3)^4 ] & = (a^2 - x^2) [4 (5a - 9x^2)(5ax - 3x^3)^3] + (5ax - 3x^3)^4 (-2x) \\ & = 4 (a^2 - x^2) (5a - 9x^2)(5ax - 3x^3)^3 - 2x (5ax - 3x^3)^4 \\ & = 2(5ax - 3x^3)^3 [ 2(a^2 - x^2)(5a - 9x^2) - x (5ax - 3x^3) ] \\ & = 2(5ax - 3x^3)^3 [2 (5a^3 - 9a^2 x^2 - 5a x^2 + 9x^4) - 5ax^2 + 3x^4 ] \\ & = 2(5ax - 3x^3)^3 (10a^3 - 18a^2 x^2 - 10a x^2 + 18x^4 - 5ax^2 + 3x^4) \\ & = 2(5ax - 3x^3)^3 (10a^3 - 18a^2 x^2 - 15a x^2 + 21x^4) \end{align}

(a)

\begin{align} u & = 2x \phantom{00000000} {du \over dx} = 2 \\ v & = 4x - 3 \phantom{0000.} {dv \over dx} = 4 \\ \\ {d \over dx} \left(2x \over 4x - 3\right) & = {(4x - 3)(2) - (2x)(4) \over (4x - 3)^2} \\ & = {2(4x - 3) - 8x \over (4x - 3)^2 } \\ & = {8x - 6 - 8x \over (4x - 3)^2} \\ & = {-6 \over (4x - 3)^2} \end{align}

(b)

\begin{align} u & = x \phantom{00000000} {du \over dx} = 1 \\ v & = 7 - 3x \phantom{0000} {dv \over dx} = -3 \\ \\ {d \over dx} \left(x \over 7 - 3x\right) & = {(7 - 3x)(1) - (x)(-3) \over (7 - 3x)^2} \\ & = {7 - 3x + 3x \over (7 - 3x)^2} \\ & = {7 \over (7 - 3x)^2} \end{align}

(a)

\begin{align} u & = 2x^3 - 1 \phantom{00000000} {du \over dx} = 6x^2 \\ v & = (3x + 1)^2 \phantom{0000000} {dv \over dx} = 2(3x + 1)(3) = 6(3x + 1) \\ \\ {d \over dx} \left[ 2x^3 - 1 \over (3x +1)^2 \right] & = { (3x + 1)^2 (6x^2) - (2x^3 - 1)[6(3x + 1)] \over [(3x + 1)^2]^2 } \\ & = { 6x^2 (3x + 1)^2 - 6(2x^3 - 1)(3x + 1) \over (3x + 1)^4 } \\ & = { 6(3x + 1) [x^2(3x + 1) - (2x^3 - 1)] \over (3x + 1)^4 } \\ & = { 6(3x + 1)(3x^3 + x^2 - 2x^3 + 1) \over (3x + 1)^4 } \\ & = { 6(3x + 1)(x^3 + x^2 + 1) \over (3x + 1)^4} \\ & = { 6(x^3 + x^2 + 1) \over (3x + 1)^3 } \\ & = { 6x^3 + 6x^2 + 6 \over (3x + 1)^3 } \end{align}

(b)

\begin{align} u & = 2x - 7 \phantom{00000000} {du \over dx} = 2 \\ \\ v & = \sqrt[3]{2x - 5} \\ & = (2x - 5)^{1 \over 3} \phantom{0000.0} {dv \over dx} = {1 \over 3}(2x - 5)^{-{2 \over 3}} (2) = {2 \over 3} (2x - 5)^{-{2 \over 3}} \\ \\ {d \over dx} \left[ 2x - 7 \over (2x - 5)^{1 \over 3} \right] & = { (2x - 5)^{1 \over 3} (2) - (2x - 7) {2 \over 3} (2x - 5)^{-{2 \over 3}} \over \left[ (2x - 5)^{1 \over 3} \right]^2 } \\ & = { 2 (2x - 5)^{1 \over 3} - {2 \over 3} (2x - 7) (2x - 5)^{-{2 \over 3}} \over (2x - 5)^{2 \over 3} } \times { (2x - 5)^{2 \over 3} \over (2x - 5)^{2 \over 3}} \\ & = { 2(2x - 5) - {2 \over 3}(2x - 7)(2x - 5)^0 \over (2x - 5)^{4 \over 3}} \\ & = { 2(2x - 5) - {2 \over 3}(2x - 7)(1) \over (2x - 5)^{4 \over 3}} \times {3 \over 3} \\ & = { 6(2x - 5) - 2(2x - 7) \over 3(2x - 5)^{4 \over 3}} \\ & = { 12x - 30 - 4x + 14 \over 3 \sqrt[3]{(2x - 5)^4} } \\ & = { 8x - 16 \over 3 \sqrt[3]{(2x - 5)^4} } \end{align}

(c)

\begin{align} \sqrt{1 + 4x \over 3x^2 - 7} & = { \sqrt{1 + 4x} \over \sqrt{3x^2 - 7} } \\ & = { (1 + 4x)^{1 \over 2} \over (3x^2 - 7)^{1 \over 2} } \\ \\ u & = (1 + 4x)^{1 \over 2} \phantom{00000000} {du \over dx} = {1 \over 2} (1 + 4x)^{-{1 \over 2}} (4) = 2(1 + 4x)^{-{1 \over 2}} \\ \\ v & = (3x^2 - 7)^{1 \over 2} \phantom{0000000} {dv \over dx} = {1 \over 2} (3x^2 - 7)^{-{1 \over 2}} (6x) = 3x (3x^2 - 7)^{-{1 \over 2}} \\ \\ \\ {d \over dx} \left[ (1 + 4x)^{1 \over 2} \over (3x^2 - 7)^{1 \over 2} \right] & = { (3x^2 - 7)^{1 \over 2} 2(1 + 4x)^{-{1 \over 2}} - (1 + 4x)^{1 \over 2} 3x (3x^2 - 7)^{-{1 \over 2}} \over \left[ (3x^2 - 7)^{1 \over 2} \right]^2 } \\ & = { 2(3x^2 - 7)^{1 \over 2} (1 + 4x)^{-{1 \over 2}} - 3x (1 + 4x)^{1 \over 2} (3x^2 - 7)^{-{1 \over 2}} \over 3x^2 - 7} \times { (3x^2 - 7)^{1 \over 2} (1 + 4x)^{1 \over 2} \over (3x^2 - 7)^{1 \over 2} (1 + 4x)^{1 \over 2} } \\ & = { 2(3x^2 - 7)(1 + 4x)^0 - 3x (1 + 4x)(3x^2 - 7)^0 \over (3x^2 - 7)^{3 \over 2} (1 + 4x)^{1 \over 2} } \\ & = { 2(3x^2 - 7) (1) - 3x (1 + 4x)(1) \over \sqrt{(3x^2 - 7)^3} \sqrt{1 + 4x} } \\ & = { 6x^2 - 14 - 3x - 12x^2 \over \sqrt{ (3x^2 - 7)^3 (1 + 4x) } } \\ & = { -6x^2 - 3x - 14 \over \sqrt{ (3x^2 - 7)^3 (1 + 4x) } } \\ & = { - (6x^2 + 3x + 14) \over \sqrt{ (3x^2 - 7)^3 (1 + 4x) } } \\ & = - {6x^2 + 3x + 14 \over \sqrt{ (3x^2 - 7)^3 (1 + 4x) } } \end{align}

\begin{align} y & = {\sqrt{x^2 + 25} \over x + 5} = {(x^2 + 25)^{1 \over 2} \over x + 5} \\ \\ \\ u & = (x^2 + 25)^{1 \over 2} \phantom{00000000} {du \over dx} = {1 \over 2}(x^2 + 25)^{-{1 \over 2}} (2x) = x(x^2 + 25)^{-{1 \over 2}} \\ \\ v & = x + 5 \phantom{000000000000.} {dv \over dx} = 1 \\ \\ \\ {dy \over dx} & = { (x + 5)x (x^2 + 25)^{-{1 \over 2}} - (x^2 + 25)^{1 \over 2} (1) \over (x + 5)^2 } \\ & = {x (x + 5)(x^2 + 25)^{-{1 \over 2}} - (x^2 + 25)^{1 \over 2} \over (x + 5)^2 } \\ \\ \text{When } & x = 12, \\ {dy \over dx} & = { 12(12 + 5)[(12)^2 + 25]^{-{1 \over 2}} - [(12)^2 + 25]^{1 \over 2} \over (12 + 5)^2 } \\ & = {35 \over 3757} \end{align}

(i)

\begin{align} y & = x^2 (5x - 3) \\ & = 5x^3 - 3x^2 \\ \\ {dy \over dx} & = 5(3)x^2 - 3(2)x \\ & = 15x^2 - 6x \\ \\ {d^2 y \over dx^2} & = 15(2)x - 6 \\ & = 30x - 6 \end{align}

(ii)

\begin{align} {dy \over dx} & > 0 \\ 15x^2 - 6x & > 0 \\ 3x(5x - 2) & > 0 \\ x(5x - 2) & > 0 \end{align}

\begin{align} x < 0 \text{ or } & x > {2 \over 5} \\ \\ \\ {d^2 y \over dx^2} & > 0 \\ 30x - 6 & > 0 \\ 30x & > 6 \\ x & > {6 \over 30} \\ x & > {1 \over 5} \\ \\ \\ \text{For } {dy \over dx} > 0 &, x < 0 \text{ or } x > {2 \over 5} \\ \text{For } {d^2 y \over dx^2} > 0 &, x > {1 \over 5} \\ \\ \therefore x & > {2 \over 5} \end{align}

(iii)

\begin{align} \left(dy \over dx\right)^2 & = (15x^2 - 6x)^2 \\ & = (15x^2)^2 - 2(15x^2)(6x) + (6x)^2 \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ & = 225x^4 - 180x^3 + 36x^2 \\ \\ \text{Since } {d^2 y \over dx^2} & = 30x - 6, \phantom{.} {d^2 y \over dx^2} \ne \left(dy \over dx\right)^2 \end{align}

\begin{align} v & = t^3 + 2t^2 + 3t + 1 \\ \\ {dv \over dt} & = 3t^2 + 2(2)t + 3 \\ & = 3t^2 + 4t + 3 \\ \\ {d^2 v \over dt^2} & = 3(2)t + 4 \\ & = 6t + 4 \\ \\ {dv \over dt} & = {d^2 v \over dt^2} \\ 3t^2 + 4t + 3 & = 6t + 4 \\ 3t^2 - 2t - 1 & = 0 \\ (t - 1)(3t + 1) & = 0 \end{align} \begin{align} t - 1 & = 0 && \text{ or } & 3t + 1 & = 0 \\ t & = 1 &&& 3t & = -1 \\ & &&& t & = -{1 \over 3} \end{align}