(a)

\begin{align} u & = x^2 + 2 &&& v & = 3 - x^3 \\ {du \over dx} & = 2x &&& {dv \over dx} & = -3x^2 \end{align} \begin{align} {dy \over dx} & = { (3 - x^3)(2x) - (x^2 + 2)(-3x^2) \over (3 - x^3)^2} \\ & = { 2x(3 - x^3) + 3x^2 (x^2 + 2) \over (3 - x^3)^2} \\ & = { 6x - 2x^4 + 3x^4 + 6x^2 \over (3 - x^3)^2 } \\ & = { x^4 + 6x^2 + 6x \over (3 - x^3)^2} \end{align}

(b)

\begin{align} u & = 1 - \sqrt{t} &&& v & = 1 + \sqrt{t} \\ & = 1 - t^{1 \over 2} &&& & = 1 + t^{1 \over 2} \\ {du \over dt} & = -{1 \over 2}t^{-{1 \over 2}} &&& {dv \over dt} & = {1 \over 2}t^{-{1 \over 2}} \\ & = -{1 \over 2\sqrt{t}} &&& & = {1 \over 2\sqrt{t}} \end{align} \begin{align} g'(t) & = { (1 + \sqrt{t}) \left(-{1 \over 2\sqrt{t}}\right) - (1 - \sqrt{t})\left(1 \over 2\sqrt{t}\right) \over (1 + \sqrt{t})^2} \\ & = { { -(1 + \sqrt{t}) \over 2\sqrt{t}} - {1 - \sqrt{t} \over 2\sqrt{t}} \over (1 + \sqrt{t})^2 } \\ & = { { -(1 + \sqrt{t}) - (1 - \sqrt{t}) \over 2 \sqrt{t}} \over (1 + \sqrt{t})^2} \\ & = { { -1 - \sqrt{t} - 1 + \sqrt{t} \over 2\sqrt{t}} \over (1 + \sqrt{t})^2} \\ & = { { -2 \over 2\sqrt{t}} \over (1 + \sqrt{t})^2 } \\ & = { {-1 \over \sqrt{t}} \over (1 + \sqrt{t})^2 } \\ & = { -1 \over \sqrt{t} (1 + \sqrt{t})^2 } \end{align}

(c)

\begin{align} u & = (x^2 - 1)^3 &&& v & = 3 - x^4 \\ {du \over dx} & = 3(x^2 - 1)^2 . (2x) \phantom{00000} [\text{Chain rule}] &&& {dv \over dx} & = -4x^3 \\ & = 6x(x^2 - 1)^2 \end{align} \begin{align} f'(x) & = {(3 - x^4)[6x(x^2 - 1)^2] - (x^2 - 1)^3 (-4x^3) \over (3 - x^4)^2 } \\ & = { 6x(3 - x^4)(x^2 - 1)^2 + 4x^3(x^2 - 1)^3 \over (3 - x^4)^2} \\ & = { 2x(x^2 - 1)^2 [ 3(3 - x^4) + 2x^2(x^2 - 1)] \over (3 - x^4)^2} \\ & = { 2x(x^2 - 1)^2 (9 - 3x^4 + 2x^4 - 2x^2) \over (3 - x^4)^2} \\ & = { 2x(x^2 - 1)^2 (9 - x^4 - 2x^2) \over (3 - x^4)^2} \end{align}

(d)

\begin{align} u & = a^2 + 2a - 7 &&& v & = \sqrt{a} \\ & &&& & = a^{1 \over 2} \\ {du \over da} & = 2a + 2 &&& {dv \over da} & = {1 \over 2} a^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{a}} \end{align} \begin{align} {dr \over da} & = { (\sqrt{a})(2a + 2) - (a^2 + 2a - 7)\left(1 \over 2\sqrt{a}\right) \over (\sqrt{a})^2} \\ & = { \sqrt{a} (2a + 2) - {a^2 + 2a - 7 \over 2\sqrt{a}} \over a} \\ & = { \sqrt{a} (2a + 2) - {a^2 + 2a - 7 \over 2\sqrt{a}} \over a} \times {2\sqrt{a} \over 2\sqrt{a}} \\ & = { \sqrt{a} (2a + 2)(2\sqrt{a}) - (a^2 + 2a - 7) \over 2a \sqrt{a}} \\ & = { 2a(2a + 2) - a^2 - 2a + 7 \over 2a\sqrt{a}} \\ & = {4a^2 + 4a - a^2 - 2a + 7 \over 2a \sqrt{a}} \\ & = {3a^2 + 2a + 7 \over 2a \sqrt{a}} \\ & = {3a^2 + 2a + 7 \over 2a (a^{1 \over 2})} \\ & = {3a^2 + 2a + 7 \over 2a^{3 \over 2}} \end{align}

(a)

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

(b)

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

(c)

\begin{align} u & = 3x &&& v & = \sqrt{3x + 1} \\ & &&& & = (3x + 1)^{1 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = {1 \over 2} (3x + 1)^{-{1 \over 2}} . (3) \\ & &&& & = {3 \over 2}(3x + 1)^{-{1 \over 2}} \\ & &&& & = {3 \over 2\sqrt{3x + 1}} \end{align} \begin{align} {d \over dx} \left( 3x \over \sqrt{3x + 1} \right) & = {(\sqrt{3x + 1})(3) - (3x) \left({3 \over 2\sqrt{3x + 1}}\right) \over \left(\sqrt{3x + 1}\right)^2} \\ & = {3\sqrt{3x + 1} - {9x \over 2\sqrt{3x + 1}} \over 3x + 1} \\ & = {3\sqrt{3x + 1} - {9x \over 2\sqrt{3x + 1}} \over 3x + 1} \times {2\sqrt{3x + 1} \over 2\sqrt{3x + 1}} \\ & = {3\sqrt{3x + 1} (2\sqrt{3x + 1}) - 9x \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {6(3x + 1) - 9x \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {18x + 6 - 9x \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {9x + 6 \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {3(3x + 2) \over 2(3x + 1)\sqrt{3x + 1}} \end{align}

(d)

\begin{align} u & = 5x^2 &&& v & = \sqrt{3x^2 - 1} \\ & &&& & = (3x^2 - 1)^{1 \over 2} \\ {du \over dx} & = 10x &&& {dv \over dx} & = {1 \over 2} (3x^2 - 1)^{-{1 \over 2}} . (6x) \\ & &&& & = 3x(3x^2 - 1)^{-{1 \over 2}} \\ & &&& & = {3x \over \sqrt{3x^2 - 1}} \end{align} \begin{align} {d \over dx} \left( 5x^2 \over \sqrt{3x^2 - 1} \right) & = {(\sqrt{3x^2 - 1}) (10x) - (5x^2) \left(3x \over \sqrt{3x^2 - 1}\right) \over \left(\sqrt{3x^2 - 1}\right)^2 } \\ & = {10x\sqrt{3x^2 - 1} - {15x^3 \over \sqrt{3x^2 - 1}} \over 3x^2 - 1} \\ & = {10x\sqrt{3x^2 - 1} - {15x^3 \over \sqrt{3x^2 - 1}} \over 3x^2 - 1} \times {\sqrt{3x^2 - 1} \over \sqrt{3x^2 - 1}} \\ & = {10x\sqrt{3x^2 - 1}(\sqrt{3x^2 - 1}) - 15x^3 \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {10x(3x^2 - 1) - 15x^3 \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {30x^3 - 10x - 15x^3 \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {15x^3 - 10x \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {5x(3x^2 - 2) \over (3x^2 - 1)\sqrt{3x^2 - 1}} \end{align}

\begin{align} u & = x^2 + x + 1 &&& v & = x + 1 \\ {du \over dx} & = 2x + 1 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1) (2x + 1) - (x^2 + x + 1)(1) \over (x + 1)^2} \\ & = {2x^2 + x + 2x + 1 - (x^2 + x + 1) \over (x + 1)^2} \\ & = {2x^2 + 3x + 1 - x^2 - x - 1 \over (x + 1)^2} \\ & = {x^2 + 2x \over (x + 1)^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {(1)^2 + 2(1) \over (1 + 1)^2} \\ & = {3 \over 4} \end{align}

The point where the curve crosses the x-axis is the x-intercept, where y = 0.

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

\begin{align} u & = \sqrt{2x - 1} &&& v & = x \\ & = (2x - 1)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(2x - 1)^{-{1 \over 2}} .(2) \phantom{000} [\text{Chain rule}] &&& {dv \over dx} & = 1 \\ & = (2x - 1)^{-{1 \over 2}} \\ & = {1 \over \sqrt{2x - 1}} \end{align} \begin{align} {dy \over dx} & = { (x)\left(1 \over \sqrt{2x - 1}\right) - (\sqrt{2x - 1})(1) \over (x)^2} \\ & = { {x \over \sqrt{2x - 1}} - \sqrt{2x - 1} \over x^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = { {(1) \over \sqrt{2(1) - 1}} - \sqrt{2(1) - 1} \over (1)^2} \\ & = {1 - 1 \over 1} \\ & = 0 \end{align}

(i)

\begin{align} u & = 3x^5 - x^3 &&& v & = x^2 \\ {du \over dx} & = 15x^4 - 3x^2 &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2)(15x^4 - 3x^2) - (3x^5 - x^3)(2x) \over (x^2)^2} \\ & = { x^2(15x^4 - 3x^2) - 2x(3x^5 - x^3) \over x^4} \\ & = { 15x^6 - 3x^4 - 6x^6 + 2x^4 \over x^4 } \\ & = { 9x^6 - x^4 \over x^4 } \\ & = { 9x^6 \over x^4 } - {x^4 \over x^4} \\ & = 9x^2 - 1 \end{align}

(ii)

\begin{align} y & = {3x^5 - x^3 \over x^2} \\ & = {3x^5 \over x^2} - {x^3 \over x^2} \\ & = 3x^3 - x \\ \\ {dy \over dx} & = 9x^2 - 1 \\ \\ \therefore \text{Ans} & \text{wers are the same} \end{align}

\begin{align} u & = \sqrt{x^2 + 1} &&& v & = 3x + 4 \\ & = (x^2 + 1)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(x^2 + 1)^{-{1 \over 2}} . (2x) &&& {dv \over dx} & = 3 \\ & = x(x^2 + 1)^{-{1 \over 2}} \\ & = {x \over \sqrt{x^2 + 1}} \end{align} \begin{align} {dy \over dx} & = {(3x + 4) \left(x \over \sqrt{x^2 + 1}\right) - (\sqrt{x^2 + 1})(3) \over (3x + 4)^2} \\ & = { {x(3x + 4) \over \sqrt{x^2 + 1}} - 3\sqrt{x^2 + 1} \over (3x + 4)^2} \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = { {(0)[3(0) + 4] \over \sqrt{(0)^2 + 1}} - 3\sqrt{(0)^2 + 1} \over [3(0) + 4]^2 } \\ & = {0 - 3 \over 16} \\ & = -{3 \over 16} \end{align}

(i) Use product rule for the differentiation

\begin{align} u & = 1 - x^2 &&& v & = \sqrt{x + 3} \\ & &&& & = (x + 3)^{1 \over 2} \\ {du \over dx} & = -2x &&& {dv \over dx} & = {1 \over 2}(x + 3)^{-{1 \over 2}} . (1) \\ & &&& & = {1 \over 2} (x + 3)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x + 3}} \end{align} \begin{align} {d \over dx} \left[ (1 - x^2) \sqrt{x + 3} \right] & = (1 - x^2) \left(1 \over 2\sqrt{x + 3}\right) + (\sqrt{x + 3})(-2x) \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - 2x \sqrt{x + 3} \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - {2x \sqrt{x + 3} \over 1} \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - {2x \sqrt{x + 3} (2\sqrt{x + 3}) \over 2\sqrt{x + 3}} \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - {4x (x + 3) \over 2\sqrt{x + 3}} \\ & = {1 - x^2 - 4x(x + 3) \over 2\sqrt{x + 3}} \\ & = {1 - x^2 - 4x^2 - 12x \over 2\sqrt{x + 3}} \\ & = {1 - 5x^2 - 12x \over 2\sqrt{x + 3}} \end{align}

(ii) The rate of change of $y$ with respect to $x$ is denoted by ${dy \over dx}$

\begin{align} u & = (1 - x^2)\sqrt{x + 3} &&& v & = x^3 \\ {du \over dx} & = {1 - 5x^2 - 12x \over 2\sqrt{x + 3}} \phantom{00000} [\text{From (i)}] &&& {dv \over dx} & = 3x^2 \end{align} \begin{align} {dy \over dx} & = { (x^3) \left({1 - 5x^2 - 12x \over 2\sqrt{x + 3}}\right) - (1 - x^2)\sqrt{x + 3} (3x^2) \over (x^3)^2} \\ & = { {x^3(1 - 5x^2 - 12x) \over 2\sqrt{x + 3}} - 3x^2 (1 - x^2) \sqrt{x + 3} \over x^6} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = { {(1)^3 [1 - 5(1)^2 - 12(1)] \over 2\sqrt{1 + 3}} - 3(1)^2 [1 - (1)^2] \sqrt{1 + 3} \over (1)^6} \\ & = {-4 - 0 \over 1} \\ & = -4 \end{align}

Method 1: By quotient rule

\begin{align} u & = \sqrt{x + 1} &&& v & = 2x^2 - 1 \\ & = (x + 1)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) &&& {dv \over dx} & = 4x \\ & = {1 \over 2} (x + 1)^{-{1 \over 2}} \\ & = {1 \over 2\sqrt{x + 1}} \end{align} \begin{align} {dy \over dx} & = {(2x^2 - 1) \left(1 \over 2\sqrt{x + 1}\right) - (\sqrt{x + 1})(4x) \over (2x^2 - 1)^2} \\ & = { {2x^2 - 1 \over 2\sqrt{x + 1}} - 4x \sqrt{4x + 1} \over (2x^2 - 1)^2} \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = { {2(0)^2 - 1 \over 2\sqrt{0 + 1}} - 4(0) \sqrt{4(0) + 1} \over [2(0)^2 - 1]^2} \\ & = -{1 \over 2} \end{align}

Method 2: By product rule

$$ y = {\sqrt{x + 1} \over 2x^2 - 1} = (x + 1)^{1 \over 2} (2x^2 - 1)^{-1} $$
\begin{align} u & = (x + 1)^{1 \over 2} &&& v & = (2x^2 - 1)^{-1} \\ {du \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) &&& {dv \over dx} & = (-1)(2x^2 - 1)^{-2} . (4x) \\ & = {1 \over 2} (x + 1)^{-{1 \over 2}} &&& & = -4x(2x^2 - 1)^{-2}\\ & = {1 \over 2\sqrt{x + 1}} &&& & = -{4x \over (2x^2 - 1)^2} \end{align} \begin{align} {dy \over dx} & = (x + 1)^{1 \over 2} \left[-{4x \over (2x^2 - 1)^2}\right] + (2x^2 - 1)^{-1} \left(1 \over 2\sqrt{x + 1}\right) \\ & = -{4x \sqrt{x + 1} \over (2x^2 - 1)^2} + {1 \over 2 (2x^2 - 1) \sqrt{x + 1}} \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = -{4(0) \sqrt{0 + 1} \over [2(0)^2 - 1]^2} + {1 \over 2[2(0)^2 - 1] \sqrt{0 + 1}} \\ & = -{1 \over 2} \end{align}

\begin{align} u & = 5x + 3 &&& v & = 10x - 6 \\ {du \over dx} & = 5 &&& {dv \over dx} & = 10 \end{align} \begin{align} {dy \over dx} & = {d \over dx}\left( {5x + 3 \over 10x - 6} \right)^4 \\ & = (4)\left( {5x + 3 \over 10x - 6} \right)^3 \underbrace{{d \over dx}\left({5x + 3 \over 10x - 6}\right)}_{\text{Quotient rule}} \\ & = 4\left( {5x + 3 \over 10x - 6} \right)^3 \left[ {(10x - 6)(5) - (5x + 3)(10) \over (10x - 6)^2} \right] \\ & = 4\left( {5x + 3 \over 10x - 6} \right)^3 \left[ {50x - 30 - 50x - 30 \over (10x - 6)^2} \right] \\ & = 4 \left[{(5x + 3)^3 \over (10x - 6)^3}\right] \left[{-60 \over (10x - 6)^2} \right] \\ & = 4 \left[ -60(5x + 3)^3 \over (10x - 6)^5 \right] \\ & = {-240(5x + 3)^3 \over (10x - 6)^5} \end{align}
To find the gradient, we need the x-coordinates of the points where y = 16
\begin{align} \text{Substitute } & y = 16 \text{ into eqn of curve,} \\ 16 & = \left( {5x + 3 \over 10x - 6} \right)^4 \\ \pm\sqrt[4]{16} & = {5x + 3 \over 10x - 6} \\ \pm2 & = {5x + 3 \over 10x -6} \end{align} \begin{align} {5x + 3 \over 10x - 6} & = 2 && \text{ or } & {5x + 3 \over 10x - 6} & = -2 \\ 5x + 3 & = 2(10x - 6) &&& 5x + 3 & = -2(10x - 6) \\ 5x + 3 & = 20x - 12 &&& 5x + 3 & = -20x + 12 \\ -15x & = -15 &&& 25x & = 9 \\ x & = 1 &&& x & = {9 \over 25} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = {-240[5(1) + 3]^3 \over [10(1) - 6]^5} &&& {dy \over dx} & = {-240\left[ 5 \left(9 \over 25 \right) + 3\right]^3 \over \left[ 10\left(9 \over 25 \right) - 6 \right]^5} \\ & = -120 &&& & = {1000 \over 3} \end{align}

$$y = \sqrt{x - a \over b - x} = \left( {x - a \over b - x} \right)^{1 \over 2}$$
\begin{align} u & = x - a &&& v & = b - x \\ {du \over dx} & = 1 &&& {dv \over dx} & = -1 \end{align} \begin{align} \require{cancel} {dy \over dx} & = {1 \over 2}\left( {x - a \over b - x} \right)^{-{1 \over 2}} \underbrace{{d \over dx}\left( {x - a \over b - x} \right)}_{\text{Quotient Rule}} \\ & = {1 \over 2}\left( {x - a \over b - x} \right)^{-{1 \over 2}} \left[ { (b - x)(1) - (x - a)(-1) \over (b - x)^2} \right] \\ & = {1 \over 2}\left( {x - a \over b - x} \right)^{-{1 \over 2}} \left[ {b - x + x - a \over (b - x)^2} \right] \\ & = {1 \over 2}\sqrt{b - x \over x - a} \left[ {b - a \over (b - x)^2} \right] \\ \\ \text{When } & x = {a + b \over 2}, \\ {dy \over dx} & = {1 \over 2}\sqrt{b - \left({a + b \over 2}\right) \over \left({a + b \over 2}\right) - a} \left[{b - a \over \left(b - \left({a + b \over 2}\right)\right)^2}\right] \\ & = {1 \over 2}\sqrt{{2b \over 2} - {a + b \over 2} \over {a + b \over 2} - {2a \over 2}} \left[{b - a \over \left({2b \over 2} - {a + b \over 2}\right)^2}\right] \\ & = {1 \over 2}\sqrt{{2b - a - b \over 2} \over {a + b - 2a \over 2}} \left[{b - a \over \left({2b - a - b \over 2}\right)^2}\right] \\ & = {1 \over 2}\sqrt{{b - a \over 2} \over {b - a \over 2}} \left[{b - a \over \left({b - a \over 2}\right)^2}\right] \\ & = {1 \over 2}\sqrt{1} \left[{(b - a) \over \left({(b - a)^2 \over 4}\right)}\right] \\ & = {1 \over 2}(1)\left[(b - a) \div \left( {(b - a)^2 \over 4} \right)\right] \\ & = {1 \over 2}\left[ \cancel{(b - a)} \times {4 \over (b - a)^\cancel{2}} \right] \\ & = {1 \over \cancel{2}}\left( \cancel{4}^2 \over b - a \right) \\ & = {2 \over b - a} \text{ (Shown)} \end{align}

(i) $y = {25 \over 4}$ is a horizontal line with gradient = 0. The gradient of the curve at A is equals to 0

\begin{align} y & = x\left(k \over \sqrt{x} - 1 \right) \\ & = {kx \over \sqrt{x}} - x \\ & = {kx \over x^{1 \over 2}} - x \\ & = kx^{1 \over 2} - x \\ \\ {dy \over dx} & = k \left(1 \over 2\right) x^{-{1 \over 2}} - 1 \\ & = {k \over 2\sqrt{x}} - 1 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {k \over 2\sqrt{x}} - 1 \\ 1 & = {k \over 2\sqrt{x}} \\ \sqrt{x} & = {k \over 2} \\ x & = \left(k \over 2\right)^2 \\ & = {k^2 \over 4} \\ \\ \therefore & \phantom{.} A \left({k^2 \over 4}, {25 \over 4}\right) \\ \\ \text{Substitute } & A \left({k^2 \over 4}, {25 \over 4}\right) \text{ into eqn of curve,} \\ {25 \over 4} & = k \left(k^2 \over 4\right)^{1 \over 2} - {k^2 \over 4} \\ {25 \over 4} & = k \left(k \over 2\right) - {k^2 \over 4} \\ {25 \over 4} & = {k^2 \over 2} - {k^2 \over 4} \\ {25 \over 4} & = {2k^2 \over 4} - {k^2 \over 4} \\ {25 \over 4} & = {2k^2 - k^2 \over 4} \\ 25 & = 2k^2 - k^2 \\ 25 & = k^2 \\ \pm \sqrt{25} & = k \\ \\ k & = 5 \text{ or } k = -5 \text{ (Reject, since } k > 0) \end{align}

(ii)

\begin{align} {dy \over dx} & = {k \over 2\sqrt{x}} - 1 \\ \\ \text{As } x & \rightarrow \infty, {k \over 2\sqrt{x}} \rightarrow 0 \\ \therefore \text{As } x & \rightarrow \infty, {dy \over dx} \rightarrow -1 \\ \\ \therefore \text{Gradient of } l & = -1 \\ \\ \\ y & = mx + c \\ y & = (-1)x + c \\ y & = -x + c \\ \\ \text{Using } & O(0, 0), \\ 0 & = -0 + c \\ 0 & = c \\ \\ \therefore \text{Eqn of } l: & \phantom{0} y = -x \end{align}

(i)

\begin{align} u & = x^{n + 1} - 1 &&& v & = x - 1 \\ {du \over dx} & = (n + 1)x^n &&& {dv \over dx} & = 1 \end{align} \begin{align} {d \over dx} \left( {x^{n + 1} - 1 \over x - 1} \right) & = {(x - 1)[(n + 1)x^n] - (x^{n + 1} - 1)(1) \over (x - 1)^2} \\ & = { (n + 1)(x^{n + 1} - x^n) - (x^{n + 1} - 1) \over (x - 1)^2} \\ & = { (n + 1)x^{n + 1} - (n + 1)x^n - x^{n + 1} + 1 \over (x - 1)^2} \\ & = { n x^{n + 1} - (n + 1)x^n + 1 \over (x - 1)^2 } \\ \\ {d \over dx} (1 + x + x^2 + ... + x^{n - 1} + x^n) & = 0 + 1 + 2x + ... + (n - 1)x^{n - 2} + n x^{n - 1} \\ \\ \therefore 1 + 2x + ... +(n - 1)x^{n - 2} + nx^{n - 1} & = { n x^{n + 1} - (n + 1)x^n + 1 \over (x - 1)^2 } \end{align}

(ii)

\begin{align} \text{From } & \text{(i),} \\ 1 + 2x + ... +(n - 1)x^{n - 2} + nx^{n - 1} & = { n x^{n + 1} - (n + 1)x^n + 1 \over (x - 1)^2 } \\ \\ \text{Let } & x = 3 \text{ and } n = 10, \\ 1 + 2(3) + ... + 9(3)^8 + 10 (3)^9 & = {(10)(3)^{11} - (11)(3)^{10} + 1 \over (3 - 1)^2} \\ & = 280 \phantom{.} 483 \end{align}