(a)

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

(b)

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

(c)

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

(d)

\begin{align} u & = 3x^2 &&& v & = 1 - 4x \\ {du \over dx} & = 6x &&& {dv \over dx} & = -4 \end{align} \begin{align} {d \over dx} \left( 3x^2 \over 1 - 4x \right) & = {(1 - 4x)(6x) - (3x^2)(-4) \over (1 - 4x)^2} \\ & = {6x(1 - 4x) + 4(3x^2) \over (1 - 4x)^2} \\ & = {6x - 24x^2 + 12x^2 \over (1 - 4x)^2} \\ & = {6x - 12x^2 \over (1 - 4x)^2} \end{align}

(e)

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

(f)

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

(c)

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

(d)

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

(e)

\begin{align} u & = 3x^2 &&& v & = \sqrt{2x^2 - 3} \\ & &&& & = (2x^2 - 3)^{1 \over 2} \\ {du \over dx} & = 6x &&& {dv \over dx} & = {1 \over 2} (2x^2 - 3)^{-{1 \over 2}} . (4x) \\ & &&& & = 2x(2x^2 - 3)^{-{1 \over 2}} \\ & &&& & = {2x \over \sqrt{2x^2 - 3}} \end{align} \begin{align} {d \over dx} \left( 3x^2 \over \sqrt{2x^2 - 3} \right) & = {(\sqrt{2x^2 - 3}) (6x) - (3x^2) \left(2x \over \sqrt{2x^2 - 3}\right) \over \left(\sqrt{2x^2 - 3}\right)^2 } \\ & = {(6x)\sqrt{2x^2 - 3} - {6x^3 \over \sqrt{2x^2 - 3}} \over 2x^2 - 3} \\ & = {(6x)\sqrt{2x^2 - 3} - {6x^3 \over \sqrt{2x^2 - 3}} \over 2x^2 - 3} \times {\sqrt{2x^2 - 3} \over \sqrt{2x^2 - 3}} \\ & = {(6x)\sqrt{2x^2-3}(\sqrt{2x^2 - 3}) - 6x^3 \over (2x^2 - 3)\sqrt{2x^2 - 3}} \\ & = {(6x)(2x^2 - 3) - 6x^3 \over (2x^2 - 3)\sqrt{2x^2 - 3}} \\ & = {12x^3 - 18x - 6x^3 \over (2x^2 - 3)\sqrt{2x^2 - 3}} \\ & = {6x^3 - 18x \over (2x^2 - 3)\sqrt{2x^2 - 3}} \end{align}

(f)

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

\begin{align} u & = 3x^2 &&& v & = 1 - 4x^2 \\ {du \over dx} & = 6x &&& {dv \over dx} & = -8x \end{align} \begin{align} {dy \over dx} & = {(1 - 4x^2)(6x) - (3x^2)(-8x) \over (1 - 4x^2)^2} \\ & = {6x(1 - 4x^2) + 8x(3x^2) \over (1 - 4x^2)^2} \\ & = {6x - 24x^3 + 24x^3 \over (1 - 4x^2)^2} \\ & = {6x \over (1 - 4x^2)^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {6(1) \over [1 - 4(1)^2]^2} \\ & = {2 \over 3} \end{align}

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

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

\begin{align} u & = x + 2 &&& v & = \sqrt{3x + 1} \\ & &&& & = (3x + 1)^{1 \over 2} \\ {du \over dx} & = 1 &&& {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} {dy \over dx} & = {(\sqrt{3x + 1})(1) - (x + 2) \left({3 \over 2\sqrt{3x + 1}}\right) \over \left( \sqrt{3x + 1}\right)^2} \\ & = {\sqrt{3x + 1} - {3(x + 2) \over 2\sqrt{3x + 1}} \over 3x + 1} \\ & = {\sqrt{3x + 1} - {3(x + 2) \over 2\sqrt{3x + 1}} \over 3x + 1} \times {2\sqrt{3x + 1} \over 2\sqrt{3x + 1}} \\ & = {2(3x + 1) - 3(x + 2) \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {6x + 2 - 3x - 6 \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {3x - 4 \over 2(3x + 1)\sqrt{3x + 1}} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {3(1) - 4 \over 2[3(1) + 1]\sqrt{3(1) + 1}} \\ & = -{1 \over 16} \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}

$$ y = \sqrt{1 - x \over x^2 + 3} = \left(1 - x \over x^2 + 3\right)^{1 \over 2} $$
\begin{align} u & = 1 - x &&& v & = x^2 + 3 \\ {du \over dx} & = -1 &&& {dv \over dx} & = 2x \end{align} \begin{align} \require{cancel} {dy \over dx} & = {1 \over 2}\left( {1 - x \over x^2 + 3} \right)^{-{1 \over 2}}\underbrace{{d \over dx}\left( {1 - x \over x^2 + 3} \right)}_{\text{Quotient rule}} \\ & = {1 \over 2}\left( {1 - x \over x^2 + 3} \right)^{-{1 \over 2}} \left[ { (x^2 + 3)(-1) - (1 - x)(2x) \over (x^2 + 3)^2} \right] \\ & = {1 \over 2}\left( {1 - x \over x^2 + 3} \right)^{-{1 \over 2}} \left[ { -x^2 - 3 - 2x + 2x^2 \over (x^2 + 3)^2} \right] \\ & = {1 \over 2}\left( 1 - x \over x^2 + 3 \right)^{-{1 \over 2}} \left[ {x^2 -2x - 3 \over (x^2 + 3)^2} \right] \\ & = {1 \over 2}\left( x^2 + 3 \over 1 - x \right)^{1 \over 2} \left[ {x^2 -2x - 3 \over (x^2 + 3)^2} \right] \\ & = {1 \over 2} \left[ {(x^2 + 3)^{1 \over 2} \over (1 - x)^{1 \over 2}} \right] \left[ {x^2 - 2x - 3 \over (x^2 + 3)^2} \right] \\ & = {1 \over 2} \left[ { \cancel{(x^2 + 3)^{1 \over 2}} (x^2 - 2x - 3) \over (1 - x)^{1 \over 2}\cancel{(x^2 + 3)^2}^{3 \over 2} } \right] \\ & = {1 \over 2} \left[ { x^2 - 2x - 3 \over (1 - x)^{1 \over 2} (x^2 + 3)^{3 \over 2} } \right] \\ & = {1 \over 2} \left[ { x^2 - 2x - 3 \over \sqrt{1 - x} (x^2 + 3)^{3 \over 2} }\right] \\ & = {x^2 - 2x - 3 \over 2\sqrt{1 - x} (x^2 + 3)^{3 \over 2} } \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = {x^2 - 2x - 3 \over 2\sqrt{1 - x}\sqrt{(x^2 + 3)^3}} \\ \\ \therefore 0 & = x^2 - 2x - 3 \\ 0 & = (x - 3)(x + 1) \\ \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x + 1 & = 0 \\ x & = 3 &&& x & = -1 \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)

$$ y = \sqrt[3]{x + h \over x - k} = \left(x + h \over x - k\right)^{1 \over 3} $$
\begin{align} u & = x + h &&& v & = x - k \\ {du \over dx} & = 1 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {1 \over 3} \left(x + h \over x - k\right)^{-{2 \over 3}} \left[(x - k)(1) - (x + h)(1) \over (x - k)^2 \right] \\ & = {1 \over 3} \left(x + h \over x - k\right)^{-{2 \over 3}} \left[x - k - x - h \over (x - k)^2 \right] \\ & = {1 \over 3} \left(x + h \over x - k\right)^{-{2 \over 3}} \left[- k - h \over (x - k)^2 \right] \\ & = {1 \over 3} (x + h)^{-{2 \over 3}} (x - k)^{2 \over 3} (- k - h) (x - k)^{-2} \\ & = {1 \over 3} (x + h)^{-{2 \over 3}} (x - k)^{-{4 \over 3}} (- k - h) \\ \\ \text{To show } & 3y^2 (x - k)^2 {dy \over dx} = -(k + h), \\ \text{L.H.S} & = 3y^2 (x - k)^2 {dy \over dx} \\ & = 3 \left[\left(x + h \over x - k\right)^{1 \over 3}\right]^2 (x - k)^2 \left[ {1 \over 3} (x + h)^{-{2 \over 3}} (x - k)^{-{4 \over 3}} (- k - h) \right] \\ & = \left(x + h \over x - k\right)^{2 \over 3} (x + h)^{-{2 \over 3}} (x - k)^{2 \over 3} (- k - h) \\ & = (x + h)^{2 \over 3} (x - k)^{-{2 \over 3}} (x + h)^{-{2 \over 3}} (x - k)^{2 \over 3} (- k - h) \\ & = (-k - h) \\ & = -(k + h) \\ & = \text{R.H.S} \end{align}

(ii) The y-axis is a vertical line, thus the gradient of the tangent is undefined

\begin{align} 3y^2 (x - k)^2 {dy \over dx} & = -(k + h) \\ {dy \over dx} & = {-(k + h) \over 3y^2 (x - k)^2} \\ \\ \text{For } {dy \over dx} & \text{ to be undefined,} \\ 3y^2 (x - k)^2 & = 0 \end{align} \begin{align} 3y^2 & = 0 && \text{ or } & x - k & = 0 \\ y^2 & = 0 &&& x & = k \text{ (Reject, since } h < x < k )\\ y & = 0 \\ \\ \text{Substitute } & \text{into eqn of curve,} \\ 0 & = \sqrt[3]{x + h \over x - k} \\ 0 & = {x + h \over x - k} \\ 0 & = x + h \\ -h & = x \\ \\ \therefore & \phantom{.} (-h, 0) \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}

\begin{align} "u" & = u^2 \phantom{000000} {d"u" \over dx} = 2u.u' \\ "v" & = v^3 \phantom{000000} {d"v" \over dx} = 3v^2 . v' \\ \\ {dy \over dx} & = {(v^3)(2u.u') - (u^2)(3v^2.v') \over (v^3)^2} \\ y' & = {2 (u)(u')(v^3) - 3 (u^2)(v^2)(v') \over v^6} \end{align}