(a)

\begin{align} u & = x &&& v & = (2x - 1)^3 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 3(2x - 1)^2 . (2) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 6(2x - 1)^2 \end{align} \begin{align} {d \over dx} [x(2x - 1)^3] & = (x)[6(2x - 1)^2] + (2x - 1)^3 (1) \\ & = 6x(2x - 1)^2 + (2x - 1)^3 \\ & = (2x - 1)^2 [6x + (2x - 1)] \\ & = (2x - 1)^2(6x + 2x - 1) \\ & = (2x - 1)^2(8x - 1) \end{align}

(b)

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

(c)

\begin{align} u & = 1 - 2x &&& v & = (3x + 2)^4 \\ {du \over dx} & = -2 &&& {dv \over dx} & = 4(3x + 2)^3 . (3) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 12(3x + 2)^3 \end{align} \begin{align} {d \over dx} [(1 - 2x)(3x + 2)^4] & = (1 - 2x)[12(3x + 2)^3] + (3x + 2)^4 (-2) \\ & = 12(1 - 2x)(3x + 2)^3 - 2(3x + 2)^4 \\ & = (3x + 2)^3 [12(1 - 2x) - 2(3x + 2)] \\ & = (3x + 2)^3 (12 - 24x - 6x - 4) \\ & = (3x + 2)^3 (8 - 30x) \\ & = (3x + 2)^3 (2)(4 - 15x) \\ & = 2(3x + 2)^3 (4 - 15x) \end{align}

(d)

\begin{align} u & = x^2 + 1 &&& v & = (1 + x)^2 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 2(1 + x)^1 .(1) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 2(1 + x) \end{align} \begin{align} {d \over dx} [(x^2 + 1)(1 + x)^2] & = (x^2 + 1)[2(1 + x)] + (1 + x)^2 (2x) \\ & = 2(x^2 + 1)(1 + x) + 2x(1 + x)^2 \\ & = 2(1 + x)[(x^2 + 1) + x(1 + x)] \\ & = 2(1 + x)(x^2 + 1 + x + x^2) \\ & = 2(1 + x)(2x^2 + x + 1) \end{align}

(a)

\begin{align} u & = \sqrt{x} &&& v & = (1 - x)^2 \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} &&& {dv \over dx} & = 2(1 - x)^1 .(-1) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2\sqrt{x}} &&& & = -2(1 - x) \end{align} \begin{align} {d \over dx} [\sqrt{x} (1 - x)^2] & = (\sqrt{x}) [-2(1 - x)] + (1 - x)^2 \left(1 \over 2\sqrt{x}\right) \\ & = -2 \sqrt{x} (1 - x) + {(1 - x)^2 \over 2\sqrt{x}} \\ & = {-2 \sqrt{x} (1 - x) \over 1} + {(1 - x)^2 \over 2\sqrt{x}} \\ & = {-2 \sqrt{x} (1 - x)(2\sqrt{x}) \over 2\sqrt{x}} + {(1 - x)^2 \over 2\sqrt{x}} \\ \\ & = {-4x(1 - x) \over 2\sqrt{x}} + {(1 - x)^2 \over 2\sqrt{x}} \\ \\ & = {-4x(1 - x) + (1 - x)^2 \over 2\sqrt{x}} \\ & = {(1 - x) [-4x + (1 - x)] \over 2\sqrt{x}} \\ & = {(1 - x) (-4x + 1 - x) \over 2 \sqrt{x}} \\ & = {(1 - x) (1 - 5x) \over 2\sqrt{x}} \end{align}

(b)

\begin{align} u & = x &&& v & = \sqrt{1 + 2x} \\ &&&& & = (1 + 2x)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(1 + 2x)^{-{1 \over 2}} .(2) \phantom{00000} [\text{Chain rule}] \\ &&&& & = (1 + 2x)^{-{1 \over 2}} \\ &&&& & = {1 \over \sqrt{1 + 2x}} \end{align} \begin{align} {d \over dx} [x \sqrt{1 + 2x}] & = (x) \left(1 \over \sqrt{1 + 2x}\right) + (\sqrt{1 + 2x})(1) \\ & = {x \over \sqrt{1 + 2x}} + \sqrt{1 + 2x} \\ & = {x \over \sqrt{1 + 2x}} + {\sqrt{1 + 2x} \over 1} \\ & = {x \over \sqrt{1 + 2x}} + {\sqrt{1 + 2x}(\sqrt{1 + 2x}) \over \sqrt{1 + 2x}} \\ & = {x \over \sqrt{1 + 2x}} + {1 + 2x \over \sqrt{1 + 2x}} \\ & = {x + 1 + 2x \over \sqrt{1 + 2x}} \\ & = {3x + 1 \over \sqrt{1 + 2x}} \end{align}

(c)

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

(d)

\begin{align} u & = 4x - 1 &&& v & = \sqrt{3x^2 + 1} \\ &&&& & = (3x^2 + 1)^{1 \over 2} \\ {du \over dx} & = 4 &&& {dv \over dx} & = {1 \over 2}(3x^2 + 1)^{-{1 \over 2}} . (6x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 3x(3x^2 + 1)^{-{1 \over 2}} \\ &&&& & = {3x \over \sqrt{3x^2 + 1}} \end{align} \begin{align} {d \over dx} \left[ (4x - 1) \sqrt{3x^2 + 1} \right] & = (4x - 1) \left(3x \over \sqrt{3x^2 + 1}\right) + \left( \sqrt{3x^2 + 1} \right) (4) \\ & = {3x(4x - 1) \over \sqrt{3x^2 + 1}} + 4 \sqrt{3x^2 + 1} \\ & = {3x(4x - 1) \over \sqrt{3x^2 + 1}} + { 4 \sqrt{3x^2 + 1} \over 1} \\ & = {3x(4x - 1) \over \sqrt{3x^2 + 1}} + { 4 \sqrt{3x^2 + 1} \left(\sqrt{3x^2 + 1}\right) \over \sqrt{3x^2 + 1}} \\ & = {3x(4x - 1) \over \sqrt{3x^2 + 1}} + {4 (3x^2 + 1) \over \sqrt{3x^2 + 1}} \\ & = {3x(4x - 1) + 4(3x^2 + 1) \over \sqrt{3x^2 + 1}} \\ & = {12x^2 - 3x + 12x^2 + 4 \over \sqrt{3x^2 + 1}} \\ & = {24x^2 - 3x + 4 \over \sqrt{3x^2 + 1}} \end{align}

(e)

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

(f)

\begin{align} u & = 2\sqrt{x} + 3 &&& v & = \sqrt{1 - 4x} \\ & = 2x^{1 \over 2} + 3 &&& & = (1 - 4x)^{1 \over 2} \\ {du \over dx} & = 2\left({1 \over 2}\right)x^{-{1 \over 2}} &&& {dv \over dx} & = {1 \over 2}(1 - 4x)^{-{1 \over 2}} . (-4) \phantom{00000} [\text{Chain rule}] \\ & = x^{-{1 \over 2}} &&& & = -2(1 - 4x)^{-{1 \over 2}} \\ & = {1 \over \sqrt{x}} &&& & = -{2 \over \sqrt{1 - 4x}} \end{align} \begin{align} {d \over dx} \left[ \left( 2\sqrt{x} + 3 \right) \sqrt{1 - 4x} \right] & = \left( 2\sqrt{x} + 3 \right) \left( -{2 \over \sqrt{1 - 4x}} \right) + \sqrt{1 - 4x} \left(1 \over \sqrt{x}\right) \\ & = {-2(2\sqrt{x} + 3) \over \sqrt{1 - 4x}} + {\sqrt{1 - 4x} \over \sqrt{x}} \\ & = {-2(2\sqrt{x} + 3)(\sqrt{x}) \over \sqrt{x}\sqrt{1 - 4x}} + {\sqrt{1 - 4x} \sqrt{1 - 4x} \over \sqrt{x}\sqrt{1 - 4x}} \\ & = { -2\sqrt{x} (2\sqrt{x} + 3) + 1 - 4x \over \sqrt{x}\sqrt{1 - 4x}} \\ & = { -4x - 6\sqrt{x} + 1 - 4x \over \sqrt{x(1 - 4x)}} \\ & = { 1 - 6\sqrt{x} - 8x \over \sqrt{x(1 - 4x)}} \end{align}

(a)

\begin{align} u & = {1 \over x} (x + 1) &&& v & = (x + 2)^3 \\ & = 1 + {1 \over x} \\ & = 1 + x^{-1} \\ {du \over dx} & = (-1)x^{-2} &&& {dv \over dx} & = 3(x + 2)^2 . (1) \phantom{00000} [\text{Chain rule}] \\ & = -{1 \over x^2} &&& & = 3(x + 2)^2 \end{align} \begin{align} {d \over dx} \left[\left( 1 + {1 \over x} \right)(x + 2)^3\right] & = \left( 1 + {1 \over x} \right) [3(x + 2)^2 ] + (x + 2)^3 \left( -{1 \over x^2} \right) \\ & = 3 \left( 1 + {1 \over x} \right) (x + 2)^2 - {1 \over x^2} (x + 2)^3 \\ & = (x + 2)^2 \left[ 3 \left(1 + {1 \over x}\right) - {1 \over x^2}(x + 2) \right] \\ & = (x + 2)^2 \left( 3 + {3 \over x} - {1 \over x} - {2 \over x^2} \right) \\ & = (x + 2)^2 \left(3 + {2 \over x} - {2 \over x^2} \right) \end{align}

(b)

\begin{align} u & = x^2 (x - 1) &&& v & = \sqrt{5 + 6x} \\ & = x^3 - x^2 &&& & = (5 + 6x)^{1 \over 2} \\ {du \over dx} & = 3x^2 - 2x &&& {dv \over dx} & = {1 \over 2}(5 + 6x)^{-{1 \over 2}} . (6) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 3(5 + 6x)^{-{1 \over 2}} \\ &&&& & = {3 \over \sqrt{5 + 6x}} \end{align} \begin{align} {d \over dx} \left[ (x^3 - x^2)\sqrt{5 + 6x} \right] & = (x^3 - x^2) \left( {3 \over \sqrt{5 + 6x}} \right) + \left( \sqrt{5 + 6x} \right)(3x^2 - 2x) \\ & = {3(x^3 - x^2) \over \sqrt{5 + 6x}} + (3x^2 - 2x)\sqrt{5 + 6x} \\ & = {3(x^3 - x^2) \over \sqrt{5 + 6x}} + { (3x^2 - 2x)\sqrt{5 + 6x} \over 1} \\ & = {3(x^3 - x^2) \over \sqrt{5 + 6x}} + {(3x^2 - 2x)\sqrt{5 + 6x}(\sqrt{5 + 6x}) \over \sqrt{5 + 6x}} \\ & = {3x^3 - 3x^2 \over \sqrt{5 + 6x}} + {(3x^2 - 2x)(5 + 6x) \over \sqrt{5 + 6x}} \\ & = {3x^3 - 3x^2 + (3x^2 - 2x)(5 + 6x) \over \sqrt{5 + 6x}} \\ & = {3x^3 - 3x^2 + 15x^2 + 18x^3 - 10x - 12x^2 \over \sqrt{5 + 6x}} \\ & = {21x^3 - 10x \over \sqrt{5 + 6x}} \\ & = {x(21x^2 - 10) \over \sqrt{5 + 6x}} \end{align}

\begin{align} u & = x &&& v & = \sqrt{3 - x^2} \\ &&&& & = (3 - x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(3 - x^2)^{-{1 \over 2}} . (-2x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = -x (3 - x^2)^{-{1 \over 2}} \\ &&&& & = {-x \over \sqrt{3 - x^2}} \end{align} \begin{align} {dy \over dx} & = (x)\left({-x \over \sqrt{3 - x^2}}\right) + \left(\sqrt{3 - x^2} \right) (1) \\ & = {-x^2 \over \sqrt{3 - x^2}} + \sqrt{3 - x^2} \\ & = {-x^2 \over \sqrt{3 - x^2}} + {\sqrt{3 - x^2} \over 1} \\ & = {-x^2 \over \sqrt{3 - x^2}} + {\sqrt{3 - x^2}\sqrt{3 - x^2} \over \sqrt{3 - x^2}} \\ & = {-x^2 \over \sqrt{3 - x^2}} + {3 - x^2 \over \sqrt{3 - x^2}} \\ & = {-x^2 + 3 - x^2 \over \sqrt{3 - x^2}} \\ & = {3 - 2x^2 \over \sqrt{3 - x^2}} \phantom{0} \text{ (Shown)} \end{align}

\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) \phantom{00000} [\text{Chain rule}] \\ &&&& & = {1 \over 2} (x - 1)^{-{1 \over 2}} \\ &&&& & = {1 \over 2\sqrt{x - 1}} \end{align} \begin{align} {dy \over dx} & = (x) \left( {1 \over 2\sqrt{x - 1}} \right) + \left( \sqrt{x - 1} \right) (1) \\ & = {x \over 2\sqrt{x - 1}} + \sqrt{x - 1} \\ & = {x \over 2\sqrt{x - 1}} + {\sqrt{x - 1} \over 1} \\ & = {x \over 2\sqrt{x - 1}} + {\sqrt{x - 1} (2\sqrt{x - 1}) \over 2\sqrt{x - 1}} \\ & = {x \over 2\sqrt{x - 1}} + {2(x - 1) \over 2\sqrt{x - 1}} \\ & = {x + 2x - 2 \over 2\sqrt{x - 1}} \\ & = {3x - 2 \over 2\sqrt{x - 1}} \\ \\ \text{When } & x = 5, \\ {dy \over dx} & = {3(5) - 2 \over 2\sqrt{(5) - 1}} \\ & = 3{1 \over 4} \end{align}

The points where the curve the x-axis are the x-intercepts, where y = 0.

\begin{align} u & = (x + 1)^3 &&& v & = x - 1 \\ {du \over dx} & = 3(x + 1)^2 . (1) &&& {dv \over dx} & = 1 \\ & = 3(x + 1)^2 \end{align} \begin{align} {dy \over dx} & = (x + 1)^3 (1) + (x - 1)[3(x + 1)^2] \\ & = (x + 1)^3 + 3(x - 1)(x + 1)^2 \\ & = (x + 1)^2 [(x + 1) + 3(x - 1)] \\ & = (x + 1)^2 (x + 1 + 3x - 3) \\ & = (x + 1)^2 (4x - 2) \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = (x + 1)^3 (x - 1) \end{align} \begin{align} (x + 1)^3 & = 0 && \text{ or } & x - 1 & =0 \\ x + 1 & = 0 &&& x & = 1 \\ x & = -1 \\ \\ \text{Substitute} & \text{ into } {dy \over dx}, &&& \text{Substitute} & \text{ into } {dy \over dx}, \\ {dy \over dx} & = (-1 + 1)^2 [4(-1) - 2] &&& {dy \over dx} & = (1 + 1)^2 [4(1) - 2] \\ & = 0 &&& & = 8 \end{align}

\begin{align} u & = \sqrt{x} &&& v & = (x - 4)^4 \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} &&& {dv \over dx} & = 4(x - 4)^3 . (1) \\ & = {1 \over 2\sqrt{x}} &&& & = 4(x - 4)^3 \end{align} \begin{align} {dy \over dx} & = (\sqrt{x}) [4(x - 4)^3] + (x - 4)^4 \left(1 \over 2\sqrt{x}\right) \\ & = 4\sqrt{x} (x - 4)^3 + {1 \over 2\sqrt{x}}(x - 4)^4 \\ & = (x - 4)^3 \left[ 4\sqrt{x} + {1 \over 2\sqrt{x}}(x - 4) \right] \\ & = (x - 4)^3 \left[ {4\sqrt{x} \over 1} + {x - 4 \over 2\sqrt{x}} \right] \\ & = (x - 4)^3 \left[ 4\sqrt{x} + {x - 4 \over 2\sqrt{x}} \right] \\ & = (x - 4)^3 \left[ {4\sqrt{x}(2\sqrt{x}) \over 2\sqrt{x}} + {x - 4 \over 2\sqrt{x}} \right] \\ & = (x - 4)^3 \left( {8x + x - 4 \over 2\sqrt{x}} \right) \\ & = (x - 4)^3 \left( {9x - 4 \over 2\sqrt{x}} \right) \\ & = {(x - 4)^3 (9x - 4) \over 2\sqrt{x}} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {(x - 4)^3 (9x - 4) \over 2\sqrt{x}} \\ \therefore 0 & = (x - 4)^3 (9x - 4) \\ \end{align} \begin{align} (x - 4)^3 & = 0 && \text{ or } & 9x - 4 & = 0 \\ x - 4 & = 0 &&& 9x & = 4 \\ x & = 4 &&& x & = {4 \over 9} \end{align}

\begin{align} y & = {x^2 - 1 \over x^2 + 1} \\ & = (x^2 - 1)\left(1 \over x^2 + 1\right) \\ & = (x^2 - 1)(x^2 + 1)^{-1} \end{align}
\begin{align} u & = x^2 - 1 &&& v & = (x^2 + 1)^{-1} \\ {du \over dx} & = 2x &&& {dv \over dx} & = (-1)(x^2 + 1)^{-2} . (2x) \\ & &&& & = -2x (x^2 + 1)^{-2} \\ & &&& & = {-2x \over (x^2 + 1)^2} \end{align} \begin{align} {dy \over dx} & = (x^2 - 1) \left[ {-2x \over (x^2 + 1)^2} \right] + (x^2 + 1)^{-1} (2x) \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + \left(1 \over x^2 + 1\right)(2x) \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + {2x \over x^2 + 1} \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + {2x (x^2 + 1) \over (x^2 + 1)(x^2 + 1)} \\ & = {-2x^3 + 2x \over (x^2 + 1)^2} + {2x^3 + 2x \over (x^2 + 1)^2} \\ & = {-2x^3 + 2x + 2x^3 + 2x \over (x^2 + 1)^2} \\ & = {4x \over (x^2 + 1)^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {4x \over (x^2 + 1)^2} \\ \therefore 0 & = 4x \\ 0 & = x \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = {0^2 - 1 \over 0^2 + 1} \\ & = -1 \end{align}

\begin{align} u & = x &&& v & = \sqrt{4 - x^2} \\ & &&& & = (4 - x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (4 - x^2)^{-{1 \over 2}} . (-2x) \\ & &&& & = -x (4 - x^2)^{-{1 \over 2}} \\ & &&& & = {-x \over \sqrt{4 - x^2}} \end{align} \begin{align} {dy \over dx} & = (x) \left( {-x \over \sqrt{4 - x^2}} \right) + \left( \sqrt{4 - x^2}\right) (1) \\ & = { -x^2 \over \sqrt{4 - x^2}} + \sqrt{4 - x^2} \\ \\ \\ y & = x\phantom{00}\text{--- (1)} \\ \\ y & = x\sqrt{4 - x^2}\phantom{00}\text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x & = x\sqrt{4 - x^2} \\ x - x\sqrt{4 - x^2} & = 0 \\ (x)\left(1 - \sqrt{4 - x^2}\right) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & 1 - \sqrt{4 - x^2} & = 0 \\ & &&& 1 & = \sqrt{4 - x^2} \\ & &&& 1^2 & = 4 - x^2 \\ & &&& 1 & = 4 - x^2 \\ & &&& x^2 & = 4 - 1 \\ & &&& x^2 & = 3 \\ & &&& x & = \pm \sqrt{3} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = { -(0)^2 \over \sqrt{4 - (0)^2} } + \sqrt{4 - (0)^2} &&& {dy \over dx} & = { -(\pm \sqrt{3})^2 \over \sqrt{4 - (\pm \sqrt{3})^2} } + \sqrt{4 - (\pm \sqrt{3})^2} \\ & = 2 &&& & = -2 \end{align}

Recall that parallel lines have the same gradient

\begin{align} 5x - y & = 0 \\ y & = 5x \\ y & = 5x + 0 \\ \\ \text{Comparing with } & y = mx + c, \\ m & = 5 \\ \\ \therefore \text{Gradient of tangent} & = 5 \end{align}
\begin{align} u & = 3x - 1 &&& v & = x - 2 \\ {du \over dx} & = 3 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = (3x - 1)(1) + (x - 2)(3) \\ & = 3x - 1 + 3x - 6 \\ & = 6x - 7 \\ \\ \text{When } & {dy \over dx} = 5, \\ 5 & = 6x - 7 \\ 12 & = 6x \\ {12 \over 6} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = [3(2) - 1](2 - 2) \\ & = 0 \\ \\ \therefore & \phantom{.} (2, 0) \end{align}

Since the curve meets the x-axis at point A, the y-coordinate of A is 0.

\begin{align} u & = x - a &&& v & = \sqrt{x - b} \\ & &&& & = (x - b)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (x - b)^{-{1 \over 2}} . (1) \\ & &&& & = {1 \over 2} (x - b)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x - b}} \end{align} \begin{align} {dy \over dx} & = (x - a) \left(1 \over 2\sqrt{x - b}\right) + \left( \sqrt{x - b} \right) (1) \\ & = {x - a \over 2\sqrt{x - b}} + \sqrt{x - b} \\ & = {x - a \over 2\sqrt{x - b}} + {\sqrt{x - b} \over 1} \\ & = {x - a \over 2\sqrt{x - b}} + {\sqrt{x - b} (2\sqrt{x - b}) \over 2\sqrt{x - b}} \\ & = {x - a \over 2\sqrt{x - b}} + {2(x - b) \over 2\sqrt{x - b}} \\ & = {x - a + 2(x - b) \over 2\sqrt{x - b}} \\ & = {x - a + 2x - 2b \over 2\sqrt{x - b}} \\ & = {3x - a - 2b \over 2\sqrt{x - b}} \\ \\ \\ \text{When } & x = b + 1, \\ {dy \over dx} & = {3(b + 1) - a - 2b \over 2\sqrt{b + 1 - b} } \\ & = {3b + 3 - a - 2b \over 2\sqrt{1}} \\ & = {b + 3 - a \over 2} \\ \\ \text{Substitute } & A(b + 1, 0) \text{ into eqn of curve,} \\ 0 & = (b + 1 - a)\sqrt{b + 1 - b} \\ 0 & = (b + 1 - a) \sqrt{1} \\ 0 & = (b + 1 - a)(1) \\ 0 & = b + 1 - a \\ a & = b + 1 \\ \\ \therefore {dy \over dx} & = {b + 3 - (b + 1) \over 2} \\ & = {b + 3 - b - 1 \over 2} \\ & = {2 \over 2} \\ & = 1 \phantom{0} \text{ (Shown)} \end{align}

\begin{align} {d \over dx} [f(x) g(x) h(x)] & = {d \over dx} [f(x) g(x)] [h(x)] \end{align} \begin{align} \text{Consider } \phantom{00} u & = f(x) &&& v & = g(x) \\ {du \over dx} & = f'(x) &&& {dv \over dx} & = g'(x) \\ \\ \\ \therefore u' & = f(x) g(x) &&& v' & = h(x) \\ {du' \over dx} & = f(x)g'(x) + g(x)f'(x) &&& {dv' \over dx} & = h'(x) \end{align} \begin{align} \therefore {d \over dx} [f(x) g(x) h(x)] & = [f(x) g(x)][h'(x)] + [h(x)][f(x)g'(x) + g(x)f'(x)] \\ & = f(x) g(x) h'(x) + f(x) h(x) g'(x) + g(x) h(x) f'(x) \\ & = fg {dh \over dx} + fh {dg \over dx} + gh {df \over dx} \end{align}

(i)

$$ x - a $$

(ii)

\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ f(x) & = (x - a) Q(x) + 0 \\ f(x) & = (x - a) Q(x), \text{ where } Q(x) \text{ denotes a polynomial} \end{align} \begin{align} u & = x - a &&& v & = Q(x) \\ {du \over dx} & = 1 &&& {dv \over dx} & = Q'(x) \end{align} \begin{align} f'(x) & = (x - a)Q'(x) + Q(x) . (1) \\ & = (x - a)Q'(x) + Q(x) \\ \\ f'(a) & = (a - a)Q'(a) + Q(a) \\ & = 0[Q'(a)] + Q(a) \\ & = 0 + Q(a) \\ & = Q(a) \\ \\ \text{If } & f'(a) = 0, \\ Q(a) & = 0 \\ \\ \implies (x - a) & \text{ is a factor of } Q(x) \\ \therefore (x - a) & \text{ is a repeated factor of } f(x) \end{align}