(a)

\begin{align} u & = x &&& v & = (4x - 3)^5 \\ {du \over dx} & = 1 &&& {dv \over dx} & = \underbrace{ 5(4x - 3)^4 . (4) }_\text{Chain rule} \\ & &&& & = 20(4x - 3)^4 \end{align} \begin{align} {d \over dx} [x (4x - 3)^5 ] & = (x)[20 (4x - 3)^4] + (4x - 3)^5 (1) \\ & = 20x (4x - 3)^4 + (4x - 3)^5 \\ & = (4x - 3)^4 [20x + (4x - 3)] \\ & = (4x - 3)^4 (20x + 4x - 3) \\ & = (4x - 3)^4 (24x - 3) \\ & = 3(4x - 3)^4 (8x - 1) \end{align}

(b)

\begin{align} u & = 7 - x^2 &&& v & = 1 - x \\ {du \over dx} & = - 2x &&& {dv \over dx} & = -1 \end{align} \begin{align} {d \over dx} [(7 - x^2)(1 - x)] & = (7 - x^2)(-1) + (1 - x)(-2x) \\ & = - 7 + x^2 - 2x + 2x^2 \\ & = 3x^2 - 2x - 7 \end{align}

(c)

\begin{align} u & = 10x^2 + 9 &&& v & = (10x + 9)^2 \\ {du \over dx} & = 20x &&& {dv \over dx} & = \underbrace{ 2(10x + 9)(10)}_\text{Chain rule} \\ & &&& & = 20(10x + 9) \end{align} \begin{align} {d \over dx} [(10x^2 + 9)(10x + 9)^2] & = (10x^2 + 9)[20(10x + 9)] + (10x + 9)^2 (20x) \\ & = 20(10x^2 + 9)(10x + 9) + 20x (10x + 9)^2 \\ & = 20(10x + 9) [ (10x^2 + 9) + x(10x + 9) ] \\ & = 20(10x + 9) (10x^2 + 9 + 10x^2 + 9x) \\ & = 20(10x + 9) (20x^2 + 9x + 9) \end{align}

(d)

\begin{align} u & = 6x^3 &&& v & = \sqrt{1 - x} \\ & &&& & = (1 - x)^{1 \over 2} \\ {du \over dx} & = 18x^2 &&& {dv \over dx} & = \underbrace{{1 \over 2}(1 - x)^{-{1 \over 2}} . (-1)}_\text{Chain rule} \\ & &&& & = -{1 \over 2} \left(1 \over \sqrt{1 - x}\right) \\ & &&& & = - {1 \over 2 \sqrt{1 - x}} \end{align} \begin{align} {d \over dx} (6x^3 \sqrt{1 - x}) & = (6x^3) \left(-{1 \over 2\sqrt{1 - x}}\right) + (\sqrt{1 - x})(18x^2) \\ & = - {6 x^3 \over 2 \sqrt{1 - x} }+ {18 x^2 \sqrt{1 - x} \over 1} \times {\sqrt{1 - x} \over \sqrt{1 - x}} \\ & = { - 3x^3 \over \sqrt{1 - x} } + {18x^2 (1 - x) \over \sqrt{1 - x}}\\ & = {-3x^3 + 18x^2 - 18x^3 \over \sqrt{1 - x}} \\ & = {18x^2 - 21x^3 \over \sqrt{1 - x}} \end{align}

(e)

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

(f)

\begin{align} u & = 8x - 5 &&& v & = \sqrt[3]{(5x^3 + 8)^2} \\ & &&& & = (5x^3 + 8)^{2 \over 3} \\ \\ {du \over dx} & = 8 &&& {dv \over dx} & = \underbrace{{2 \over 3}(5x^3 + 8)^{-{1 \over 3}}.(15x^2)}_\text{Chain rule} \\ & &&& & = 10x^2 (5x^3 + 8)^{-{1 \over 3}} \\ & &&& & = {10x^2 \over \sqrt[3]{5x^3 + 8}} \end{align} \begin{align} {d \over dx} \left[ (8x - 5) \sqrt[3]{ (5x^3 + 8)^2 } \right] & = (8x - 5) \left(10x^2 \over \sqrt[3]{5x^3 + 8}\right) + \sqrt[3]{5x^3 + 8)^2} (8) \\ & = {10x^2 (8x - 5) \over \sqrt[3]{5x^3 + 8} } + 8 \sqrt[3]{ (5x^3 + 8)^2 } \\ & = {80x^3 - 50x^2 \over (5x^3 + 8)^{1 \over 3} } + 8(5x^3 + 8)^{2 \over 3} \times { (5x^3 + 8)^{1 \over 3} \over (5x^3 + 8)^{1 \over 3} } \\ & = {80x^3 - 50x^2 \over (5x^3 + 8)^{1 \over 3} } + { 8(5x^3 + 8) \over (5x^3 + 8)^{1 \over 3} } \\ & = {80x^3 - 50x^2 + 40x^3 + 64 \over (5x^3 + 8)^{1 \over 3} } \\ & = {120x^3 - 50x^2 + 64 \over \sqrt[3]{5x^3 + 8} } \end{align}

(a)

\begin{align} t^3 (t + 1) (t - 7)^8 & = (t^4 + t^3)(t - 7)^8 \end{align}
\begin{align} u & = t^4 + t^3 &&& v & = (t - 7)^8 \\ {du \over dt} & = 4t^3 + 3t^2 &&& {dv \over dt} & = \underbrace{8(t - 7)^7. (1)}_\text{Chain rule} \\ & &&& & = 8(t - 7)^7 \end{align} \begin{align} {d \over dt} [ (t^4 + t^3) (t - 7)^8 ] & = (t^4 + t^3)[8(t - 7)^7] + (t - 7)^8 (4t^3 + 3t^2) \\ & = 8(t^4 + t^3) (t - 7)^7 + (t - 7)^8 (4t^3 + 3t^2) \\ & = 8t^3 (t + 1) (t - 7)^7 + t^2 (t - 7)^8 (4t + 3) \\ & = t^2 (t - 7)^7 [ 8t(t + 1) + (t - 7)(4t + 3) ] \\ & = t^2 (t - 7)^7 ( 8t^2 + 8t + 4t^2 + 3t - 28t - 21 ) \\ & = t^2 (t - 7)^7 ( 12t^2 -17t - 21) \end{align}

(b)

\begin{align} {1 \over t}(5t - 2)(6 - t^2)^9 & = t^{-1} (5t - 2) (6 - t^2)^9 \\ & = (5 - 2t^{-1}) (6 - t^2)^9 \end{align}
\begin{align} u & = 5 - 2t^{-1} &&& v & = (6 - t^2)^9 \\ {du \over dt} & = - 2(-1)t^{-2} &&& {dv \over dt} & = \underbrace{ 9(6 - t^2)^8 . (-2t) }_\text{Chain rule} \\ & = 2 t^{-2} &&& & = - 18t (6 - t^2)^8 \end{align} \begin{align} {d \over dt} [(5 - 2t^{-1})(6 - t^2)^9] & = (5 - 2t^{-1}) [-18t (6 -t^2)^8 ] + (6 - t^2)^9 (2t^{-2}) \\ & = - 18t (5 - 2t^{-1}) (6 - t^2)^8 + (6 - t^2)^9 (2t^{-2}) \\ & = (6 - t^2)^8 [ -18t (5 - 2t^{-1}) + (6 - t^2)(2t^{-2}) ] \\ & = (6 - t^2)^8 ( -90t + 36 + 12t^{-2} - 2 ) \\ & = (6 - t^2)^8 ( -90t + 34 + 12t^{-2} ) \\ & = 2(6 - t^2)^8 (-45t + 17 + 6t^{-2}) \\ & = 2(6 - t^2)^8 \left(-45t + 17 + {6 \over t^2} \right) \\ & = 2(6 - t^2)^8 \left( {- 45t^3 + 17t^2 + 6 \over t^2} \right) \\ & = {2 \over t^2} (6 - t^2)^8 (-45t^3 + 17t^2 + 6) \end{align}

\begin{align} u & = 4x + 1 &&& v & = \sqrt{2x - 1} \\ & &&& & = (2x - 1)^{1 \over 2} \\ {du \over dx} & = 4 &&& {dv \over dx} & = \underbrace{ {1 \over 2} (2x - 1)^{-{1 \over 2}} . (2) }_\text{Chain rule} \\ & &&& & = (2x - 1)^{-{1 \over 2}} \\ & &&& & = {1 \over \sqrt{2x - 1}} \end{align} \begin{align} {dy \over dx} & = (4x + 1) \left(1 \over \sqrt{2x - 1}\right) + (\sqrt{2x - 1})(4) \\ & = {4x + 1 \over \sqrt{2x - 1}} + 4 \sqrt{2x - 1} \times {\sqrt{2x - 1} \over \sqrt{2x - 1}} \\ & = {4x + 1 \over \sqrt{2x - 1}} + {4 (2x - 1) \over \sqrt{2x - 1}} \\ & = {4x + 1 + 4(2x - 1) \over \sqrt{2x - 1}} \\ & = {4x + 1 + 8x - 4 \over \sqrt{2x - 1}} \\ & = {12x - 3 \over \sqrt{2x - 1}} \end{align}

\begin{align} y & = a(5x - 1)^3 (bx + 1) \\ & = (5x - 1)^3 (abx + a) \\ \\ \text{Using } & (0, - 1), \phantom{000000} [y \text{-intercept of } - 1] \\ -1 & = (0 - 1)^3 (0 + a) \\ -1 & = (-1)(a) \\ -1 & = - a \\ 1 & = a \\ \\ y & = (5x - 1)^3 (bx + 1) \end{align}
\begin{align} u & = (5x - 1)^3 &&& v & = bx + 1 \\ {du \over dx} & = \underbrace{ 3(5x - 1)^2 . (5)}_\text{Chain rule} &&& {dv \over dx} & = b \\ & = 15(5x - 1)^2 \end{align} \begin{align} {dy \over dx} & = (5x - 1)^3 (b) + (bx + 1)[15(5x - 1)^2] \\ & = b (5x - 1)^3 + 15(bx + 1) (5x - 1)^2 \\ \\ \text{When } & x = 0 \text{ and } {dy \over dx} = 10, \\ 10 & = b (0 - 1)^3 + 15(0 + 1) (-1)^2 \\ 10 & = b(-1) + 15(1)(1) \\ 10 & = -b + 15 \\ b & = 15 - 10 \\ b & = 5 \\ \\ \\ \therefore a & = 1, b = 5 \end{align}

\begin{align} u & = {1 \over x} &&& v & = (x^2 - 2)^4 \\ & = x^{-1} \\ {du \over dx} & = -x^{-2} &&& {dv \over dx} & = \underbrace{4(x^2 - 2)^3 . (2x)}_\text{Chain rule} \\ & = - {1 \over x^2} &&& & = 8x(x^2 - 2)^3 \end{align} \begin{align} {dy \over dx} & = \left(1 \over x\right)[8x(x^2 - 2)^3] + (x^2 - 2)^4 \left(-{1 \over x^2}\right) \\ & = {8x(x^2 - 2)^3 \over x} - {(x^2 - 2)^4 \over x^2} \\ & = {8x^2(x^2 - 2)^3 \over x^2} - {(x^2 - 2)^4 \over x^2} \\ & = {8x^2 (x^2 - 2)^3 - (x^2 - 2)^4 \over x^2 } \\ & = {(x^2 - 2)^3 [8x^2 - (x^2 - 2)] \over x^2} \\ & = {(x^2 - 2)^3 (8x^2 - x^2 + 2) \over x^2} \\ & = {(x^2 - 2)^3 (7x^2 + 2) \over x^2} \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = {(x^2 - 2)^3 (7x^2 + 2) \over x^2} \\ 0 & = (x^2 - 2)^3 (7x^2 + 2) \end{align}
\begin{align} (x^2 - 2)^3 & = 0 && \text{ or } & 7x^2 + 2 & = 0 \\ x^2 - 2 & = \sqrt[3]{0} &&& 7x^2 & = -2 \\ x^2 - 2 & = 0 &&& x^2 & = -{2 \over 7} \\ x^2 & = 2 &&& x & = \pm \sqrt{-{2 \over 7}} \phantom{0} \text{ (No real roots}) \\ x & = \pm \sqrt{2} \end{align} \begin{align} \text{Substitute } & x = \sqrt{2} \text{ into eqn of curve,} &&& \text{Substitute } & x = -\sqrt{2} \text{ into eqn of curve,} \\ y & = {1 \over \sqrt{2}} (2 - 2)^4 &&& y & = {1 \over -\sqrt{2}} (2- 2)^4 \\ y & = 0 &&& y & = 0 \\ \\ \therefore & \phantom{.} (\sqrt{2}, 0) &&& \therefore & \phantom{.} (-\sqrt{2}, 0) \end{align}

\begin{align} u & = x &&& v & = (5x - 2)^7 \\ {du \over dx} & = 1 &&& {dv \over dx} & = \underbrace{7(5x - 2)^6 . (5)}_\text{Chain rule} \\ & &&& & = 35(5x - 2)^6 \\ \\ \\ u & = 3x^2 - 1 &&& v & = (1 - 4x^2)^5 \\ {du \over dx} & = 6x &&& {dv \over dx} & = \underbrace{5(1 - 4x^2)^4 . (-8x) }_\text{Chain rule} \\ & &&& & = -40x(1 - 4x^2)^4 \end{align}
\begin{align} f'(x) & = (x)[35(5x - 2)^6] + (5x - 2)^7 (1) - \{ (3x^2 - 1)[-40x(1 - 4x^2)^4 ] + (1 - 4x^2)^5 (6x) \} \\ & = 35x (5x - 2)^6 + (5x - 2)^7 - [ - 40x (3x^2 - 1)(1 - 4x^2)^4 + 6x(1 - 4x^2)^5 ] \\ & = 35x (5x - 2)^6 + (5x - 2)^7 + 40x (3x^2 - 1)(1 - 4x^2)^4 + 6x(1 - 4x^2)^5 \\ \\ f'(0) & = 35(0)(0 - 2)^6 + (0 - 2)^7 + 40(0)(0 - 1)(1 - 0)^4 + 6(0)(1 - 0)^5 \\ & = -128 \phantom{0} \text{(Shown)} \end{align}

(i)

\begin{align} u & = ax + b &&& v & = (cx + d)^n \\ {du \over dx} & = a &&& {dv \over dx} & = \underbrace{n (cx + d)^{n - 1} . (c) }_\text{Chain rule} \\ & &&& & = cn (cx + d)^{n - 1} \end{align} \begin{align} {dy \over dx} & = (ax + b)[cn (cx + d)^{n - 1}] + (cx + d)^n (a) \\ & = cn (ax + b)(cx + d)^{n - 1} + a (cx + d)^n \\ & = (cx + d)^{n - 1} [ cn (ax + b) + a(cx + d) ] \\ & = (cx + d)^{n - 1} ( acn x + bcn + acx + ad ) \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = (cx + d)^{n - 1} ( acn x + bcn + acx + ad ) \end{align}
\begin{align} (cx + d)^{n - 1} & = 0 && \text{ or } & acn x + bcn + acx + ad & = 0 \\ cx + d & = 0 &&& acn x + ac x & = - bcn - ad \\ cx & = -d &&& x(acn + ac) & = -bcn - ad \\ x & = -{d \over c} &&& x & = {-bcn - ad \over acn + ac} \phantom{0} (\text{Shown}) \end{align}

(ii)

\begin{align} \text{If } c = 0, \text{ then } x & = -{d \over c} \text{ is not defined} \\ \\ \\ \text{For } x = {-bcn - ad \over acn + ac} & \text{ to be not defined, } acn + ac = 0 \\ \\ acn + ac & = 0 \\ c(an + a) & = 0 \end{align} \begin{align} c & = 0 && \text{ or } & an + a & = 0 \\ & &&& an & = -a \\ & &&& n & = {-a \over a} \\ & &&& n & = -1 \end{align}
$$ \therefore c \ne 0 \text{ and } n \ne -1 $$