(a)

\begin{align} {d \over dx} (x^{20}) & = 20x^{19} \end{align}

(b)

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

(c)

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

(d)

\begin{align} {d \over dx} x^{-{3 \over 5}} & = -{3 \over 5} x^{-{8 \over 5}} \\ & = -{3 \over 5} \left(1 \over \sqrt[5]{x^8}\right) \\ & = - {3 \over 5 \sqrt[5]{x^8}} \end{align}

(e)

\begin{align} {d \over dx} (88) & = 0 \end{align}

(f)

\begin{align} {d \over dx} (-2.9) & = 0 \end{align}

(g)

\begin{align} {d \over dx} \left( {4 \over 5}x^3 + {2 \over \sqrt{x}} - 7 \right) & = {d \over dx} \left( {4 \over 5}x^3 + 2 x^{-{1 \over 2}} - 7 \right) \\ & = {4 \over 5} (3) x^2 + 2 \left(-{1 \over 2}\right) x^{-{3 \over 2}} - 0 \\ & = {12 \over 5} x^2 - x^{-{3 \over 2}} \\ & = {12 \over 5} x^2 - {1 \over \sqrt{x^3}} \end{align}

(h)

\begin{align} {d \over dx} \left( 6 \sqrt[3]{x} - {\pi \over x^2} + x \right) & = {d \over dx} \left( 6 x^{1 \over 3} - \pi x^{-2} + x \right) \\ & = 6 \left(1 \over 3\right) x^{-{2 \over 3}} - \pi (-2) x^{-3} + 1 \\ & = 2 x^{-{2 \over 3}} + 2 \pi x^{-3} + 1 \\ & = \left(2 \over 1\right) \left(1 \over \sqrt[3]{x^2}\right) + \left(2\pi \over 1\right)\left(1 \over x^3\right) + 1 \\ & = {2 \over \sqrt[3]{x^2}} + {2\pi \over x^3} + 1 \end{align}

(i)

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

(j)

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

(a)

\begin{align} {dy \over dx} & = {1 \over 3}(2)x - 1 \\ & = {2 \over 3}x - 1 \\ \\ \text{When } & x = 6, \\ {dy \over dx} & = {2 \over 3}(6) - 1 \\ & = 3 \end{align}

(b)

\begin{align} y & = {4 \over x} \\ y & = 4 x^{-1} \\ \\ {dy \over dx} & = 4(-1)x^{-2} \\ & = -4 x^{-2} \\ & = \left(-{4 \over 1}\right)\left(1 \over x^2\right) \\ & = - {4 \over x^2} \phantom{000000} [\text{Need to find value of } x \text{ to substitute in}] \\ \\ \text{Substitute } & y = -2 \text{ into eqn of curve,} \\ -2 & = {4 \over x} \\ -2x & = 4 \\ x & = {4 \over -2} \\ x & = -2 \\ \\ \text{Substitute } & x = -2 \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{4 \over (-2)^2} \\ & = -1 \end{align}

(c)

\begin{align} y & = (5x - 1)(2x + 1) \\ y & = 10x^2 + 5x - 2x - 1 \\ y & = 10x^2 + 3x - 1 \\ \\ {dy \over dx} & = 10(2)x + 3 \\ & = 20x + 3 \\ \\ \text{When } & x = 0, \phantom{000000} [y \text{-intercept}] \\ {dy \over dx} & = 20(0) + 3 \\ & = 3 \end{align}

(d)

\begin{align} y & = {x + 1 \over x^2} \\ y & = {x \over x^2} + {1 \over x^2} \\ y & = {1 \over x} + {1 \over x^2} \\ y & = x^{-1} + x^{-2} \\ \\ {dy \over dx} & = (-1)x^{-2} + (-2)x^{-3} \\ & = - {1 \over x^2} - {2 \over x^3} \phantom{000000} [\text{Need to find value of } x \text{ to substitute in}] \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \phantom{000000} [x \text{-intercept}] \\ 0 & = {x + 1 \over x^2} \\ 0 & = x + 1 \\ -1 & = x \\ \\ \text{Substitute } & x = -1 \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{1 \over (-1)^2} - {2 \over (-1)^3} \\ & = 1 \end{align}

\begin{align} y & = (10 - x^2)^2 \\ y & = 10^2 - 2(10)(x^2) + (x^2)^2 \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ y & = 100 - 20x^2 + x^4 \\ \\ {dy \over dx} & = -20(2)x + 4x^3 \\ & = -40x + 4x^3 \\ \\ \text{When } & {dy \over dx} = 0 , \\ 0 & = -40x + 4x^3 \\ 0 & = 4x(-10 + x^2) \end{align}
\begin{align} 4x & = 0 && \text{ or } & -10 + x^2 & = 0 \\ x & = 0 &&& x^2 & = 10 \\ & &&& x & = \pm \sqrt{10} \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = [10 - (0)^2]^2 &&& y & = [10 - (\pm \sqrt{10})^2]^2 \\ y & = 100 &&& y & = 0 \\ \\ \therefore & \phantom{.} (0, 100) &&& \therefore & \phantom{.} (\sqrt{10}, 0),(-\sqrt{10}, 0) \end{align}

\begin{align} y & = ax + {b \over x^2} \\ \\ \text{Using } & (-1, -9), \\ -9 & = a(-1) + {b \over (-1)^2} \\ -9 & = -a + {b \over 1} \\ -9 & = -a + b \\ a & = b + 9 \phantom{00} \text{--- (1)} \\ \\ \\ y & = ax + {b \over x^2} \\ y & = ax + bx^{-2} \\ \\ {dy \over dx} & = a + b(-2)x^{-3} \\ & = a - 2b x^{-3} \\ & = a - \left(2b \over 1\right)\left(1 \over x^3\right) \\ & = a - {2b \over x^3} \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = 12, \\ 12 & = a - {2b \over (1)^3} \\ 12 & = a - {2b \over 1} \\ 12 & = a - 2b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 12 & = b + 9 - 2b \\ -b + 2b & = 9 -12 \\ b & = -3 \\ \\ \text{Substitute } & \text{into (1),} \\ a & = -3 + 9 \\ a & = 6 \\ \\ \\ \therefore a & = 6, b = -3 \end{align}

\begin{align} y & = {a \over x} + {b \over x^2} \\ y & = ax^{-1} + b x^{-2} \\ \\ {dy \over dx} & = a(-1)x^{-2} + b(-2)x^{-3} \\ & = -a x^{-2} -2 b x^{-3} \\ & = - \left(a \over 1\right)\left(1 \over x^2\right) - \left(2b \over 1\right)\left(1 \over x^3\right) \\ & = - {a \over x^2} - {2b \over x^3} \\ \\ \text{When } & x = {1 \over 4} \text{ and } {dy \over dx} = -3, \\ -3 & = -{a \over \left(1 \over 4\right)^2} - {2b \over \left(1 \over 4\right)^3 } \\ -3 & = -{a \over {1 \over 16}} - {2b \over {1 \over 64}} \\ -3 & = -16a - 128b \\ 3 & = 16a + 128b \\ \\ \text{Let } & a = {19 \over 16}, \\ 3 & = 16\left(19 \over 16\right) + 128b \\ 3 & = 19 + 128b \\ -16 & = 128b \\ {-16 \over 128} & = b \\ -{1 \over 8} & = b \\ \\ \\ \therefore a & = {19 \over 16}, b = -{1 \over 8} \end{align}

(i)

\begin{align} y & = 3 - {1 \over 4}x^2 \\ \\ {dy \over dx} & = -{1 \over 4}(2)x \\ & = -{1 \over 2}x \\ \\ \text{When } & x = -4, \\ {dy \over dx} & = -{1 \over 2}(-4) \\ & = 2 \end{align}

(ii)

\begin{align} \text{Gradient at new point} & = {-1 \over 2} \phantom{000000} [m_1 \times m_2 = -1] \\ & = -{1 \over 2} \\ \\ \text{Substitute } & {dy \over dx} = -{1 \over 2} \text{ into } {dy \over dx} = -{1 \over 2}x, \\ -{1 \over 2} & = -{1 \over 2}x \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = 3 - {1 \over 4}(1)^2 \\ y & = {11 \over 4} \\ \\ \therefore & \phantom{.} \left(1, 2{3 \over 4}\right) \end{align}

(i)

\begin{align} y & = ax^3 - x^2 + b \\ \\ \text{Using } & \left(1, {1 \over 4}\right), \\ {1 \over 4} & = a(1)^3 - (1)^2 + b \\ {1 \over 4} & = a(1) - 1 + b \\ {1 \over 4} & = a - 1 + b \\ {5 \over 4} & = a + b \phantom{00} \text{--- (1)} \\ \\ {dy \over dx} & = a(3)x^2 - 2x \\ & = 3a x^2 - 2x \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = -{5 \over 4}, \\ -{5 \over 4} & = 3a(1)^2 - 2(1) \\ -{5 \over 4} & = 3a(1) - 2 \\ {3 \over 4} & = 3a \\ {1 \over 4} & = a \\ \\ \text{Substitute } & a = {1 \over 4} \text{ into (1),} \\ {5 \over 4} & = {1 \over 4} + b \\ 1 & = b \\ \\ \therefore a & = {1 \over 4}, b = 1 \end{align}

(ii)

\begin{align} {dy \over dx} & = 3a x^2 - 2x \\ & = 3 \left(1 \over 4\right)x^2 - 2x \\ & = {3 \over 4}x^2 - 2x \\ \\ \text{When } & {dy \over dx} = -{5 \over 4}, \\ -{5 \over 4} & = {3 \over 4}x^2 - 2x \\ 0 & = {3 \over 4}x^2 - 2x + {5 \over 4} \\ 0 & = 3x^2 - 8x + 5 \\ 0 & = (x - 1)(3x - 5) \end{align}
\begin{align} x - 1 & = 0 && \text{ or } & 3x - 5 & = 0 \\ x & = 1 \text{ (Provided point)} &&& 3x & = 5 \\ & &&& x & = {5 \over 3} \end{align}
\begin{align} y & = ax^3 - x^2 + b \\ y & = {1 \over 4}x^3 - x^2 + 1 \\ \\ \text{When } & x = {5 \over 3}, \\ y & = {1 \over 4}\left(5 \over 3\right)^3 - \left(5 \over 3\right)^2 + 1 \\ y & = -{67 \over 108} \\ \\ \therefore & \phantom{.} \left(1{2 \over 3}, - {67 \over 108}\right) \end{align}

(i)

\begin{align} y & = x^2 (5x - 1) \\ y & = 5x^3 - x^2 \\ \\ {dy \over dx} & = 5(3)x^2 - 2x \\ & = 15x^2 - 2x \\ \\ \\ y - x & = 0.3 \\ y & = x + 0.3 \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of line} & = 1 \\ \\ \\ \text{Substitute } & {dy \over dx} = 1 \text{ into } {dy \over dx} = 15x^2 - 2x, \\ 1 & = 15x^2 - 2x \\ 0 & = 15x^2 - 2x - 1 \\ 0 & = (5x + 1)(3x - 1) \end{align}
\begin{align} 5x + 1 & = 0 && \text{ or } & 3x - 1 & =0 \\ 5x & = -1 &&& 3x & = 1 \\ x & = -{1 \over 5} &&& x & = {1 \over 3} \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = \left(-{1 \over 5}\right)^2 \left[5 \left(-{1 \over 5}\right) - 1 \right] &&& y & = \left(1 \over 3\right)^2 \left[ 5 \left(1 \over 3\right) - 1 \right] \\ y & = -{2 \over 25} &&& y & = {2 \over 27} \\ \\ \therefore & \phantom{.} \left(-{1 \over 5}, -{2 \over 25}\right) &&& \therefore & \phantom{.} \left({1 \over 3}, {2 \over 27} \right) \end{align}

(ii)

\begin{align} {dy \over dx} & = 15x^2 - 2x \\ \\ \text{When } & {dy \over dx} = -2, \\ -2 & = 15x^2 - 2x \\ 0 & = 15x^2 - 2x + 2 \\ \\ b^2 - 4ac & = (-2)^2 - 4(15)(2) \\ & = -116 < 0 \\ \\ \implies 0 & = 15x^2 - 2x + 2 \text{ has no solutions} \\ \\ \therefore \text{No points } & \text{on curve such that gradient of tangent} = -2 \end{align}

(iii)

\begin{align} {dy \over dx} & = 15x^2 - 2x \\ & = 15 \left(x^2 - {2 \over 15}x\right) \\ & = 15 \left[ x^2 - {2 \over 15}x + \left(1 \over 15\right)^2 - \left(1 \over 15\right)^2 \right] \phantom{000000} [\text{Complete the square}] \\ & = 15 \left[ \left(x - {1 \over 15}\right)^2 - {1 \over 225} \right] \\ & = 15 \left(x - {1 \over 15}\right)^2 - {1 \over 15} \\ \\ \text{Minimum } & \text{value of } {dy \over dx} = - {1 \over 15} \end{align}

Note: I don't think this is relevant for O levels

(i)

\begin{align} \text{Let } f(x) & = x^3 \\ \\ f'(x) & = \lim_{h \rightarrow 0} {f(x + h) - f(x) \over h} \\ & = \lim_{h \rightarrow 0} {(x + h)^3 - x^3 \over h} \\ & = \lim_{h \rightarrow 0} {(x^2 + 2hx + h^2)(x + h) - x^3 \over h} \\ & = \lim_{h \rightarrow 0} {x^3 + hx^2 + 2hx^2 + 2h^2 x + h^2 x + h^3 - x^3 \over h} \\ & = \lim_{h \rightarrow 0} {3hx^2 + 3h^2 x + h^3 \over h} \\ & = \lim_{h \rightarrow 0} 3x^2 + 3hx + h^2 \\ & = 3x^2 + 3(0)x + 0^2 \\ & = 3x^2 \end{align}

(ii)

\begin{align} \text{Let } f(x) & = (x + 1)(x - 1) \\ & = x^2 - 1 \\ \\ f'(x) & = \lim_{h \rightarrow 0} {f(x + h) - f(x) \over h} \\ & = \lim_{h \rightarrow 0} {(x + h)^2 - 1 - (x^2 - 1) \over h} \\ & = \lim_{h \rightarrow 0} {x^2 + 2hx + h^2 - 1 - x^2 + 1 \over h} \\ & = \lim_{h \rightarrow 0} {2hx + h^2 \over h} \\ & = \lim_{h \rightarrow 0} 2x + h \\ & = 2x + 0 \\ & = 2x \end{align}