(a)

\begin{align} {d \over dx} (x^{10}) & = 10x^9 \end{align}

(b)

\begin{align} y & = \sqrt[4]{x} \\ y & = x^{1 \over 4} \\ \\ {dy \over dx} & = {1 \over 4} x^{-{3 \over 4}} \\ & = {1 \over 4} \left({1 \over x^{3 \over 4}}\right) \\ & = {1 \over 4 \sqrt[4]{x^3}} \end{align}

(c)

\begin{align} f'(x) & = 0 \end{align}

(d)

\begin{align} {d \over dx}\left({1 \over x^3}\right) & = {d \over dx}(x^{-3}) \\ & = -3x^{-4} \\ & = -3 \left(1 \over x^4\right) \\ & = -{3 \over x^4} \end{align}

(e)

\begin{align} {dy \over dx} & = -{2 \over 3} x^{-{5 \over 3}} \\ & = -{2 \over 3} \left(1 \over x^{5 \over 3}\right) \\ & = -{2 \over 3 \sqrt[3]{x^5}} \end{align}

(f)

\begin{align} f'(x) & = 0 \end{align}

(a) Note that $\sqrt{7}$ is a constant

\begin{align} 20x^5 - {4 \over x^3} + \sqrt{7} & = 20x^5 - 4x^{-3} + \sqrt{7} \\ \\ {d \over dx} (20x^5 - 4x^{-3} + \sqrt{7}) & = 20(5)x^4 - 4(-3)x^{-4} + 0 \\ & = 100x^4 + 12\left(1 \over x^4\right) \\ & = 100x^4 + {12 \over x^4} \end{align}

(b) Note that $\pi$ is a constant

\begin{align} y & = \pi x^4 + {3 \over x} - 5\sqrt[3]{x} \\ & = \pi x^4 + 3x^{-1} - 5 x^{1 \over 3} \\ \\ {dy \over dx} & = \pi (4) x^3 + 3(-1) x^{-2} - 5 \left(1 \over 3\right) x^{-{2 \over 3}} \\ & = 4\pi x^3 - 3 \left(1 \over x^2\right) - {5 \over 3} \left(1 \over x^{2 \over 3}\right) \\ & = 4\pi x^3 - {3 \over x^2} - {5 \over 3\sqrt[3]{x^2}} \end{align}

(c)

\begin{align} f(x) & = {7 \over 2}x^4 + {5 \over 3x^4} + {3 \over 8} \\ & = {7 \over 2}x^4 + {5 \over 3} x^{-4} + {3 \over 8} \\ \\ f'(x) & = {7 \over 2}(4)x^3 + {5 \over 3}(-4) x^{-5} + 0 \\ & = 14x^3 - {20 \over 3} \left(1 \over x^5\right) \\ & = 14x^3 - {20 \over 3x^5} \end{align}

(a)

\begin{align} {dA \over dr} & = \pi (2) r \\ & = 2\pi r \end{align}

(b)

\begin{align} {d \over dt} (at^m) & = a(m)t^{m - 1} \\ & = am t^{m - 1} \end{align}

(c)

\begin{align} g'(y) & = \pi (n) y^{n - 1} + 0 \\ & = n \pi y^{n - 1} \end{align}

(a)

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

(b)

\begin{align} y & = 4x^2 - 5\sqrt{x} - {1 \over x} \\ & = 4x^2 - 5x^{1 \over 2} - x^{-1} \\ \\ {dy \over dx} & = 4(2)x - 5 \left(1 \over 2\right)x^{-{1 \over 2}} - (-1)x^{-2} \\ & = 8x - {5 \over 2} \left(1 \over x^{1 \over 2}\right) + x^{-2} \\ & = 8x - {5 \over 2 \sqrt{x}} + {1 \over x^2} \\ \\ \text{When } & x = 4, \\ {dy \over dx} & = 8(4) - {5 \over 2 \sqrt{4}} + {1 \over (4)^2} \\ & = 30{13 \over 16} \end{align}

(c)

\begin{align} y & = (2x + 1)(3x - 4) \\ & = 6x^2 - 8x + 3x - 4 \\ & = 6x^2 - 5x - 4 \\ \\ {dy \over dx} & = 6(2)x - 5(1) - 0 \\ & = 12x - 5 \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = 12(3) - 5 \\ & = 31 \end{align}

(d)

\begin{align} y & = {x^2 + x + 4 \over \sqrt{x}} \\ & = {x^2 + x + 4 \over x^{1 \over 2}} \\ & = {x^2 \over x^{1 \over 2}} + {x \over x^{1 \over 2} } + {4 \over x^{1 \over 2}} \\ & = x^{3 \over 2} + x^{1 \over 2} + 4x^{-{1 \over 2}} \\ \\ {dy \over dx} & = {3 \over 2}x^{1 \over 2} + {1 \over 2}x^{-{1 \over 2}} + 4\left(-{1 \over 2}\right)x^{-{3 \over 2}} \\ & = {3 \over 2}\sqrt{x} + {1 \over 2} \left(1 \over \sqrt{x}\right) - 2 \left(1 \over x^{{3 \over 2}}\right) \\ & = {3 \over 2}\sqrt{x} + {1 \over 2\sqrt{x}} - {2 \over \sqrt{x^3}} \\ \\ \text{When } & x = 9, \\ {dy \over dx} & = {3 \over 2}\sqrt{9} + {1 \over 2\sqrt{9}} - {2 \over \sqrt{9^3}} \\ & = 4 {16 \over 27} \end{align}

(a)

\begin{align} y & = 3x^2 - 4x + 3 \\ \\ {dy \over dx} & = 3(2)x - 4(1) + 0 \\ & = 6x - 4 \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 6(2) - 4 \\ & = 8 \end{align}

(b)

\begin{align} y & = 5x + {1 \over x^2} \\ & = 5x + x^{-2} \\ \\ {dy \over dx} & = 5(1) + (-2)x^{-3} \\ & = 5 - 2 \left(1 \over x^3\right) \\ & = 5 - {2 \over x^3} \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = 5 - {2 \over (3)^3} \\ & = 4{25 \over 27} \end{align}

(c)

\begin{align} y & = 4x(x - \sqrt{x}) \\ & = 4x(x - x^{1 \over 2}) \\ & = 4x^2 - 4x^{3 \over 2} \\ \\ {dy \over dx} & = 4(2)x - 4\left(3 \over 2\right)x^{1 \over 2} \\ & = 8x - 6 \sqrt{x} \\ \\ \text{When } & x = 1, \\ {dy \over dx} 7 & = 8(1) - 6 \sqrt{1} \\ & = 2 \end{align}

(d) For this part, we need the value $x$ (when $y = 3$) to substitute into ${dy \over dx}$

\begin{align} y & = {x - 4 \over x} \\ & = {x \over x} - {4 \over x} \\ & = 1 - 4x^{-1} \\ \\ {dy \over dx} & = 0 - 4(-1) x^{-2} \\ & = 4 \left(1 \over x^2\right) \\ & = {4 \over x^2} \\ \\ \text{Substitute } & y = 3 \text{ into eqn of curve,} \\ 3 & = 1 - 4x^{-1} \\ 3 - 1 & = -4 \left(1 \over x\right) \\ 2 & = -{4 \over x} \\ 2x & = -4 \\ x & = {-4 \over 2} \\ & = -2 \\ \\ \text{Substitute } & x = -2 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {4 \over (-2)^2} \\ & = 1 \end{align}

(a) The point where the curve crosses the y-axis is the y-intercept, where x = 0

\begin{align} y & = 3x^2 - 5x - 1 \\ \\ {dy \over dx} & = 3(2)x - 5(1) - 0 \\ & = 6x - 5 \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = 6(0) - 5 \\ & = - 5 \end{align}

(b) The point where the curve crosses the x-axis is the x-intercept, where y = 0

\begin{align} y & = {5x - 4 \over x^2} \\ & = {5x \over x^2} - {4 \over x^2} \\ & = {5 \over x} - 4x^{-2} \\ & = 5x^{-1} - 4x^{-2} \\ \\ {dy \over dx} & = 5(-1) x^{-2} - 4(-2) x^{-3} \\ & = -5 \left(1 \over x^2\right) + 8 \left(1 \over x^3\right) \\ & = -{5 \over x^2} + {8 \over x^3} \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = {5x - 4 \over x^2} \\ 0 & = 5x - 4 \\ 4 & = 5x \\ {4 \over 5} & = x \\ \\ \text{Substitute } & x = {4 \over 5} \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{5 \over \left(4 \over 5\right)^2} + {8 \over \left(4 \over 5\right)^3} \\ & = 7{13 \over 16} \end{align}

\begin{align} y & = x^2 - 5x + 7 \\ \\ {dy \over dx} & = 2x - 5(1) + 0 \\ & = 2x - 5 \\ \\ \text{When } & {dy \over dx} = 3, \\ 3 & = 2x - 5 \\ 8 & = 2x \\ {8 \over 2} & = x \\ 4 & = x \\ \\ \text{Substitute } & x = 4 \text{ into eqn of curve,} \\ y & = (4)^2 - 5(4) + 7 \\ & = 3 \\ \\ \text{Coordinates} & \text{ of point: } (4, 3) \end{align}

\begin{align} v & = 3t^2 + 5t - 7 \\ \\ {dv \over dt} & = 3(2)t + 5(1) - 0 \\ & = 6t + 5 \\ \\ {dv \over dt} & < 0 \\ 6t + 5 & < 0 \\ 6t & < - 5 \\ t & < - {5 \over 6} \end{align}

(a)

\begin{align} y & = {3t^2 - t\sqrt{t} + 6t \over \sqrt{t}} \\ & = {3t^2 - t (t^{1 \over 2}) + 6t \over t^{1 \over 2}} \\ & = {3t^2 \over t^{1 \over 2}} - {t(t^{1 \over 2}) \over t^{1 \over 2}} + {6t \over t^{1 \over 2}} \\ & = 3t^{3 \over 2} - t + 6t^{1 \over 2} \\ \\ {dy \over dt} & = 3\left(3 \over 2\right)t^{1 \over 2} - 1 + 6 \left(1 \over 2\right) t^{-{1 \over 2}} \\ & = {9 \over 2}\sqrt{t} - 1 + 3 \left(1 \over \sqrt{t} \right) \\ & = {9 \over 2}\sqrt{t} - 1 + {3 \over \sqrt{t}} \end{align}

(b)

\begin{align} (t^3 + 2)(t^2 - 1) & = t^5 - t^3 + 2t^2 - 2 \\ \\ {d \over dt} (t^5 - t^3 + 2t^2 - 2) & = 5t^4 - 3t^2 + 2(2)t - 0 \\ & = 5t^4 - 3t^2 + 4t \end{align}

(c) Use the identity: $(a + b)^2 = a^2 + ab + b^2$

\begin{align} y & = \left(t^2 + {2 \over t}\right)^2 \\ & = (t^2)^2 + 2(t^2)\left(2 \over t\right) + \left(2 \over t\right)^2 \\ & = t^4 + {4t^2 \over t} + {4 \over t^2} \\ & = t^4 + 4t + 4t^{-2} \\ \\ {dy \over dt} & = 4t^3 + 4(1) + 4(-2)t^{-3} \\ & = 4t^3 + 4 - 8 \left(1 \over t^3\right) \\ & = 4t^3 + 4 - {8 \over t^3} \end{align}

(i)

\begin{align} y & = ax^3 + x + c \\ \\ \text{Using } & (1, 2), \\ 2 & = a(1)^3 + (1) + c \\ 2 & = a(1) + 1 + c \\ 2 & = a + 1 + c \\ 1 & = a + c \phantom{000} \text{--- (1)} \\ \\ {dy \over dx} & = a(3)x^2 + (1) + 0 \\ & = 3ax^2 + 1 \\ \\ \text{Substitute } & x = 1 \text{ and } {dy \over dx} = -2, \\ -2 & = 3a(1)^2 + 1 \\ -2 & = 3a(1) + 1 \\ -2 & = 3a + 1 \\ -3 & = 3a \\ {-3 \over 3} & = a \\ -1 & = a \\ \\ \text{Substitute } & a = -1 \text{ into (1),} \\ 1 & = -1 + c \\ 2 & = c \\ \\ \therefore a & = -1, c = 2 \end{align}

(ii)

\begin{align} {dy \over dx} & = 3(-1)x^2 + 1 \\ & = -3x^2 + 1 \\ \\ \text{When } & {dy \over dx} = -2, \\ -2 & = -3x^2 + 1 \\ -3 & = -3x^2 \\ {-3 \over -3} & = x^2 \\ 1 & = x^2 \\ \pm \sqrt{1} & = x \\ \pm 1 & = x \phantom{00000} [\text{Other point has } x \text{-coordinate -1}] \\ \\ y & = (-1)x^3 + x + (2) \\ & = -x^3 + x + 2 \\ \\ \text{When } & x = -1, \\ y & = -(-1)^3 + (-1) + 2 \\ & = -(-1) - 1 + 2 \\ & = 2 \\ \\ \therefore \text{Other } & \text{point is (-1, 2)} \end{align}

\begin{align} y & = ax^2 + bx \\ \\ {dy \over dx} & = a(2)x + b(1) \\ & = 2ax + b \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = -2, \\ -2 & = 2a(1) + b \\ -2 & = 2a + b \phantom{000} \text{--- (1)} \\ \\ \text{When } & x = 3 \text{ and } {dy \over dx} = 10, \\ 10 & = 2a(3) + b \\ 10 & = 6a + b \\ -b & = 6a - 10 \\ b & = 10 - 6a \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ -2 & = 2a + 10 - 6a \\ -2 - 10 & = 2a - 6a \\ -12 & = -4a \\ {-12 \over -4} & = a \\ 3 & = a \\ \\ \text{Substitute } & a = 3 \text{ into (2),} \\ b & = 10 - 6(3) \\ & = - 8 \\ \\ \therefore a & = 3, b = - 8 \end{align}

(a)

\begin{align} y & = {h \over x^2} + {k \over x} \\ \\ \text{Using } & (-1, 5), \\ 5 & = {h \over (-1)^2} + {k \over -1} \\ 5 & = {h \over 1} - k \\ 5 & = h - k \\ k + 5 & = h \phantom{000} \text{--- (1)} \\ \\ \\ y & = h x^{-2} + k x^{-1} \\ \\ {dy \over dx} & = h(-2)x^{-3} + k(-1) x^{-2} \\ & = -2h \left(1 \over x^3\right) - k \left(1 \over x^2\right) \\ & = -{2h \over x^3} - {k \over x^2} \\ \\ \text{Substitute } & x = -1 \text{ and } {dy \over dx} = 4, \\ 4 & = -{2h \over (-1)^3} - {k \over (-1)^2} \\ 4 & = -{2h \over -1} - {k \over 1} \\ 4 & = -(-2h) - k \\ 4 & = 2h - k \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4 & = 2(k + 5) - k \\ 4 & = 2k + 10 - k \\ 4 - 10 & = 2k - k \\ -6 & = k \\ \\ \text{Substitute } & k = -6 \text{ into (1),} \\ h & = (-6) + 5 \\ & = -1 \\ \\ \therefore h & = -1, k = -6 \end{align}

(b)

\begin{align} y & = mx - {n \over x} \\ & = mx - nx^{-1} \\ \\ {dy \over dx} & = m(1) - n(-1)x^{-2} \\ & = m + n\left(1 \over x^2\right) \\ & = m + {n \over x^2} \\ \\ \text{Substitute } & x = {1 \over 2} \text{ and } {dy \over dx} = 3, \\ 3 & = m + {n \over \left(1 \over 2\right)^2} \\ 3 & = m + {n \over {1 \over 4}} \\ 3 & = m + 4n \\ \\ \text{Let } & m = -1, \\ 3 & = -1 + 4n \\ 4 & = 4n \\ {4 \over 4} & = n \\ 1 & = n \\ \\ \text{Possible ans: } & m = -1, n = 1 \end{align}

(i)

\begin{align} \text{Eqn of tangent: } y - 19x & = 2 \\ y & = 19x + 2 \\ \\ \text{Gradient of tangent} & = 19 \\ \\ \\ y & = x(x^2 + 7) - 3 \\ & = x^3 + 7x - 3 \\ \\ {dy \over dx} & = 3x^2 + 7(1) - 0 \\ & = 3x^2 + 7 \\ \\ \text{When } & {dy \over dx} = 19, \\ 19 & = 3x^2 + 7 \\ 12 & = 3x^2 \\ {12 \over 3} & = x^2 \\ 4 & = x^2 \\ \pm \sqrt{4} & = x \\ \pm 2 & = x \end{align} \begin{align} \text{Substitute } & x = 2 \text{ into eqn of curve, } & & & \text{Substitute } & x = -2 \text{ into eqn of curve, } \\ y & = (2)^3 + 7(2) - 3 & & & y & = (-2)^3 + 7(-2) - 3 \\ & = 19 & & & & = -25 \end{align}
$$ \text{Coordinates are } (2, 19) \text{ and } (-2, -25) $$

(ii)

\begin{align} {dy \over dx} & = 3x^2 + 7 \\ \\ \text{When } & {dy \over dx} = 5, \\ 5 & = 3x^2 + 7 \\ -2 & = 3x^2 \\ -{2 \over 3} & = x^2 \\ \\ \text{No real} & \text{ roots to solution} \\ \implies 0 \text{ points} & \text{ with } {dy \over dx} = 5 \\ \therefore \text{0 points} & \text{ that tangent to curve has gradient 5} \end{align}

(iii)

\begin{align} {dy \over dx} & = 3x^2 + 7 \\ & = 3(x)^2 + 7 \\ \\ \text{Minimum value of } {dy \over dx} & = 7 \\ \\ \therefore \text{Minimum value of gradient} & = 7 \end{align}

Note: I don't think this question would be tested in O levels

(a)

\begin{align} f'(x) & = \lim_{h \rightarrow 0} {f(x + h) - f(x) \over h} \\ & = \lim_{h \rightarrow 0} { \sqrt{x + h} - \sqrt{x} \over h} \\ & = \lim_{h \rightarrow 0} \left( { \sqrt{x + h} - \sqrt{x} \over h} \times {\sqrt{x + h} + \sqrt{x} \over \sqrt{x + h} + \sqrt{x}} \right) \\ & = \lim_{h \rightarrow 0} { (\sqrt{x + h})^2 - (\sqrt{x})^2 \over h(\sqrt{x + h} + \sqrt{x}) } \\ & = \lim_{h \rightarrow 0} {x + h - x \over h (\sqrt{x + h} + \sqrt{x})} \\ & = \lim_{h \rightarrow 0} {h \over h (\sqrt{x + h} + \sqrt{x})} \\ & = \lim_{h \rightarrow 0} {1 \over \sqrt{x + h} + \sqrt{x}} \\ & = {1 \over \sqrt{x} + \sqrt{x} } \\ & = {1 \over 2\sqrt{x}} \\ & = {1 \over 2} \left(1 \over \sqrt{x}\right) \\ & = {1 \over 2} x^{-{1 \over 2}} \end{align}

(b) From Book A Chapter 4.3 (pp. 68), $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

\begin{align} \text{Let } g(x) & = \sqrt[3]{x} \\ & = x^{1 \over 3} \\ \\ g'(x) & = \lim_{h \rightarrow 0} {g(x + h) - g(x) \over h} \\ & = \lim_{h \rightarrow 0} {(x + h)^{1 \over 3} - x^{1 \over 3} \over h} \\ \\ \text{Let } & a = (x + h)^{1 \over 3} \text{ and } b = x^{1 \over 3} \\ \\ g'(x) & = \lim_{h \rightarrow 0} {a - b \over h} \\ & = \lim_{h \rightarrow 0} \left( {a - b \over h} \times {a^2 + ab + b^2 \over a^2 + ab + b^2} \right) \\ & = \lim_{h \rightarrow 0} {(a - b)(a^2 + ab + b^2) \over h(a^2 + ab + b^2)} \\ & = \lim_{h \rightarrow 0} {a^3 - b^3 \over h(a^2 + ab + b^2)} \\ \\ \text{Since } & a = (x + h)^{1 \over 3} \text{ and } b = x^{1 \over 3}, \\ g'(x) & = \lim_{h \rightarrow 0} {x + h - x \over h[ (x + h)^{2 \over 3} + (x + h)^{1 \over 3}(x)^{1 \over 3} + x^{2 \over 3}]} \\ & = \lim_{h \rightarrow 0} {h \over h[ (x + h)^{2 \over 3} + (x + h)^{1 \over 3}(x)^{1 \over 3} + x^{2 \over 3}]} \\ & = \lim_{h \rightarrow 0} {1 \over (x + h)^{2 \over 3} + (x + h)^{1 \over 3}(x)^{1 \over 3} + x^{2 \over 3}} \\ & = {1 \over (x)^{2 \over 3} + (x)^{1 \over 3}(x)^{1 \over 3} + x^{2 \over 3}} \\ & = {1 \over x^{2 \over 3} + x^{2 \over 3} + x^{2 \over 3} } \\ & = {1 \over 3x^{2 \over 3}} \\ & = {1 \over 3} x^{-{2 \over 3}} \end{align}