(a)

$$\begin{align} {d \over dx} (2x + 5)^7 & = 7(2x + 5)^6 . [2(1) + 0] \\ & = 7(2x + 5)^6 (2) \\ & = 14(2x + 5)^6 \end{align}$$

(b)

$$\begin{align} {d \over dx} (3 - 4x)^8 & = 8(3 - 4x)^7 . [0 - 4(1)] \\ & = 8(3 - 4x)^7 (-4) \\ & = -32(3 - 4x)^7 \end{align}$$

(c)

$$\begin{align} {d \over dx} \left({1 \over 2}x - 4\right)^5 & = 5\left({1 \over 2}x - 4\right)^4 . \left[ {1 \over 2}(1) - 0 \right] \\ & = 5\left({1 \over 2}x - 4\right)^4 \left(1 \over 2\right) \\ & = {5 \over 2} \left({1 \over 2}x - 4\right)^4 \end{align}$$

(d)

$$\begin{align} {d \over dx} [3(x + 4)^9] & = 3(9)(x + 4)^8 . (1 + 0) \\ & = 27(x + 4)^8 (1) \\ & = 27(x + 4)^8 \end{align}$$

(e)

$$\begin{align} {d \over dx} \left[ {(5x - 3)^6 \over 8 } \right] & = {d \over dx} \left[ {1 \over 8} (5x - 3)^6 \right] \\ & = {1 \over 8}(6) (5x - 3)^5 . (5 - 0) \\ & = {3 \over 4} (5x - 3)^5 (5) \\ & = {15 \over 4} (5x - 3)^5 \end{align}$$

(f)

$$\begin{align} {d \over dx} \left[ {2 \over 3} \left({x \over 6} - 1\right)^4 \right] & = {2 \over 3}(4)\left({x \over 6} - 1\right)^3 . \left({1 \over 6} - 0\right) \\ & = {8 \over 3} \left({x \over 6} - 1\right)^3 \left(1 \over 6\right) \\ & = {4 \over 9} \left({x \over 6} - 1\right)^3 \end{align}$$

(a)

$$\begin{align} {d \over dx} \left(1 \over 3x + 2\right) & = {d \over dx} (3x + 2)^{-1} \\ & = (-1)(3x + 2)^{-2} . (3 + 0) \\ & = - (3x + 2)^{-2} (3) \\ & = -{3 \over (3x + 2)^2} \end{align}$$

(b)

$$\begin{align} {d \over dx} \left(6 \over 2 - 5x\right) & = {d \over dx} [6(2 - 5x)^{-1}] \\ & = 6(-1)(2 - 5x)^{-2} . (0 - 5) \\ & = -6 (2 - 5x)^{-2} (-5) \\ & = 30 (2 - 5x)^{-2} \\ & = {30 \over (2 - 5x)^2} \end{align}$$

(c)

$$\begin{align} {d \over dx} \left(12 \over 2 + 3x^2\right) & = {d \over dx} [12(2 + 3x^2)^{-1}] \\ & = 12(-1)(2 + 3x^2)^{-2} . (0 + 6x) \\ & = -12 (2 + 3x^2)^{-2} (6x) \\ & = -72x (2 + 3x^2)^{-2} \\ & = -{72x \over (2 + 3x^2)^2} \end{align}$$

(d)

$$\begin{align} {d \over dx} \left[ 1 \over (x^2 + 2)^3 \right] & = {d \over dx} [(x^2 + 2)^{-3}] \\ & = (-3)(x^2 + 2)^{-4} . (2x + 0) \\ & = -3 (x^2 + 2)^{-4} (2x) \\ & = -6x (x^2 + 2)^{-4} \\ & = -{6x \over (x^2 + 2)^4} \end{align}$$

(e)

$$\begin{align} {d \over dx} \left[ 5 \over (3 - 4x)^3 \right] & = {d \over dx} [5(3 - 4x)^{-3}] \\ & = 5(-3)(3 - 4x)^{-4} . (0 - 4) \\ & = -15 (3 - 4x)^{-4} (-4) \\ & = 60 (3 - 4x)^{-4} \\ & = {60 \over (3 - 4x)^4} \end{align}$$

(f)

$$\begin{align} {d \over dx} \left[ 3 \over 4(5 - 3x^2) \right] & = {d \over dx} \left[{3 \over 4} (5 - 3x^2)^{-1} \right] \\ & = {3 \over 4}(-1) (5 - 3x^2)^{-2} . (0 - 6x) \\ & = -{3 \over 4} (5 - 3x^2)^{-2} (-6x) \\ & = {9x \over 2}\left[ 1 \over (5 - 3x^2)^2\right] \\ & = {9x \over 2(5 - 3x^2)^2} \end{align}$$

(a)

$$\begin{align} {d \over dx} \sqrt{2x + 3} & = {d \over dx} (2x + 3)^{1 \over 2} \\ & = {1 \over 2}(2x + 3)^{-{1 \over 2}} . (2 + 0) \\ & = {1 \over 2}(2x + 3)^{-{1 \over 2}} (2) \\ & = (2x + 3)^{-{1 \over 2}} \\ & = {1 \over \sqrt{2x + 3}} \end{align}$$

(b)

$$\begin{align} {d \over dx} \sqrt{6 - 5x^2} & = {d \over dx} (6 - 5x^2)^{1 \over 2} \\ & = {1 \over 2}(6 - 5x^2)^{-{1 \over 2}} . (0 - 10x) \\ & = {1 \over 2} (6 - 5x^2)^{-{1 \over 2}} (-10x) \\ & = -5x (6 - 5x^2)^{-{1 \over 2}} \\ & = -{5x \over \sqrt{6 - 5x^2}} \end{align}$$

(c)

$$\begin{align} {d \over dx} \sqrt{3x^3 - 4} & = {d \over dx} (3x^3 - 4)^{1 \over 2} \\ & = {1 \over 2}(3x^3 - 4)^{-{1 \over 2}} . (9x^2 - 0 ) \\ & = {1 \over 2}(3x^3 - 4)^{-{1 \over 2}} (9x^2) \\ & = {9x^2 \over 2} \left(1 \over \sqrt{3x^3 - 4} \right) \\ & = {9x^2 \over 2\sqrt{3x^3 - 4}} \end{align}$$

(d)

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

(e)

$$\begin{align} {d \over dx} \sqrt[3]{27 - x^6} & = {d \over dx} (27 - x^6)^{1 \over 3} \\ & = {1 \over 3} (27 - x^6)^{-{2 \over 3}} . (0 - 6x^5) \\ & = {1 \over 3} (27 - x^6)^{-{2 \over 3}} (-6x^5) \\ & = -2x^5 \left(1 \over \sqrt[3]{(27 - x^6)^2}\right) \\ & = -{2x^5 \over \sqrt[3]{(27 - x^6)^2} } \end{align}$$

(f)

$$\begin{align} {d \over dx} \left(2 \over \sqrt{x - 3}\right) & = {d \over dx} \left[ 2(x - 3)^{-{1 \over 2}} \right] \\ & = 2 \left(-{1 \over 2}\right) (x - 3)^{-{3 \over 2}} . (1 - 0) \\ & = -(x - 3)^{-{3 \over 2}} (1) \\ & = -(x - 3)^{-{3 \over 2}} \\ & = - {1 \over \sqrt{(x - 3)^3}} \end{align}$$

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

$$\begin{align} y & = 2t + 5 + {4 \over 5t - 9} \\ & = 2t + 5 + 4(5t - 9)^{-1} \\ \\ {dy \over dt} & = 2(1) + 0 + 4(-1)(5t - 9)^{-2} . (5 - 0) \\ & = 2 - 4(5t - 9)^{-2} (5) \\ & = 2 - 20(5t - 9)^{-2} \\ & = 2 - {20 \over (5t - 9)^2} \\ \\ \text{When } & t = 2, \\ {dy \over dt} & = 2 - {20 \over [5(2) - 9]^2} \\ & = -18 \end{align}$$

$$\begin{align} y & = 2 + {12 \over (3x - 4)^2} \\ & = 2 + 12(3x - 4)^{-2} \\ \\ {dy \over dx} & = 0 + 12(-2)(3x - 4)^{-3} . (3 - 0) \\ & = -24(3x - 4)^{-3} (3) \\ & = -72 (3x - 4)^{-3} \\ & = -{72 \over (3x - 4)^3} \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = -{72 \over [3(2) - 4]^3 } \\ & = -9 \end{align}$$

(a)

$$\begin{align} {d \over dx} \left(x^2 + {3 \over x}\right)^5 & = {d \over dx} (x^2 + 3x^{-1})^5 \\ & = 5(x^2 + 3x^{-1})^4 . [2x + 3(-1)x^{-2}] \\ & = 5 \left(x^2 + {3 \over x}\right)^4 \left[ 2x - 3 \left(1 \over x^2\right) \right] \\ & = 5 \left(x^2 + {3 \over x}\right)^4 \left( 2x - {3 \over x^2} \right) \end{align}$$

(b)

$$\begin{align} {d \over dx} \left(2 \over 3\sqrt{2x^2 - 5}\right) & = {d \over dx} \left[ {2 \over 3} (2x^2 - 5)^{-{1 \over 2}} \right] \\ & = {2 \over 3} \left(-{1 \over 2}\right)(2x^2 - 5)^{-{3 \over 2}} . (4x - 0) \\ & = -{1 \over 3} (2x^2 - 5)^{-{3 \over 2}} (4x) \\ & = -{4x \over 3} \left(1 \over \sqrt{(2x^2 - 5)^3}\right) \\ & = -{4x \over 3\sqrt{(2x^2 - 5)^3}} \end{align}$$

(c)

$$\begin{align} {d \over dx} \left(4 - \sqrt{x} \right)^{10} & = {d \over dx} \left(4 - x^{1 \over 2} \right)^{10} \\ & = 10(4 - x^{1 \over 2})^9 . \left(0 - {1 \over 2}x^{-{1 \over 2}}\right) \\ & = 10(4 - \sqrt{x})^9 \left(- {1 \over 2\sqrt{x}} \right) \\ & = -{10(4 - \sqrt{x})^9 \over 2\sqrt{x}} \\ & = -{5(4 - \sqrt{x})^9 \over \sqrt{x}} \end{align}$$

(d)

$$\begin{align} {d \over dx} \left(x + 5 - 6\sqrt[3]{x}\right)^6 & = {d \over dx} \left(x + 5 - 6 x^{1 \over 3}\right)^6 \\ & = 6 \left(x + 5 - 6 x^{1 \over 3}\right)^5 . \left[ 1 + 0 - 6 \left(1 \over 3\right) x^{-{2 \over 3}} \right] \\ & = 6 \left(x + 5 - 6 \sqrt[3]{x}\right)^5 \left[ 1 - 2 \left(1 \over \sqrt[3]{x^2}\right) \right] \\ & = 6 \left(x + 5 - 6 \sqrt[3]{x}\right)^5 \left( 1 - {2 \over \sqrt[3]{x^2}} \right) \end{align}$$

(e)

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

(f)

$$\begin{align} (3x - 1)^2 \sqrt{3x - 1} & = (3x - 1)^2 (3x - 1)^{1 \over 2} \\ & = (3x - 1)^{5 \over 2} \phantom{00000000} [a^m \times a^n = a^{m + n}] \\ \\ {d \over dx} (3x - 1)^{5 \over 2} & = {5 \over 2} (3x - 1)^{3 \over 2} . (3) \\ & = {15 \over 2} (3x - 1)^{3 \over 2} \\ & = {15 \over 2} \sqrt{(3x - 1)^3} \end{align}$$

$$\begin{align} y & = (5 - \sqrt{x})^4 \\ & = \left(5 - x^{1 \over 2}\right)^4 \\ \\ {dy \over dx} & = 4\left(5 - x^{1 \over 2}\right)^3 . \left[ 0 - {1 \over 2}x^{-{1 \over 2}} \right] \\ & = 4\left(5 - \sqrt{x}\right)^3 \left[{1 \over 2} \left(1 \over \sqrt{x}\right) \right] \\ & = 4\left(5 - \sqrt{x}\right)^3 \left(1 \over 2\sqrt{x}\right) \\ & = { 4\left(5 - \sqrt{x}\right)^3 \over 2\sqrt{x} } \\ & = { 2\left(5 - \sqrt{x}\right)^3 \over \sqrt{x} } \\ \\ \text{When } & x = 16, \\ {dy \over dx} & = { 2 \left(5 - \sqrt{16}\right)^3 \over \sqrt{16}} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Let } & c = 0, \\ \text{Eqn of line:} & \phantom{00} y = {1 \over 2}x \end{align}$$

If the tangent to the curve is parallel to the x-axis (horizontal line), the gradient of the tangent is equals to 0

$$\begin{align} y & = \sqrt{x^2 - 4x + 8} \\ & = (x^2 - 4x + 8)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(x^2 - 4x + 8)^{-{1 \over 2}} . (2x - 4 + 0) \\ & = {1 \over 2} \left(1 \over \sqrt{x^2 - 4x + 8} \right) (2x - 4) \\ & = {2x - 4 \over 2 \sqrt{x^2 - 4x + 8} } \\ & = {2(x - 2) \over 2\sqrt{x^2 - 4x + 8}} \\ & = {x - 2 \over \sqrt{x^2 - 4x + 8}} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {x - 2 \over \sqrt{x^2 - 4x + 8}} \\ 0 & = x - 2 \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = \sqrt{ (2)^2 - 4(2) + 8 } \\ & = 2 \\ \\ \therefore \text{Coordi} & \text{nates is (2, 2)} \end{align}$$

(i)

$$\begin{align} {10x - 11 \over 2x^2 - 3x - 5} & = {10x - 11 \over (2x - 5)(x + 1)} \\ & = {A \over 2x - 5} + {B \over x + 1} \\ & = {A(x + 1) + B(2x - 5) \over (2x - 5)(x + 1)} \\ \\ \therefore 10x - 11 & = A(x + 1) + B(2x - 5) \\ \\ \text{Let } & x = 2.5, \\ 10(2.5) - 11 & = A(2.5 + 1) + B(0) \\ 14 & = 3.5A \\ {14 \over 3.5} & = A \\ 4 & = A \\ \\ 10x - 11 & = 4(x + 1) + B(2x - 5) \\ \\ \text{Let } & x = -1, \\ 10(-1) - 11 & = 4(0) + B[2(-1) - 5] \\ -21 & = B(-7) \\ -21 & = -7B \\ {-21 \over -7} & = B \\ 3 & = B \\ \\ \therefore {10x - 11 \over (2x - 5)(x + 1)} & = {4 \over 2x - 5} + {3 \over x + 1} \end{align}$$

(ii) Apply the result from part (i)

$$\begin{align} {d \over dx} \left({10x - 11 \over 2x^2 - 3x - 5}\right) & = {d \over dx} \left( {4 \over 2x - 5} + {3 \over x + 1} \right) \\ & = {d \over dx} \left[ 4(2x - 5)^{-1} + 3(x + 1)^{-1} \right] \\ & = 4(-1)(2x - 5)^{-2} . (2 - 0) + 3(-1)(x + 1)^{-2} . (1 + 0) \\ & = -4(2x - 5)^{-2} (2) - 3 (x + 1)^{-2} (1) \\ & = -8(2x - 5)^{-2} - 3(x + 1)^{-2} \\ & = -{8 \over (2x - 5)^2} - {3 \over (x + 1)^2} \end{align}$$

$$\begin{align} y & = {a \over 2 + bx} \\ \\ \text{Using } & (1, 1), \\ 1 & = {a \over 2 + b(1)} \\ 1 & = {a \over 2 + b} \\ 2 + b & = a \phantom{000} \text{--- (1)} \\ \\ y & = a(2 + bx)^{-1} \\ \\ {dy \over dx} & = a(-1)(2 + bx)^{-2} . (0 + b) \\ & = -a(2 + bx)^{-2} (b) \\ & = -ab (2 + bx)^{-2} \\ & = -{ab \over (2 + bx)^2} \\ \\ \text{Substitute } & x = 1 \text{ and } {dy \over dx} = {3 \over 5}, \\ {3 \over 5} & = -{ab \over [2 + b(1)]^2} \\ {3 \over 5} & = -{ab \over (2 + b)^2} \\ 3(2 + b)^2 & = 5(-ab) \\ 3[2^2 + 2(2)(b) + b^2] & = -5ab \\ 3(4 + 4b + b^2) & = -5ab \\ 12 + 12b + 3b^2 & = -5ab \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 12 + 12b + 3b^2 & = -5(2 + b)(b) \\ 12 + 12b + 3b^2 & = -5(2b + b^2) \\ 12 + 12b + 3b^2 & = -10b - 5b^2 \\ 8b^2 + 22b + 12 & = 0 \\ 4b^2 + 11b + 6 & = 0 \\ (4b + 3)(b + 2) & = 0 \end{align}$$ $$\begin{align} 4b + 3 & = 0 & & \text{ or } & b + 2 & = 0 \\ 4b & = -3 & & & b & = -2 \text{ (Reject, since } y = {a \over 2 + bx} \text{ is undefined at (1, 1)}) \\ b & = -{3 \over 4} \\ \\ \text{Substitute } & b = -{3 \over 4} \text{ into (1),} \\ a & = -{3 \over 4} + 2 \\ & = {5 \over 4} \\ \\ \therefore a & = {5 \over 4}, b = -{3 \over 4} \end{align}$$

More on why we reject $b = -2$: $$\begin{align} \text{When } & b = -2, \\ y & = {a \over 2 + (-2)x} \\ y & = {a \over 2 - 2x} \\ \\ \text{Substitute } & x = 1, \phantom{000000} [\text{For the point (1,1)}] \\ y & = {a \over 2 - 2(1)} \\ y & = {a \over 0} \phantom{0000000} [\text{Undefined}] \end{align}$$

$$\begin{align} y & = {1 \over \sqrt{x^2 - 2} } \\ & = (x^2 - 2)^{-{1 \over 2}} \\ \\ {dy \over dx} & = -{1 \over 2}(x^2 - 2)^{-{3 \over 2}} . (2x - 0) \\ & = -{1 \over 2} \left(1 \over \sqrt{(x^2 - 2)^3} \right) (2x) \\ & = -x \left(1 \over \sqrt{(x^2 - 2)^3} \right) \\ & = -{x \over \sqrt{(x^2 -2 )^3} } \\ \\ \\ \text{To show that } & (x^2 - 2){dy \over dx} + xy = 0, \\ \\ \text{L.H.S} & = (x^2 - 2){dy \over dx} + xy \\ & = (x^2 - 2) \left[ -{x \over \sqrt{(x^2 -2 )^3} } \right] + x \left(1 \over \sqrt{x^2 - 2}\right) \\ & = - {x (x^2 - 2) \over (x^2 - 2)^{3 \over 2}} + {x \over (x^2 - 2)^{1 \over 2}} \times {(x^2 - 2) \over (x^2 - 2)} \\ & = - {x (x^2 - 2) \over (x^2 - 2)^{3 \over 2}} + {x (x^2 - 2) \over (x^2 - 2)^{3 \over 2}} \\ & = {-x(x^2 - 2) + x(x^2 - 2) \over (x^2 - 2)^{3 \over 2}} \\ & = {0 \over (x^2 - 2)^{3 \over 2}} \\ & = 0 \\ & = \text{R.H.S} \end{align}$$

If the graph of f'(x) lies entirely below the x-axis, the gradient of the graph y = f(x) is always negative

$$\begin{align} f(x) & = \sqrt{4 - \sqrt{x}} \\ & = \left( 4 - x^{1 \over 2} \right)^{1 \over 2} \\ \\ f'(x) & = {1 \over 2} \left(4 - x^{1 \over2}\right)^{-{1 \over 2}} . \left( 0 - {1 \over 2}x^{-{1 \over 2}}\right) \\ & = {1 \over 2} \left(1 \over \sqrt{4 - \sqrt{x}}\right) \left[ -{1 \over 2} \left(1 \over \sqrt{x}\right) \right] \\ & = {1 \over 2\sqrt{4 - \sqrt{x}}} \left(-{1 \over 2\sqrt{x}}\right) \\ & = -{1 \over 4 \sqrt{x} \sqrt{4 - \sqrt{x}}} \end{align}$$ $$\begin{align} \text{For } f'(x) & < 0, \\ \\ 4 \sqrt{x} \sqrt{4 - \sqrt{x}} & > 0 \phantom{00000} [\text{I.e. denominator of } f'(x) \text{ must be positive}] \\ \\ \text{Since } 4\sqrt{x} \text{ is alw} & \text{ays positive for } x > 0, \\ \sqrt{4 - \sqrt{x}} & > 0 \\ 4 - \sqrt{x} & > 0 \\ -\sqrt{x} & > -4 \\ \sqrt{x} & < 4 \\ x & < 16 \\ \\ \text{Since } x > 0, \text{ range } & \text{of values: } 0 < x < 16 \end{align}$$