(a)

\begin{align} u & = 3x + 2 & & & v & = 2 - x^2 \\ {du \over dx} & = 3 & & & {dv \over dx} & = -2x \end{align} \begin{align} {d \over dx}(3x + 2)(2 - x^2) & = (3x + 2)(-2x) + (2 - x^2)(3) \\ & = -6x^2 - 4x + 6 - 3x^2 \\ & = -9x^2 - 4x + 6 \end{align}

(b)

\begin{align} u & = x & & & v & = (5x + 1)^7 \\ {du \over dx} & = 1 & & & {dv \over dx} & = 7(5x + 1)^6. (5) \phantom{00000} [\text{Chain rule}] \\ & & & & & = 35(5x + 1)^6 \end{align} \begin{align} {d \over dx}[x(5x + 1)^7] & = (x)[35(5x + 1)^6] + (5x + 1)^7 (1) \\ & = 35x (5x + 1)^6 + (5x + 1)^7 \\ & = (5x + 1)^6 \left[ 35x + (5x + 1) \right] \\ & = (5x + 1)^6 (35x + 5x + 1) \\ & = (5x + 1)^6 (40x + 1) \end{align}

(c)

\begin{align} u & = 2x + 3 & & & v & = (x - 3)^4 \\ {du \over dx} & = 2 & & & {dv \over dx} & = 4(x - 3)^3 . (1) \phantom{00000} [\text{Chain rule}] \\ & & & & & = 4(x - 3)^3 \end{align} \begin{align} {d \over dx}(2x + 3)(x - 3)^4 & = (2x + 3)[4(x - 3)^3] + (x - 3)^4 (2) \\ & = (8x + 12)(x - 3)^3 + 2(x - 3)^4 \\ & = (x - 3)^3 [ (8x + 12) + 2(x -3)] \\ & = (x - 3)^3 (8x + 12 + 2x - 6) \\ & = (x - 3)^3 (10x + 6) \\ & = (x - 3)^3 (2)(5x + 3) \\ & = 2(x - 3)^3 (5x + 3) \end{align}

(d)

\begin{align} u & = 6x^2 - 1 & & & v & = (1 + 5x)^3 \\ {du \over dx} & = 6(2)x & & & {dv \over dx} & = 3(1 + 5x)^2 . (5) \phantom{00000} [\text{Chain rule}] \\ & = 12x & & & & = 15(1 + 5x)^2 \end{align} \begin{align} {d \over dx}(6x^2 - 1)(1 + 5x)^3 & = (6x^2 - 1)[15(1 + 5x)^2] + (1 + 5x)^3 (12x) \\ & = (90x^2 - 15)(1 + 5x)^2 + 12x(1 + 5x)^3 \\ & = (1 + 5x)^2 [(90x^2 - 15) + 12x(1 + 5x)] \\ & = (1 + 5x)^2 (90x^2 - 15 + 12x + 60x^2) \\ & = (1 + 5x)^2 (150x^2 + 12x - 15) \\ & = 3(1 + 5x)^2 (50x^2 + 4x - 5) \end{align}

\begin{align} u & = (2x + 3)^5 & & & v & = (x + 2)^8 \\ {du \over dx} & = 5(2x + 3)^4. (2) \phantom{00000} [\text{Chain rule}] & & & {dv \over dx} & = 8(x + 2)^7. (1) \phantom{00000} [\text{Chain rule}] \\ & = 10(2x + 3)^4 & & & & = 8(x + 2)^7 \end{align} \begin{align} {dy \over dx} & = (2x + 3)^5 [8(x + 2)^7] + (x + 2)^8 [10(2x + 3)^4] \\ & = 8(2x + 3)^5(x + 2)^7 + 10(x + 2)^8 (2x + 3)^4 \\ \\ \text{When } & x = -1, \\ {dy \over dx} & = 8[2(-1)+3]^5 [(-1)+2]^7 + 10[(-1)+2]^8 [2(-1)+3]^4 \\ & = 18 \end{align}

\begin{align} u & = (x + 3)^4 & & & v & = (x - 5)^7 \\ {du \over dx} & = 4(x + 3)^3 . (1) \phantom{00000} [\text{Chain rule}] & & & {dv \over dx} & = 7(x - 5)^6 . (1) \phantom{00000} [\text{Chain rule}] \\ & = 4(x + 3)^3 & & & & = 7(x - 5)^6 \end{align} \begin{align} {dy \over dx} & = (x + 3)^4 [7(x - 5)^6] + (x - 5)^7 [4(x + 3)^3] \\ & = 7(x +3)^4 (x - 5)^6 + 4(x - 5)^7 (x + 3)^3 \\ & = (x + 3)^3 (x - 5)^6 [ 7(x + 3) + 4(x - 5)] \\ & = (x + 3)^3 (x - 5)^6 (7x + 21 + 4x - 20) \\ & = (x + 3)^3 (x - 5)^6 (11x + 1) \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = (x + 3)^3 (x - 5)^6 (11x + 1) \\ \\ x = - 3 & \phantom{0} \text{ or } \phantom{0} x = 5 \phantom{0} \text{ or } \phantom{0} 11x + 1 = 0 \\ & \phantom{0000000000000000-} 11x = -1 \\ & \phantom{0000000000000000000.} x = -{1 \over 11} \end{align}

Method 1: Use product rule

\begin{align} u & = \sqrt{x} & & & v & = 7\sqrt{x} - 1 \\ & = x^{1 \over 2} & & & & = 7x^{1 \over 2} - 1 \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} & & & {dv \over dx} & = 7\left(1 \over 2\right)x^{-{1 \over 2}} \\ & = {1 \over 2} \left(1 \over \sqrt{x}\right) & & & & = 7 \left(1 \over 2\right)\left(1 \over \sqrt{x}\right) \\ & = {1 \over 2\sqrt{x}} & & & & = {7 \over 2\sqrt{x}} \end{align} \begin{align} {d \over dx} \left[ \sqrt{x} (7\sqrt{x} - 1) \right] & = \sqrt{x} \left(7 \over 2\sqrt{x}\right) + (7\sqrt{x} - 1) \left(1 \over 2\sqrt{x}\right) \\ & = {7\sqrt{x} \over 2\sqrt{x}} + {7\sqrt{x} - 1 \over 2\sqrt{x}} \\ & = {7\sqrt{x} + 7\sqrt{x} - 1 \over 2\sqrt{x}} \\ & = {14 \sqrt{x} - 1 \over 2\sqrt{x}} \\ & = {14 \sqrt{x} \over 2\sqrt{x}} - {1 \over 2\sqrt{x}} \\ & = 7 - {1 \over 2\sqrt{x}} \end{align}

Method 2: Expand, simplify and differentiate

\begin{align} \sqrt{x}(7\sqrt{x} - 1) & = 7x - \sqrt{x} \\ & = 7x - x^{1 \over 2} \\ \\ {d \over dx} \left(7x - x^{1 \over 2}\right) & = 7 - {1 \over 2}x^{-{1 \over 2}} \\ & = 7 - {1 \over 2} \left(1 \over \sqrt{x}\right) \\ & = 7 - {1 \over 2\sqrt{x}} \end{align}

(a)

\begin{align} u & = 3x & & & v & = \sqrt{4 - 7x^3} \\ & & & & & = (4 - 7x^3)^{1 \over 2} \\ {du \over dx} & = 3 & & & {dv \over dx} & = {1 \over 2} (4 - 7x^3)^{-{1 \over 2}}. (-21x^2) \\ & & & & & = -{21x^2 \over 2} \left(1 \over \sqrt{4 - 7x^3}\right) \\ & & & & & = -{21x^2 \over 2\sqrt{4 - 7x^3}} \end{align} \begin{align} {d \over dx} \left( 3x \sqrt{4 - 7x^3} \right) & = (3x) \left(-{21x^2 \over 2\sqrt{4 - 7x^3}}\right) + \left(\sqrt{4 - 7x^3}\right) (3) \\ & = {-63x^3 \over 2\sqrt{4 - 7x^3}} + {3\sqrt{4 - 7x^3} \over 1} \\ & = {-63x^3 \over 2\sqrt{4 - 7x^3}} + {3\sqrt{4 - 7x^3} (2\sqrt{4 - 7x^3}) \over 2\sqrt{4 - 7x^3}} \\ & = {-63x^3 \over 2\sqrt{4 - 7x^3}} + {6(4 - 7x^3) \over 2\sqrt{4 - 7x^3}} \\ & = {-63x^3 + 24 - 42x^3 \over 2\sqrt{4 - 7x^3}} \\ & = {-105x^3 + 24 \over 2\sqrt{4 - 7x^3}} \end{align}

(b)

\begin{align} u & = 3x - 1 & & & v & = \sqrt{2x^2 + 3} \\ & & & & & = (2x^2 + 3)^{1 \over 2} \\ {du \over dx} & = 3 & & & {dv \over dx} & = {1 \over 2}(2x^2 + 3)^{-{1 \over 2}} . (4x) \\ & & & & & = (2x) \left(1 \over \sqrt{2x^2 + 3} \right) \\ & & & & & = {2x \over \sqrt{2x^2 + 3}} \end{align} \begin{align} {d \over dx} \left[ (3x - 1) \sqrt{2x^2 + 3} \right] & = (3x - 1)\left({2x \over \sqrt{2x^2 + 3}}\right) + (\sqrt{2x + 3})(3) \\ & = {2x(3x - 1) \over \sqrt{2x^2 + 3}} + {3\sqrt{2x^2 + 3} \over 1} \\ & = {2x(3x - 1) \over \sqrt{2x^2 + 3}} + {3\sqrt{2x^2 + 3} (\sqrt{2x^2 + 3}) \over \sqrt{2x^2 + 3}} \\ & = {2x(3x - 1) \over \sqrt{2x^2 + 3}} + {3(2x^2 +3) \over \sqrt{2x^2 + 3}} \\ & = {2x(3x - 1) + 3(2x^2 + 3) \over \sqrt{2x^2 + 3}} \\ & = {6x^2 - 2x + 6x^2 + 9 \over \sqrt{2x^2 + 3}} \\ & = {12x^2 - 2x + 9 \over \sqrt{2x^2 + 3}} \end{align}

(c)

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

(d)

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

\begin{align} u & = 3x + 2 & & & v & = (2 - x)^{-1} \\ {du \over dx} & = 3 & & & {dv \over dx} & = (-1)(2 - x)^{-2}. (-1) \\ & & & & & = (2 - x)^{-2} \\ & & & & & = {1 \over (2 - x)^2} \end{align} \begin{align} {dy \over dx} & = (3x+2) \left[{1 \over (2 - x)^2}\right] + \left(1 \over 2 - x\right)(3) \\ & = {3x+2 \over (2 - x)^2} + {3 \over 2 - x} \\ & = {3x+2 \over (2 - x)^2} + {3(2-x) \over (2-x)^2} \\ & = {3x+2 + 3(2 -x) \over (2-x)^2} \\ & = {3x+2 +6 - 3x \over (2 - x)^2} \\ & = {8 \over (2 - x)^2} \\ \\ \text{When } & {dy \over dx} = 8, \\ 8 & = {8 \over (2 - x)^2} \\ 8(2 - x)^2 & = 8 \\ (2 - x)^2 & = {8 \over 8} \\ (2 - x)^2 & = 1 \\ 2 - x & = \pm \sqrt{1} \\ 2 - x & = \pm 1 \end{align} \begin{align} 2 - x & = 1 & & \text{ or } & 2 - x & = - 1 \\ -x & = -1 & & & - x & = -3 \\ x & = 1 & & & x & = 3 \end{align}

Since the y-axis is a vertical line, the tangent to the curve must be a horizontal line (with gradient = 0)

\begin{align} u & = 2x - 3 & & & v & = (x + 5)^5 \\ {du \over dx} & = 2 & & & {dv \over dx} & = 5(x + 5)^4 . (1) \\ & & & & & = 5(x + 5)^4 \end{align} \begin{align} {dy \over dx} & = (2x - 3)[5(x + 5)^4] + (x + 5)^5 (2) \\ & = 5(2x - 3)(x + 5)^4 + 2(x + 5)^5 \\ & = (x + 5)^4 [5(2x - 3) + 2(x + 5)] \\ & = (x + 5)^4 (10x - 15 + 2x + 10) \\ & = (x + 5)^4 (12x - 5) \\ \\ \text{When } & {dy \over dx} = 0, \phantom{00000} [\text{Gradient of tangent = 0}] \\ 0 & = (x + 5)^4 (12x - 5) \end{align} \begin{align} (x + 5)^4 & = 0 & & \text{ or } & 12x - 5 & =0 \\ x + 5 & = 0 & & & 12x & = 5 \\ x & = -5 & & & x & = {5 \over 12} \end{align}

\begin{align} u & = x + a & & & v & = \sqrt{b - x} \\ & & & & & = (b - x)^{1 \over 2} \\ {du \over dx} & = 1 & & & {dv \over dx} & = {1 \over 2}(b - x)^{-{1 \over 2}} . (-1) \\ & & & & & = -{1 \over 2} \left(1 \over \sqrt{b - x}\right) \\ & & & & & = -{1 \over 2\sqrt{b - x}} \end{align} \begin{align} {dy \over dx} & = (x + a) \left(-{1 \over 2\sqrt{b - x}} \right) + (\sqrt{b - x}) (1) \\ & = {-(x + a) \over 2\sqrt{b - x}} + {\sqrt{b - x} \over 1} \\ & = {-x - a \over 2\sqrt{b - x}} + {\sqrt{b - x}(2\sqrt{b - x}) \over 2\sqrt{b - x}} \\ & = {-x - a \over 2\sqrt{b - x}} + {2(b - x) \over 2\sqrt{b - x}} \\ & = {-x - a \over 2\sqrt{b - x}} + {2b - 2x \over 2\sqrt{b - x}} \\ & = {-3x -a + 2b \over 2\sqrt{b - x}} \\ \\ \text{Substitute } & x = {1 \over 4} \text{ and } {dy \over dx} = 0, \\ 0 & = {-3\left(1 \over 4\right) - a + 2b \over 2 \sqrt{b - \left(1 \over 4\right)} } \\ 0 & = -{3 \over 4} - a + 2b \\ \\ \text{Let } & a = {1 \over 4}, \\ 0 & = -{3 \over 4} - {1 \over 4} + 2b \\ -2b & = -1 \\ b & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ \text{Possible values: } & a = {1 \over 4}, b = {1 \over 2} \end{align}

(i) Note we cannot take the square root of a negative number. For example, $\sqrt{-1}$ has no real solution.

\begin{align} \text{For } \sqrt{9 - 2x^2} & \text{ to have real roots,} \\ 9 - 2x^2 & \ge 0 \\ -2x^2 & \ge - 9 \\ x^2 & \le {9 \over 2} \\ x^2 & \le \left({3 \over \sqrt{2}}\right)^2 \\ \\ \therefore -{3 \over \sqrt{2}} \le & \phantom{.} x \le {3 \over \sqrt{2}} \end{align}

(ii)

\begin{align} u & = x^2 & & & v & = \sqrt{9 - 2x^2} \\ & & & & & = (9 - 2x^2)^{1 \over 2} \\ {du \over dx} & = 2x & & & {dv \over dx} & = {1 \over 2}(9 - 2x^2)^{-{1 \over 2}} . (-4x) \\ & & & & & = -2x \left(1 \over \sqrt{9 - 2x^2}\right) \\ & & & & & = -{2x \over \sqrt{9 - 2x^2}} \end{align} \begin{align} {dy \over dx} & = (x^2) \left(-{2x \over \sqrt{9 - 2x^2}}\right) + \left(\sqrt{9 - x^2}\right)(2x) \\ & = -{2x^3 \over \sqrt{9 - 2x^2}} + {2x \sqrt{9 - x^2} \over 1} \\ & = {-2x^3 \over \sqrt{9 - 2x^2}} + {2x \sqrt{9 - 2x^2} \left(\sqrt{9 - 2x^2}\right) \over \sqrt{9 - 2x^2}} \\ & = {-2x^3 \over \sqrt{9 - 2x^2}} + {2x(9 - 2x^2) \over \sqrt{9 - 2x^2}} \\ & = {-2x^3 \over \sqrt{9 - 2x^2}} + {18x - 4x^3 \over \sqrt{9 - 2x^2}} \\ & = {-6x^3 + 18x \over \sqrt{9 - 2x^2}} \\ \\ \text{To show } & y{dy \over dx} + 6x^3(x^2 - 3) = 0, \\ \text{L.H.S} & = \left(x^2\sqrt{9 - 2x^2}\right) \left({-6x^3 + 18x \over \sqrt{9 - 2x^2}}\right) + 6x^3 (x^2 - 3) \\ & = x^2 (-6x^3 + 18x) + 6x^5 - 18x^3 \\ & = -6x^5 + 18x^3 + 6x^5 - 18x^3 \\ & = 0 \\ & = \text{R.H.S} \end{align}