(a)

\begin{align} {d \over dx} (4 \sin x - 3) & = 4 \cos x - 0 \\ & = 4 \cos x \end{align}

(b)

\begin{align} {d \over dx} (x^2 - 5 \cos x) & = (2)x - 5 (- \sin x) \\ & = 2x + 5 \sin x \end{align}

(c)

\begin{align} {d \over dx} (4x^2 + 3 \tan x) & = 4(2)x + 3\sec^2 x \\ & = 8x + 3\sec^2 x \end{align}

(d)

\begin{align} {d \over dx} (2\sin x + 3\cos x) & = 2\cos x + 3(- \sin x) \\ & = 2\cos x - 3\sin x \end{align}

(a)

\begin{align} u & = x &&& v & = \tan x \\ {du \over dx} & = 1 &&& {dv \over dx} & = \sec^2 x \end{align} \begin{align} {d \over dx} (x \tan x) & = (x)(\sec^2 x) + (\tan x)(1) \phantom{000000} [\text{Product rule}] \\ & = x\sec^2 x + \tan x \end{align}

(b)

\begin{align} u & = x^2 &&& v & = \cos x \\ {du \over dx} & = 2x &&& {dv \over dx} & = -\sin x \end{align} \begin{align} {d \over dx} (x^2 \cos x) & = (x^2)(-\sin x) + (\cos x)(2x) \phantom{000000} [\text{Product rule}] \\ & = -x^2\sin x + 2x\cos x \end{align}

(c)

\begin{align} u & = (x + 1)^2 &&& v & = \sin x \\ {du \over dx} & = 2(x + 1).(1) \phantom{0000} [\text{Chain rule}] &&& {dv \over dx} & = \cos x \\ & = 2(x + 1) \end{align} \begin{align} {d \over dx} \left[ (x + 1)^2 \sin x \right] & = (x + 1)^2(\cos x) + (\sin x)(2)(x + 1)(1) \phantom{000000} [\text{Product rule}] \\ & = (x + 1)^2\cos x + 2(x + 1)\sin x \end{align}

(d)

\begin{align} u & = 1 - 2 \sin x &&& v & = \cos x \\ {du \over dx} & = -2 \cos x &&& {dv \over dx} & = - \sin x \end{align} \begin{align} {d \over dx} \left( 1 - 2\sin x \over \cos x \right) & = {(\cos x)(- 2\cos x) - (1 - 2\sin x)(-\sin x) \over \cos^2 x} \phantom{000000} [\text{Quotient rule}] \\ & = { -2 \cos^2 x + \sin x(1 - 2\sin x) \over \cos^2 x } \\ & = {-2\cos^2 x + \sin x - 2\sin^2 x \over \cos^2 x} \\ & = {-2(\cos^2 x + \sin^2 x) + \sin x \over \cos^2 x} \\ & = {-2(1) + \sin x \over \cos^2 x} \phantom{000000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = {-2 + \sin x \over \cos^2 x} \end{align}

(a)

\begin{align} {d \over dx} (1 - \cos x)^3 & = 3(1 - \cos x)^2[0 -(-\sin x)] \phantom{00000} [\text{Chain rule}] \\ & = 3(1 - \cos x)^2\sin x \end{align}

(b)

\begin{align} {d \over dx} (2 + 3\sin x)^2 & = 2(2 + 3\sin x)(0 + 3\cos x) \phantom{00000} [\text{Chain rule}] \\ & = 2(2 + 3\sin x)(3\cos x) \\ & = 6\cos x (2 + 3\sin x) \end{align}

(c)

\begin{align} {d \over dx} \sqrt{2 - \tan x} & = {d \over dx} (2 - \tan x)^{1 \over 2} \\ & = {1 \over 2}(2 - \tan x)^{-{1 \over 2}}(0 -\sec^2 x) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2}\left(1 \over \sqrt{2 - \tan x}\right)(-\sec^2 x) \\ & = -{\sec^2 x \over 2\sqrt{2 - \tan x}} \end{align}

(d)

\begin{align} {d \over dx} \sqrt{\sin x + 2\cos x} & = {d \over dx} (\sin x + 2\cos x)^{1 \over 2} \\ & = {1 \over 2}(\sin x + 2\cos x)^{-{1 \over 2}}(\cos x - 2\sin x) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2}\left(1 \over \sqrt{\sin x + 2\cos x}\right)(\cos x - 2\sin x) \\ & = {\cos x - 2\sin x \over 2\sqrt{\sin x + 2\cos x}} \end{align}

(a)

\begin{align} {d \over dx} (3\tan 2x) & = 3(\sec^2 2x)(2) \\ & = 6\sec^2 2x \end{align}

(b)

\begin{align} {d \over dx} \left[ 4 \sin \left({1 \over 2}x\right) \right] & = 4\cos \left({1 \over 2}x\right)\left(1 \over 2\right) \\ & = 2\cos \left({1 \over 2}x\right) \end{align}

(c)

\begin{align} {d \over dx} \left[ \cos \left( 2x - {\pi \over 3} \right) \right] & = \left[-\sin \left( 2x - {\pi \over 3} \right) \right] (2) \\ & = -2\sin \left( 2x - {\pi \over 3} \right) \end{align}

(d)

\begin{align} {d \over dx} \left[ \sin 3x + \cos (4x - 5) \right] & = (\cos 3x)(3) + [-\sin (4x - 5)] (4) \\ & = 3\cos 3x - 4\sin (4x - 5) \end{align}

(a)

\begin{align} u & = \sin x &&& v & = \cos 3x \\ {du \over dx} & = \cos x &&& {dv \over dx} & = (-\sin 3x)(3) \\ & &&& & = -3 \sin 3x \end{align} \begin{align} {d \over dx} (\sin x \cos 3x) & = (\sin x)(-3\sin 3x) + (\cos 3x)(\cos x) \phantom{000000} [\text{Product rule}] \\ & = -3 \sin x \sin 3x + \cos x \cos 3x \end{align}

(b)

\begin{align} u & = 1 + x^2 &&& v & = \tan 5x \\ {du \over dx} & = 2x &&& {dv \over dx} & = (\sec^2 5x)(5) \\ & &&& & = 5 \sec^2 5x \end{align} \begin{align} {d \over dx} [(1 + x^2)\tan 5x] & = (1 + x^2)(5 \sec^2 5x) + (\tan 5x)(2x) \phantom{000000} [\text{Product rule}] \\ & = 5(1 + x^2)\sec^2 5x + 2x \tan 5x \end{align}

(c)

\begin{align} u & = x &&& v & = \cos 2x \\ {du \over dx} & = 1 &&& {dv \over dx} & = (- \sin 2x) (2) \\ & &&& & = -2 \sin 2x \end{align} \begin{align} {d \over dx} \left( x \over \cos 2x \right) & = {(\cos 2x)(1) - (x)(-2 \sin 2x) \over (\cos 2x)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {\cos 2x + 2x \sin 2x \over \cos^2 2x} \\ & = {1 \over \cos 2x}\left( \cos 2x + 2x \sin 2x \over \cos 2x \right) \\ & = \sec 2x \left( \cos 2x + 2x \sin 2x \over \cos 2x \right) \\ & = \sec 2x \left( {\cos 2x \over \cos 2x} + {2x \sin 2x \over \cos 2x} \right) \\ & = \sec 2x (1 + 2x\tan 2x) \end{align}

(d)

\begin{align} u & = \cos (x + \pi) &&& v & = \sin 3x \\ {du \over dx} & = [ - \sin (x + \pi) ] (1) &&& {dv \over dx} & = (\cos 3x)(3) \\ & = - \sin (x + \pi) &&& & = 3 \cos 3x \end{align} \begin{align} {d \over dx} \left[ \cos (x + \pi) \over \sin 3x \right] & = {(\sin 3x)[-\sin (x + \pi)] - [\cos (x + \pi)](3\cos 3x) \over (\sin 3x)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {-\sin (x + \pi) \sin 3x - 3\cos (x + \pi) \cos 3x \over \sin^2 3x} \\ & = {-[\sin (x + \pi) \sin 3x + 3\cos (x + \pi) \cos 3x] \over \sin^2 3x} \\ & = -{\sin (x + \pi) \sin 3x + 3\cos (x + \pi) \cos 3x \over \sin^2 3x} \end{align}

\begin{align} y & = \sin 2x \\ \\ {dy \over dx} & = (\cos 2x)(2) \\ & = 2\cos 2x \\ \\ \text{When } & x = {\pi \over 6}, \\ {dy \over dx} & = 2\cos \left[ 2\left(\pi \over 6\right) \right] \phantom{000000} [\text{Radian mode!}] \\ & = 1 \end{align}

(a)

\begin{align} {d \over dx} (2\sin^3 x) & = {d \over dx} [ 2 (\sin x)^3] \\ & = 2(3)(\sin x)^2 . (\cos x) \phantom{00000} [\text{Chain rule}] \\ & = 6(\sin x)^2(\cos x) \\ & = 6\sin^2 x \cos x \end{align}

(b)

\begin{align} {d \over dx} (\cos^2 3x) & = {d \over dx} (\cos 3x)^2 \\ & = (2)(\cos 3x). (-3 \sin 3x) \phantom{00000000} [\text{Chain rule}] \\ & = -6 \cos 3x \sin 3x \end{align}

(c)

\begin{align} {d \over dx} [4 \tan^2 (5x) + 3] & = 4{d \over dx}[\tan (5x)]^2 + {d \over dx}(3) \\ & = 4(2)[\tan (5x)] . \left[ 5\sec^2 (5x) \right] + 0 \\ & = 40 \tan (5x) \sec^2 (5x) \end{align}

(d)

\begin{align} {d \over dx} [x + 3\sin^5 (2x)] & = {d \over dx}(x) + 3{d \over dx}[\sin (2x)]^5 \\ & = 1 + 3(5)[\sin (2x)]^4 . [2 \cos (2x) ] \\ & = 1 + 30 \sin^4 (2x) \cos (2x) \end{align}

(e)

\begin{align} u & = 1 + x &&& v & = \cos^7 (2x) \\ & &&& & = \left[ \cos (2x) \right]^7 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 7 [\cos (2x)]^6 . [-2\sin (2x)] \\ & &&& & = - 14 \cos^6 (2x) \sin (2x) \end{align} \begin{align} {dy \over dx} & = (1 + x)\left[ -14 \cos^6 (2x) \sin (2x) \right] + \left[ \cos^7 (2x) \right] (1) \phantom{00000} [\text{Product rule}] \\ & = -14 (1 + x) \cos^6 (2x) \sin (2x) + \cos^7 (2x) \\ & = \cos^6 (2x) \left[ -14 (1 + x) \sin (2x) + \cos (2x) \right] \end{align}

(f)

\begin{align} u & = \cos^2 2x &&& v & = (1 - x)^2 \\ & = (\cos 2x)^2 \\ {du \over dx} & = 2(\cos 2x) . (- 2 \sin 2x) &&& {dv \over dx} & = 2(1 - x). (-1) \\ & = -4 \cos 2x \sin 2x &&& & = -2(1 - x) \end{align} \begin{align} {dy \over dx} & = { (1 - x)^2 (-4 \cos 2x \sin 2x) - \left[ \cos^2 2x \right] [-2(1 - x)] \over [(1 - x)^2]^2} \phantom{00000} [\text{Quotient rule}] \\ & = { -4(1 - x)^2 \cos 2x \sin 2x + 2(1 - x) \cos^2 2x \over (1 - x)^4 } \\ & = { 2(1 - x) \left[ -2(1 - x) \cos 2x \sin 2x + \cos^2 2x \right] \over (1 - x)^3} \\ & = { 2 \left[ -2(1 - x) \cos 2x \sin 2x + \cos^2 2x \right] \over (1 - x)^3 } \\ & = { 2 \left[ (x - 1) (2 \sin 2x \cos 2x) + \cos^2 2x \right] \over (1 - x)^3} \\ & = { 2 \left[ (x - 1) \sin 4x + \cos^2 2x \right] \over (1 - x)^3} \phantom{00000000000000000} [\sin 2A = 2 \sin A \cos A] \\ & = { 2(x - 1) \sin 4x + 2\cos^2 2x \over (1 - x)^3 } \\ & = { (2x - 2) \sin 4x + 2 \cos^2 2x \over (1 - x)^3 } \end{align}

(a)

\begin{align} {dx \over dt} & = 0.2 \text{ radians per second} \\ \\ y & = x + \cos x \\ \\ {dy \over dx} & = 1 + (-\sin x) \\ & = 1 - \sin x \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = (1 - \sin x) \times 0.2 \\ & = 0.2(1 - \sin x) \\ \\ \text{When } & x = {\pi \over 6}, \\ {dy \over dt} & = 0.2 \left(1 - \sin {\pi \over 6}\right) \phantom{00000} [\text{Radian mode!}] \\ & = 0.1 \text{ units per second} \end{align}

(b)

\begin{align} {dx \over dt} & = {\pi \over 30} \text{ radians per second} \\ \\ y & = \sqrt{1 + \sin x} \\ & = (1 + \sin x)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(1 + \sin x)^{-{1 \over 2}} . (\cos x) \phantom{0000000} [\text{Chain rule}] \\ & = {1 \over 2}\left(1 \over \sqrt{1 + \sin x}\right) (\cos x) \\ & = {\cos x \over 2\sqrt{1 + \sin x}} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {\cos x \over 2\sqrt{1 + \sin x}} \times {\pi \over 30} \\ & = {\pi \cos x \over 60 \sqrt{1 + \sin x}} \\ \\ \text{When } & x = {\pi \over 3}, \\ {dy \over dt} & = {\pi \cos {\pi \over 3} \over 60 \sqrt{1 + \sin {\pi \over 3} } } \phantom{00000000} [\text{Radian mode!}] \\ & = 0.019 \phantom{.} 165 \\ & \approx 0.0192 \text{ units per second} \end{align}

(a)

\begin{align} y & = x - 2 \sin x \\ \\ {dy \over dx} & = 1 - 2\cos x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 1 - 2\cos x \\ 2\cos x & = 1 \\ \cos x & = {1 \over 2} \phantom{000000} [\text{1st & 4th quadrants since} \cos x > 0] \end{align}

\begin{align} \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = {\pi \over 3} \\ \\ x & = {\pi \over 3}, 2\pi - {\pi \over 3} \\ & = {\pi \over 3}, {5\pi \over 3} \text{ (Reject)} \\ \\ \therefore x & = {\pi \over 3} \end{align}

(b)

\begin{align} y & = x - \tan x \\ \\ {dy \over dx} & = 1 - \sec^2 x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 1 - \sec^2 x \\ \sec^2 x & = 1 \\ {1 \over \cos^2 x} & = 1 \\ 1 & = \cos^2 x \\ \pm \sqrt{1} & = \cos x \\ \pm 1 & = \cos x \end{align}

$$ \therefore x = 0, \pi, 2\pi \text{ (Reject)} $$

\begin{align} \text{Substitute } & x = 0 \text{ into eqn of curve,} &&& \text{Substitute } & x = \pi \text{ into eqn of curve,} \\ y & = 0 - \tan 0 &&& y & = \pi - \tan \pi \\ & = 0 &&& & = \pi - 0 \\ & &&& & = \pi \\ \\ \therefore & \phantom{.} (0, 0) &&& \therefore & \phantom{.} (\pi, \pi) \end{align}

(i)

\begin{align} y & = 2 \cos x + 3 \sin x \\ \\ {dy \over dx} & = 2(-\sin x) + 3(\cos x) \\ & = -2\sin x + 3 \cos x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = -2\sin x + 3 \cos x \\ 2\sin x & = 3\cos x \\ {2\sin x \over \cos x} & = 3 \\ 2\tan x & = 3 \\ \tan x & = {3 \over 2} \phantom{000000} [\text{1st & 3rd quadrants since } \tan x > 0] \end{align}

\begin{align} \text{Basic angle, } \alpha & = \tan^{-1} \left({3 \over 2}\right) \phantom{00000} [\text{Radian mode!}] \\ & = 0.98279 \\ \\ x & = 0.98279, \pi + 0.98279 \\ & = 0.98279, 4.12438 \text{ (Reject)} \\ \\ \therefore x & \approx 0.983 \end{align}

(ii)

\begin{align} {dy \over dx} & = -2\sin x + 3 \cos x \\ \\ {d^2 y \over dx^2} & = -2(\cos x) + 3(-\sin x) \\ & = -2\cos x - 3\sin x \\ \\ \text{When } & x = 0.98279, \\ {d^2y \over dx^2} & = -2\cos (0.98279) - 3\sin (0.98279) \\ & = -3.60555 \\ \\ \therefore \text{Statio} & \text{nary pt is a maximum pt} \end{align}

\begin{align} y & = x + 4\tan^2 2x \\ \\ {dy \over dx} & = 1 + 4 (2) (\tan 2x) . (2 \sec^2 2x) \phantom{00000} [\text{Chain rule}] \\ & = 1 + 16 \tan 2x \sec^2 2x \\ & = 1 + 16 \tan 2x \left(1 \over \cos^2 2x\right) \\ & = 1 + {16 \tan 2x \over \cos^2 2x} \\ \\ \text{When } & x = {\pi \over 6}, \\ {dy \over dx} & = 1 + {16 \tan \left[2({\pi \over 6})\right] \over \cos^2 \left[2({\pi \over 6})\right]} \\ & = 1 + {16 \tan {\pi \over 3} \over \cos^2 {\pi \over 3}} \\ & = 1 + {16 (\sqrt{3}) \over ({1 \over 2})^2} \\ & = 1 + {16 (\sqrt{3}) \over {1 \over 4}} \\ & = 1 + 16 \sqrt{3} \div {1 \over 4} \\ & = 1 + 16 \sqrt{3} \times {4 \over 1} \\ & = 1 + 64\sqrt{3} \end{align}

Note: Since the tangent to the curve is parallel to the x-axis (a horizontal line), the gradient of the tangent is equals to 0

\begin{align} u & = \sin 3x &&& v & = 2 - \cos 3x \\ {du \over dx} & = 3\cos 3x &&& {dv \over dx} & = - (-3\sin 3x) \\ & &&& & = 3 \sin 3x \end{align} \begin{align} {dy \over dx} & = {(2 - \cos 3x)(3 \cos 3x) - (\sin 3x)(3 \sin 3x) \over (2 - \cos 3x)^2 } \phantom{00000} [\text{Quotient rule}] \\ & = {3\cos 3x(2 - \cos 3x) - 3 \sin^2 3x \over (2 - \cos 3x)^2 } \\ & = {6 \cos 3x - 3 \cos^2 3x - 3 \sin^2 3x \over (2 - \cos 3x)^2} \\ & = {6 \cos 3x - 3(\cos^2 3x + \sin^2 3x) \over (2 - \cos 3x)^2 } \\ & = {6 \cos 3x - 3(1) \over (2 - \cos 3x)^2 } \phantom{00000000000000000000} [\sin^2 A + \cos^2 A = 1] \\ & = {6 \cos 3x - 3 \over (2 - \cos 3x)^2 } \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {6 \cos 3x - 3 \over (2 - \cos 3x)^2 } \\ 0 & = 6 \cos 3x - 3 \\ 3 & = 6 \cos 3x \\ {3 \over 6} & = \cos 3x \\ {1 \over 2} & = \cos 3x \phantom{00000} [\text{1st & 4th quadrant since } \cos 3x > 0] \end{align}

In addition, since $0 < x < \pi$, we need to consider solutions that satisfy the inequality: $0 < 3x < 3\pi$

\begin{align} \text{Basic angle, } \alpha & = \cos^{-1} \left({1 \over 2}\right) \\ & = {\pi \over 3} \\ \\ 3x & = {\pi \over 3}, 2\pi - {\pi \over 3} \\ & = {\pi \over 3}, {5\pi \over 3} \\ & = {\pi \over 3}, {5\pi \over 3}, {\pi \over 3} + 2\pi, {5\pi \over 3} + 2\pi \\ & = {\pi \over 3}, {5\pi \over 3}, {7\pi \over 3}, {11\pi \over 3} \text{ (Reject)} \\ \\ x & = {\pi \over 9}, {5\pi \over 9}, {7\pi \over 9} \end{align}

(i) Recall that the tangent & normal to the curve (at the same point) are perpendicular, thus $m_1 \times m_2 = -1$

\begin{align} y & = 1 + \cos x \\ \\ \text{When } & x = {\pi \over 6}, \\ y & = 1 + \cos {\pi \over 6} \\ & = 1 + {\sqrt{3} \over 2} \\ \\ \text{Coordinate} & \text{ is } \left( {\pi \over 6}, 1 + {\sqrt{3} \over 2} \right) \\ \\ {dy \over dx} & = 0 + (-\sin x) \\ & = -\sin x \\ \\ \text{When } & x = {\pi \over 6}, \\ {dy \over dx} & = -\sin {\pi \over 6} \\ & = -{1 \over 2} \\ \\ \text{Gradient of tangent} & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & \left( {\pi \over 6}, 1 + {\sqrt{3} \over 2} \right), \\ 1 + {\sqrt{3} \over 2} & = -{1 \over 2} \left(\pi \over 6\right) + c \\ 1 + {\sqrt{3} \over 2} & = -{\pi \over 12} + c \\ 1 + {\sqrt{3} \over 2} + {\pi \over 12} & = c \\ \\ \text{Eqn of tangent: } & y = -{1 \over 2}x + 1 + {\sqrt{3} \over 2} + {\pi \over 12} \\ 2y & = - x + 2 + \sqrt{3} + {\pi \over 6} \\ x + 2y & = 2 + \sqrt{3} + {\pi \over 6} \\ \\ \\ \text{Gradient of normal} & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & \left( {\pi \over 6}, 1 + {\sqrt{3} \over 2} \right), \\ 1 + {\sqrt{3} \over 2} & = 2\left({\pi \over 6}\right) + c \\ 1 + {\sqrt{3} \over 2} & = {\pi \over 3} + c \\ 1 + {\sqrt{3} \over 2} - {\pi \over 3} & = c \\ \\ \text{Eqn of normal: } & y = 2x + 1 + {\sqrt{3} \over 2} - {\pi \over 3} \\ y - 2x & = + 1 + {\sqrt{3} \over 2} - {\pi \over 3} \end{align}

(ii)

\begin{align} \text{Eqn of tangent: } & y = -{1 \over 2}x + 1 + {\sqrt{3} \over 2} + {\pi \over 12} \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 2}x + 1 + {\sqrt{3} \over 2} + {\pi \over 12} \\ {1 \over 2}x & = 1 + {\sqrt{3} \over 2} + {\pi \over 12} \\ x & = 2 + \sqrt{3} + {\pi \over 6} \\ \\ \\ \text{Eqn of normal: } & y = 2x + 1 + {\sqrt{3} \over 2} - {\pi \over 3} \\ \\ \text{Let } & y = 0, \\ 0 & = 2x + 1 + {\sqrt{3} \over 2} - {\pi \over 3} \\ -2x & = 1 + {\sqrt{3} \over 2} - {\pi \over 3} \\ 2x & = -1 - {\sqrt{3} \over 2} + {\pi \over 3} \\ x & = - {1 \over 2} - {\sqrt{3} \over 4} + {\pi \over 6} \end{align}

The height of the triangle is the $y$-coordinate of the point $ \left( {\pi \over 6}, 1 + {\sqrt{3} \over 2} \right) $

\begin{align} \text{Base} & = 2 + \sqrt{3} + {\pi \over 6} - \left( {\pi \over 6} - {1 \over 2} - {\sqrt{3} \over 4} \right) \\ & = 2 + \sqrt{3} + {\pi \over 6} - {\pi \over 6} + {1 \over 2} + {\sqrt{3} \over 4} \\ & = 2 + {1 \over 2} + \sqrt{3} + {\sqrt{3} \over 4} \\ & = {5 \over 2} + {4\sqrt{3} \over 4} + {\sqrt{3} \over 4} \\ & = {5 \over 2} + {5\sqrt{3} \over 4} \\ & = {5 \over 2} \left(1 + {\sqrt{3} \over 2}\right) \\ \\ \text{Area} & = {1 \over 2} \times \text{Base } \times \text{Height} \\ & = {1 \over 2} \times {5 \over 2} \left(1 + {\sqrt{3} \over 2}\right) \times \left(1 + {\sqrt{3} \over 2} \right) \\ & = {5 \over 4}\left( 1 + {\sqrt{3} \over 2} \right) \left(1 + {\sqrt{3} \over 2} \right) \\ & = {5 \over 4}\left(1 + {\sqrt{3} \over 2}\right)^2 \phantom{000} \text{ (Shown)} \end{align}

(i)

\begin{align} y & = 2 \sin 2x - 1 \\ \\ \text{Amplitude} & = 2 \\ \text{Center line: } & y = -1 \\ \\ \text{Max. value} & = 1 \\ \text{Min. value} & = -3 \\ \\ \text{Period} & = {2\pi \over 2} \\ & = \pi \\ \text{No. of cycles} & = 2 \end{align}

(ii)

\begin{align} y & = 2 \sin 2x - 1 \\ \\ \text{When } & x = {\pi \over 4}, \\ y & = 2 \sin 2\left(\pi \over 4\right) - 1 \\ & = 2 \sin {\pi \over 2} - 1 \\ & = 2(1) - 1 \\ & = 1 \\ \\ \text{Coordinates} & \text{ of the point is } \left( {\pi \over 4}, 1 \right) \\ \\ {dy \over dx} & = 2(2 \cos 2x) \\ & = 4 \cos 2x \\ \\ \text{When } & x = {\pi \over 4}, \\ {dy \over dx} & = 4\cos \left[ 2\left( \pi \over 4 \right) \right] \\ & = 4\cos {\pi \over 2} \\ & = 0 \implies \text{Tangent is a horizontal line} \\ \\ \text{Eqn of } & \text{tangent: } y = 1 \end{align}

(iii) Referring to the sketch in (i), the tangent $y = 1$ will pass the two maximum points of the graph

\begin{align} \text{First max. point: } & \left( {\pi \over 4}, 1 \right) \\ \\ \text{Period} & = \pi \\ \\ {\pi \over 4} + \pi & = {5\pi \over 4} \\ \\ \text{Next max. point: } & \left({5\pi \over 4}, 1 \right) \end{align}

(i)

\begin{align} f(x) & = \sin x + \cos x \\ \\ f'(x) & = \cos x + (- \sin x) \\ & = \cos x - \sin x \\ \\ \text{Let } & f'(x) = 0, \\ 0 & = \cos x - \sin x \\ \sin x & = \cos x \\ {\sin x \over \cos x} & = 1 \\ \tan x & = 1 \phantom{00000000} [\text{1st & 3rd quadrants since } \tan x > 0] \end{align}

\begin{align} \text{Basic angle, } \alpha & = \tan^{-1} (1) \\ & = {\pi \over 4} \\ \\ x & = {\pi \over 4}, {\pi \over 4} + \pi \\ & = {\pi \over 4}, {5\pi \over 4} \end{align} \begin{align} \text{Substitute } & x = {\pi \over 4} \text{ into eqn of curve,} &&& \text{Substitute } & x = {5\pi \over 4} \text{ into eqn of curve,} \\ y & = \sin {\pi \over 4} + \cos {\pi \over 4} &&& y & = \sin {5\pi \over 4} + \cos {5\pi \over 4} \\ & = {1 \over \sqrt{2}} + {1 \over \sqrt{2}} &&& & = -{1 \over \sqrt{2}} - {1 \over \sqrt{2}} \\ & = {2 \over \sqrt{2}} \times {\sqrt{2} \over \sqrt{2}} &&& & = -{2 \over \sqrt{2}} \times {\sqrt{2} \over \sqrt{2}} \\ & = {2\sqrt{2} \over 2} &&& & = -{2\sqrt{2} \over 2} \\ & = \sqrt{2} &&& & = -\sqrt{2} \\ \\ \therefore & \phantom{.} \left({\pi \over 4}, \sqrt{2}\right) &&& \therefore & \phantom{.} \left({5\pi \over 4}, -\sqrt{2} \right) \end{align}

(ii)

\begin{align} f'(x) & = \cos x - \sin x \\ \\ f''(x) & = -\sin x - \cos x \end{align} \begin{align} \text{When } & x = {\pi \over 4}, &&& \text{When } & x = {5\pi \over 4}, \\ f''(x) & = -\sin {\pi \over 4} - \cos {\pi \over 4} &&& f''(x) & = -\sin {5\pi \over 4} - \cos {5\pi \over 4} \\ & = -{1 \over \sqrt{2}} - {1 \over \sqrt{2}} &&& & = -\left(-{1 \over \sqrt{2}}\right) - \left(-{1 \over \sqrt{2}}\right) \\ & = -{2 \over \sqrt{2}} &&& & = {1 \over \sqrt{2}} + {1 \over \sqrt{2}} \\ & &&& & = {2 \over \sqrt{2}} \\ \\ \therefore \left( {\pi \over 4}, \sqrt{2} \right) & \text{ is a max. pt} &&& \therefore \left( {5\pi \over 4}, -\sqrt{2} \right) & \text{ is a min. pt} \end{align}

A rough sketch of the graph using the turning points as reference:

\begin{align} \therefore \text{For } & f'(x) > 0, \\ 0 \le x < {\pi \over 4} \phantom{00} & \text{or}\phantom{00} {5\pi \over 4} < x \le 2\pi \end{align}

(iii)

$$ \therefore \text{For } f'(x) < 0, {\pi \over 4} < x < {5\pi \over 4} $$

(i)

\begin{align} \text{Let } F \text{ denote } & \text{the midpoint of } BC \\ \\ \cos \angle ABF & = {Adj \over Hyp} \\ \cos \theta & = {BF \over BA} \\ \cos \theta & = {BF \over 8} \\ \\ BF & = 8 \cos \theta \\ \\ BC & = 2 \times BF \\ & = 2 \times 8 \cos \theta \\ & = 16 \cos \theta \\ \\ \sin \angle ABF & = {Opp \over Hyp} \\ \sin \theta & = {AF \over BA} \\ \sin \theta & = {AF \over 8} \\ \\ AF & = 8 \sin \theta \\ \\ S & = \text{Area of rectangle }BCDE + \text{ Area of triangle }ABC \\ & = BE \times BC + {1 \over 2} \times BC \times AF \\ & = (4)(16\cos \theta) + {1 \over 2}(16\cos \theta)(8\sin \theta) \\ & = 64\cos \theta + 64\sin \theta \cos \theta \\ & = 64\cos \theta + 32(2\sin \theta \cos \theta) \\ & = 64\cos \theta + 32(\sin 2\theta) \phantom{000000000} [\sin 2A = 2\sin A \cos A] \\ & = 64\cos \theta + 32\sin 2\theta \end{align}

(ii)

\begin{align} S & = 64\cos \theta + 32\sin 2\theta \\ \\ {dS \over d\theta} & = 64(-\sin \theta) + 32 (2 \cos 2\theta) \\ & = -64 \sin \theta + 64 \cos 2\theta \\ \\ {d^2 S \over d\theta^2} & = -64 \cos \theta + 64(-2 \sin 2\theta) \\ & = -64 \cos \theta - 128 \sin 2\theta \\ \\ \text{Let } & {dS \over d\theta} = 0, \\ 0 & = - 64\sin \theta + 64\cos 2\theta \\ 0 & = - \sin \theta + \cos 2\theta \\ 0 & = - \sin \theta + (\cos 2\theta) \\ 0 & = - \sin \theta + (1 - 2\sin^2 \theta) \phantom{000000} [\cos 2A = 1 - 2\sin^2 A] \\ 0 & = - 2\sin^2 \theta - \sin \theta + 1 \\ 0 & = 2\sin^2 \theta + \sin \theta - 1 \\ 0 & = (\sin \theta + 1)(2 \sin \theta - 1) \end{align}

For the first equation, \begin{align} \sin \theta + 1 & = 0 \\ \sin \theta & = -1 \end{align}

$$ \theta = {3\pi \over 2} \text{ (Reject, }\theta\text{ is acute)} $$

For the second equation, \begin{align} 2 \sin \theta - 1 & = 0 \\ 2 \sin \theta & = 1 \\ \sin \theta & = {1 \over 2} \phantom{00000000} [\text{1st & 2nd quadrants since } \sin \theta > 0] \end{align}

\begin{align} \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 2\right) \\ & = {\pi \over 6} \\ \\ \theta & = {\pi \over 6}, \pi - {\pi \over 6} \\ & = {\pi \over 6}, {5\pi \over 6} \text{ (Reject, }\theta\text{ is acute)} \\ \\ \text{Substitute } & \theta = {\pi \over 6} \text{ into } {d^2 S \over d\theta^2}, \\ {d^2 S \over d\theta^2} & = - 64 \cos {\pi \over 6} - 128 \sin \left[ 2\left({\pi \over 6}\right)\right] \\ & = -166.27687 \\ \\ \therefore \text{When } \theta & = {\pi \over 6}, S \text{ is a maximum value} \end{align}

\begin{align} y & = a \tan^m (x^n) \\ & = a\left[ \tan (x^n) \right]^m \\ \\ {dy \over dx} & = a (m) \left[ \tan (x^n) \right]^{m - 1} . (n)(x^{n - 1})[\sec^2 (x^n)] \phantom{00000} [\text{Chain rule}] \\ & = a m n x^{n - 1} \tan^{m - 1} (x^n) \sec^2 (x^n) \end{align}

(i)

\begin{align} u & = k + x &&& v & = \cos x \\ {du \over dx} & = 1 &&& {dv \over dx} & = -\sin x \end{align} \begin{align} {dy \over dx} & = (k + x)(-\sin x) + (\cos x)(1) \phantom{00000} [\text{Product rule}] \\ & = -(k + x)\sin x + \cos x \end{align}
\begin{align} u & = -(k + x) &&& v & = \sin x \\ {du \over dx} & = - (1) &&& {dv \over dx} & = \cos x \\ & = -1 \end{align} \begin{align} {d^2 y \over dx^2} & = [-(k + x)](\cos x) + (\sin x)(-1) + (- \sin x) \\ & = -(k + x) \cos x - \sin x - \sin x \\ & = -(k + x) \cos x - 2 \sin x \end{align}

(ii)

\begin{align} {d^2 y \over dx^2} + y & = -(k + x) \cos x - 2 \sin x + (k + x) \cos x \\ & = -2 \sin x \\ \\ \therefore {d^2 y \over dx^2} + y & \text{ is independent of } k \end{align}

(i)

\begin{align} {d \over dx} (\sec x) & = {d \over dx} \left(1 \over \cos x\right) \\ & = {d \over dx} (\cos x)^{-1} \\ & = (-1) (\cos x)^{-2} . (-\sin x) \phantom{0000000} [\text{Chain rule}] \\ & = (\cos x)^{-2} (\sin x) \\ & = {1 \over \cos^2 x} (\sin x) \\ & = {1 \over \cos x} \left(\sin x \over \cos x\right) \\ & = \sec x \tan x \phantom{00000} \text{ (Shown)} \end{align}

(ii) Use the result from (i) to differentiate $\sec 2x$

\begin{align} y & = x - \sec 2x \\ \\ {dy \over dx} & = 1 - 2 \sec 2x \tan 2x \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = 1 - 2 \sec [2(3)] \tan [2(3)] \\ & = 1 - 2 \sec 6 \tan 6 \\ & = 1 - 2 \left(1 \over \cos 6\right) \tan 6 \phantom{00000} [\text{Radian mode!}] \\ & = 1.6061 \\ & \approx 1.61 \end{align}

(iii)

\begin{align} {dx \over dt} & = 2 \text{ radians per second} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = 1.6061 \times 2 \\ & = 3.2122 \\ & \approx 3.21 \text{ radians per second} \end{align}

(i)

\begin{align} {1 \over 1 + \sin x} + {1 \over 1 - \sin x} & = {1 - \sin x \over (1 + \sin x)(1 - \sin x) } + {1 + \sin x \over (1 + \sin x)(1 - \sin x)} \\ & = {1 - \sin x + 1 + \sin x \over (1 + \sin x)(1 - \sin x)} \\ & = {2 \over (1 + \sin x)(1 - \sin x)} \\ & = {2 \over 1 - \sin^2 x} \phantom{00000000000} [(a + b)(a - b) = a^2 - b^2] \\ & = {2 \over \cos^2 x} \phantom{00000000000000} [\sin^2 A + \cos^2 A = 1] \end{align} \begin{align} \therefore \sin x \cos x \left( {1 \over 1 + \sin x} + {1 \over 1 - \sin x} \right) & = \sin x \cos x \left(2 \over \cos^2 x\right) \\ & = {2 \sin x \cos x \over \cos^2 x } \\ & = {2 \sin x \over \cos x} \\ & = 2 \tan x \\ \\ \therefore k & = 2 \end{align}

(ii)

\begin{align} {d \over dx} \left[ \sin x \cos x \left( {1 \over 1 + \sin x} + {1 \over 1 - \sin x} \right) \right] & = {d \over dx} ( 2 \tan x) \\ & = 2 \sec^2 x \end{align}

(i)

\begin{align} \text{R.H.S} & = \tan 3x \\ & = \tan (x + 2x) \\ & = { \tan x + \tan 2x \over 1 - \tan x \tan 2x } \\ & = { \tan x + {2 \tan x \over 1 - \tan^2 x} \over 1 - \tan x \left( {2 \tan x \over 1 - \tan^2 x} \right) } \\ & = { \tan x + {2 \tan x \over 1 - \tan^2 x} \over 1 - { 2 \tan^2 x \over 1 - \tan^2 x } } \\ & = { \tan x + {2 \tan x \over 1 - \tan^2 x} \over 1 - { 2 \tan^2 x \over 1 - \tan^2 x } } \times {1 - \tan^2 x \over 1 - \tan^2 x} \\ & = { \tan x ( 1 - \tan^2 x) + 2 \tan x \over 1 - \tan ^2 x - 2\tan^2 x } \\ & = { \tan x - \tan^3 x + 2 \tan x \over 1 - 3\tan^2 x } \\ & = { 3 \tan x - \tan^3 x \over 1 - 3\tan^2 x} \phantom{000} \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{Let } y & = {3 \tan x - \tan^3 x \over 1 - 3 \tan^2 x} \\ & = \tan 3x \\ \\ {dy \over dx} & = 3 \sec^2 3x \\ \\ \text{When } & x = {\pi \over 3}, \\ {dy \over dx} & = 3 \sec^2 \left[3 \left({\pi \over 3}\right) \right] \\ & = 3 \sec^2 \pi \\ & = {3 \over \cos^2 \pi} \\ & = {3 \over (-1)^2} \\ & = 3 \end{align}

(i)

\begin{align} y & = x + \cos x \\ \\ {dy \over dx} & = 1 + (-\sin x) \\ & = 1 - \sin x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 1 - \sin x \\ \sin x & = 1 \end{align}

\begin{align} x & = {\pi \over 2}, {\pi \over 2} - 2 \pi \\ & = {\pi \over 2}, - {3\pi \over 2} \end{align}

(ii)

(iii)

\begin{align} {dy \over dx} & = 1 - \sin x \\ \\ \text{Since } - 1 \le & \phantom{.} \sin x \le 1, {dy \over dx} \ge 0 \\ \\ \implies \text{Curve is always} & \text{ increasing & has no turning points} \end{align}

(iv)

\begin{align} y & = x + \cos x \\ \\ \text{When } & x = -{3 \pi \over 2}, \\ y & = -{3\pi \over 2} + \cos \left(-{3 \pi \over 2}\right) \\ & = -{3\pi \over 2} + 0 \\ & = -{3\pi \over 2} \\ \\ \therefore \text{Point} & \text{ is } \left(-{3 \pi \over 2}, -{3\pi \over 2}\right) \\ \\ \\ \text{When } & x = {\pi \over 2}, \\ y & = {\pi \over 2} + \cos {\pi \over 2} \\ & = {\pi \over 2} + 0 \\ & = {\pi \over 2} \\ \\ \therefore \text{Point} & \text{ is } \left({\pi \over 2}, {\pi \over 2}\right) \\ \\ \text{Let } & x = 0, \\ y & = 0 + \cos 0 \\ & = 0 + 1 \\ & = 1 \\ \\ \implies y & \text{-intercept is } 1 \end{align}

The differentiation techniques applies only for angles in radian. So for this question, we need to convert to radians.

\begin{align} 180^\circ & = \pi \text{ radians} \\ \\ 1^\circ & = {\pi \over 180} \text{ radians} \\ \\ x^\circ & = {\pi \over 180}x \text{ radians} \\ \\ \\ {d \over dx} (\sin x^\circ) & = {d \over dx} \left[ \sin \left({\pi \over 180}x\right) \right] \\ & = {\pi \over 180} \cos \left({\pi \over 180}x \right) \\ & = {\pi \over 180} \cos x^\circ \end{align}

(i)

\begin{align} u & = \sin x \phantom{0000000000} {du \over dx} = \cos x \\ \\ v & = 1 + \cos x \phantom{000000.} {dv \over dx} = - \sin x \\ \\ {dy \over dx} & = (\sin x)(- \sin x) + (1 + \cos x)(\cos x) \phantom{000000} [\text{Product rule}] \\ & = - \sin^2 x + \cos x(1 + \cos x) \\ & = - \sin^2 x + \cos x + \cos^2 x \\ & = - (1 - \cos^2 x) + \cos x + \cos^2 x \phantom{000000000.} [\sin^2 A + \cos^2 A = 1 \rightarrow \sin^2 A = 1 - \cos^2 A] \\ & = - 1 + \cos^2 x + \cos x + \cos^2 x \\ & = 2 \cos^2 x + \cos x - 1 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2 \cos^2 x + \cos x - 1 \\ 0 & = (2 \cos x - 1)(\cos x + 1) \\ \\ 2 \cos x - 1 & = 0 \phantom{0} \text{ or } \cos x + 1 = 0 \end{align}

\begin{align} 2 \cos x + 1 & = 0 \\ 2 \cos x & = -1 \\ \cos x & = -{1 \over 2} \phantom{000000} [\text{1st or 4th quadrant}] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \\ & = {\pi \over 3} \end{align}

\begin{align} x & = {\pi \over 3} &&& x & = 2\pi - {\pi \over 3} \\ & &&& & = {5\pi \over 3} \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = \sin {\pi \over 3} \left(1 + \cos {\pi \over 3} \right) &&& y & = \sin {5\pi \over 3} \left(1 + \cos {5\pi \over 3}\right) \\ & = \left(\sqrt{3} \over 2\right) \left(1 + {1 \over 2}\right) &&& & = \left(-{\sqrt{3} \over 2}\right)\left(1 + {1 \over 2}\right) \\ & = {3\sqrt{3} \over 4} &&& & = -{3\sqrt{3} \over 4} \end{align}

\begin{align} \cos x + 1 & = 0 \\ \cos x & = -1 \end{align}

\begin{align} x & = \pi \\ \\ \text{Substitute } & \text{into eqn of curve,} \\ y & = \sin \pi (1 + \cos \pi) \\ & = 0 \\ \\ \therefore & \phantom{.} (\pi, 0) \\ \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = \sin 0 (1 + \cos 0) \\ & = 0 \\ \\ \text{Left limit: } & (0, 0) \\ \\ \\ \text{Substitute } & x = 2\pi \text{ into eqn of curve,} \\ y & = \sin 2\pi (1 + \cos 2\pi) \\ & = 0 \\ \\ \text{Right limit: } & (2\pi, 0) \end{align}

(ii)

Note: I assumed that the high tide starts at 6 am

\begin{align} y & = \sin x \\ \\ {dy \over dx} & = \cos x \phantom{0000000} [\text{Speed/flow of water}] \\ \\ {d^2 y \over dx} & = - \sin x \phantom{00000} \left[\text{Rate of change of } {dy \over dx}\right] \\ \\ {d^3 y \over dx^3} & = - \cos x \\ \\ \text{Let } & {d^2 y \over dx} = 0, \\ - \sin x & = 0 \\ \sin x & = 0 \end{align}

\begin{align} x & = 0, \pi, 2\pi, 3\pi \\ \\ \text{Substitute } & x = 0 \text{ into } {d^3 y \over dx^3}, \\ {d^3 y \over dx^3} & = - \cos 0 \\ & = - 1 < 0 \implies \text{Maximum} \\ \\ \text{Substitute } & x = \pi \text{ into } {d^3 y \over dx^3}, \\ {d^3 y \over dx^3} & = - \cos \pi \\ & = 1 < 0 \implies \text{Minimum} \\ \\ \text{Substitute } & x = 2\pi \text{ into } {d^3 y \over dx^3}, \\ {d^3 y \over dx^3} & = - \cos 2\pi \\ & = - 1 < 0 \implies \text{Maximum} \\ \\ \text{Substitute } & x = 3\pi \text{ into } {d^3 y \over dx^3}, \\ {d^3 y \over dx^3} & = - \cos 3\pi \\ & = 1 < 0 \implies \text{Minimum} \\ \\ \\ \therefore & \text{Timings: } 3 \text{ am}, 3 \text{ pm} \end{align}