\begin{align*} y & = 2 \sin^3 4x + 5 \\ & = 2 (\sin 4x)^3 + 5 \\ \\ {dy \over dx} & = 2 \underbrace{ (3) (\sin 4x)^2 (4 \cos 4x) }_{ \text{Chain rule} } + 0 \\ & = 24 \sin^2 4x \cos 4x \\ \\ \text{When } & x = {\pi \over 12}, \\ {dy \over dx} & = 24 \sin^2 \left[4 \left(\pi \over 12\right)\right] \cos \left[4 \left(\pi \over 12\right)\right] \\ & = 24 \left( \sin {\pi \over 3} \right)^2 \left( \cos {\pi \over 3} \right) \phantom{0000000000} [\text{Radian mode!}] \\ & = 9 \end{align*}

\begin{align*} y & = 2e^{3x + 4} \\ \\ {dy \over dx} & = 2(3)e^{3x + 4} \phantom{000000} \left[ {d \over dx} [ e^{f(x)} ] = f'(x) e^{f(x)} \right]\\ & = 6e^{3x + 4} \\ \\ \text{Let } & {dy \over dx} = 6, \\ 6 & = 6 e^{3x + 4} \\ {6 \over 6} & = e^{3x + 4} \\ 1 & = e^{3x + 4} \\ e^0 & = e^{3x + 4} \\ 0 & = 3x + 4 \\ -4 & = 3x \\ -{4 \over 3} & = x \\ \\ \text{Substitute } & x = -{4 \over 3} \text{ into eqn of curve,} \\ y & = 2 e^{3 \left(-{4 \over 3}\right) + 4 } \\ y & = 2 \\ \\ \therefore & \phantom{.} \left(-{4 \over 3}, 2 \right) \end{align*}

\begin{align*} y & = (px + q)^3 \\ \\ \text{Using } & (0, 8), \\ 8 & = [p(0) + q]^3 \\ 8 & = (0 + q)^3 \\ 8 & = q^3 \\ \sqrt[3]{8} & = q \\ 2 & = q \\ \\ y & = (px + 2)^3 \\ \\ {dy \over dx} & = 3(px + 2)^2 . (p) \phantom{00000} [\text{Chain rule}] \\ & = 3p(px + 2)^2 \\ \\ \text{When } & x = 0 \text{ and } {dy \over dx} = 6, \\ 6 & = 3p[p(0) + 2]^2 \\ 6 & = 3p(0 + 2)^2 \\ 6 & = 3p(2)^2 \\ 6 & = 3p(4) \\ 6 & = 12p \\ {6 \over 12} & = p \\ {1 \over 2} & = p \\ \\ \therefore p & = {1 \over 2}, q = 2 \end{align*}

\begin{align*} {d^2 y \over dx^2} & = x + \cos 2x \\ \\ {dy \over dx} & = \int x + \cos 2x \phantom{.} dx \\ & = {x^2 \over 2} + {\sin 2x \over 2} + c \\ & = {1 \over 2}x^2 + {1 \over 2} \sin 2x + c \\ \\ \text{When } & x = 0 \text{ and } {dy \over dx} = 2, \\ 2 & = {1 \over 2}(0)^2 + {1 \over 2} \sin [2(0)] + c \\ 2 & = {1 \over 2}(0) + {1 \over 2}(0) + c \\ 2 & = 0 + 0 + c \\ 2 & = c \\ \\ {dy \over dx} & = {1 \over 2}x^2 + {1 \over 2} \sin 2x + 2 \\ \\ y & = \int {1 \over 2}x^2 + {1 \over 2} \sin 2x + 2 \phantom{.} dx \\ & = {1 \over 2} \left(x^3 \over 3\right) + {1 \over 2}\left(-\cos 2x \over 2\right) + 2x + d \\ & = {1 \over 6} x^3 - {1 \over 4} \cos 2x + 2x + d \\ \\ \text{Using } & (0, 1), \\ 1 & = {1 \over 6}(0)^3 - {1 \over 4} \cos [2(0)] + 2(0) + d \\ 1 & = {1 \over 6}(0) - {1 \over 4} (1) + 0 + d \\ 1 & = 0 - {1 \over 4} + d \\ {5 \over 4} & = d \\ \\ \therefore y & = {1 \over 6} x^3 - {1 \over 4} \cos 2x + 2x + {5 \over 4} \end{align*}