(a)

\begin{align} y & = 3x^4 \\ \\ {dy \over dx} & = 3(4)x^3 \\ & = 12x^3 \\ \\ {d^2 y \over dx^2} & = 12(3)x^2 \\ & = 36x^2 \end{align}

(b)

\begin{align} y & = 5x - 7 \\ \\ {dy \over dx} & = 5 \\ \\ {d^2 y \over dx^2} & = 0 \end{align}

(c)

\begin{align} y & = 3x^2 + 5x - 1 \\ \\ {dy \over dx} & = 3(2)x + 5 \\ & = 6x + 5 \\ \\ {d^2 y \over dx^2} & = 6 \end{align}

(d)

\begin{align} y & = {1 \over x} \\ & = x^{-1} \\ \\ {dy \over dx} & = (-1)x^{-2} \\ & = -x^{-2} \\ & = -{1 \over x^2} \\ \\ {d^2 y \over dx^2} & = -(-2)x^{-3} \\ & = 2x^{-3} \\ & = 2 \left(1 \over x^3\right) \\ & = {2 \over x^3} \end{align}

(e)

\begin{align} y & = (3x + 2)^{10} \\ \\ {dy \over dx} & = (10)(3x + 2)^9 . (3) \phantom{00000} [\text{Chain rule}] \\ & = 30(3x + 2)^9 \\ \\ {d^2 y \over dx^2} & = 30(9)(3x + 2)^8 . (3) \phantom{00000} [\text{Chain rule}] \\ & = 810(3x + 2)^8 \end{align}

(f)

\begin{align} y & = \sqrt{x - 4} \\ & = (x - 4)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2} (x - 4)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2} (x - 4)^{-{1 \over 2}} \\ & = {1 \over 2} \left(1 \over \sqrt{x - 4}\right) \\ & = {1 \over 2\sqrt{x - 4}} \\ \\ {d^2 y \over dx^2} & = {1 \over 2} \left(-{1 \over 2}\right) (x - 4)^{-{3 \over 2}}. (1) \phantom{00000} [\text{Chain rule}] \\ & = -{1 \over 4} (x - 4)^{-{3 \over 2}} \\ & = -{1 \over 4} \left(1 \over \sqrt{(x - 4)^3} \right) \\ & = -{1 \over 4\sqrt{(x - 4)^3}} \end{align}

(a)

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

(b)

\begin{align} y & = {x - 1 \over x^2} \\ & = {x \over x^2} - {1 \over x^2} \\ & = x^{-1} - x^{-2} \\ \\ {dy \over dx} & = (-1)x^{-2} - (-2) x^{-3} \\ & = -x^{-2} + 2x^{-3} \\ & = - {1 \over x^2} + 2 \left(1 \over x^3\right) \\ & = - {1 \over x^2} + {2 \over x^3} \\ \\ {d^2 y \over dx^2} & = -(-2)x^{-3} + 2(-3)x^{-4} \\ & = 2x^{-3} - 6x^{-4} \\ & = 2 \left(1 \over x^3\right) - 6 \left(1 \over x^4\right) \\ & = {2 \over x^3} - {6 \over x^4} \end{align}

(c)

\begin{align} y & = {3x^3 - x \over \sqrt{x}} \\ & = {3x^3 \over x^{1 \over 2} } - {x \over x^{1 \over 2}} \\ & = 3x^{5 \over 2} - x^{1 \over 2} \\ \\ {dy \over dx} & = 3 \left(5 \over 2\right) x^{3 \over 2} - {1 \over 2}x^{-{1 \over 2}} \\ & = {15 \over 2} x^{3 \over 2} - {1 \over 2}\left(1 \over \sqrt{x}\right) \\ & = {15 \over 2} \sqrt{x^3} - {1 \over 2\sqrt{x}} \\ \\ {d^2 y \over dx^2} & = {15 \over 2} \left(3 \over 2\right) x^{1 \over 2} - {1 \over 2} \left(-{1 \over 2}\right) x^{-{3 \over 2}} \\ & = {45 \over 4} x^{1 \over 2} + {1 \over 4} \left(1 \over \sqrt{x^3}\right) \\ & = {45 \over 4} \sqrt{x} + {1 \over 4\sqrt{x^3}} \end{align}

(d)

\begin{align} u & = x &&& v & = x - 1 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x - 1)(1) - (x)(1) \over (x - 1)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {x - 1 - x \over (x - 1)^2} \\ & = {-1 \over (x - 1)^2} \\ & = -{1 \over (x - 1)^2} \\ & = -(x - 1)^{-2} \\ \\ {d^2 y \over dx^2} & = - (-2) (x - 1)^{-3} . 1 \phantom{000000} [\text{Chain rule}] \\ & = 2(x - 1)^{-3} \\ & = 2 \left[1 \over (x - 1)^3 \right] \\ & = {2 \over (x - 1)^3} \end{align}

(e)

\begin{align} u & = 2x + 5 &&& v & = x - 4 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x - 4)(2) - (2x + 5)(1) \over (x - 4)^2} \\ & = {2x - 8 - 2x - 5 \over (x - 4)^2} \\ & = {-13 \over (x - 4)^2 } \\ & = -{13 \over (x - 4)^2} \\ & = -13 (x - 4)^{-2} \\ \\ {d^2 y \over dx^2} & = -13(-2)(x - 4)^{-3} . (1) \phantom{000000} [\text{Chain rule}] \\ & = 26(x - 4)^{-3} \\ & = 26 \left[1 \over (x - 4)^3 \right] \\ & = {26 \over (x - 4)^3} \end{align}

(f) Need to use quotient rule for each derivative

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

(i)

\begin{align} u & = 3x + 4 &&& v & = 4x - 1 \\ {du \over dx} & = 3 &&& {dv \over dx} & = 4 \end{align} \begin{align} {dy \over dx} & = {(4x - 1)(3) - (3x + 4)(4) \over (4x - 1)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {12x - 3 - 12x - 16 \over (4x - 1)^2} \\ & = {-19 \over (4x - 1)^2} \\ & = -{19 \over (4x - 1)^2} \\ \\ \text{When } & {dy \over dx} = -{19 \over 4}, \\ -{19 \over 4} & = -{19 \over (4x - 1)^2} \\ 4 & = (4x - 1)^2 \\ \pm \sqrt{4} & = 4x - 1 \\ \pm 2 & = 4x - 1 \end{align} \begin{align} 4x - 1 & = 2 & & \text{ or } & 4x - 1 & = -2 \\ 4x & = 3 &&& 4x & = -1 \\ x & = {3 \over 4} &&& x & = -{1 \over 4} \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = {3\left(3 \over 4\right) +4 \over 4 \left(3 \over 4\right) - 1} & & & y & = {3\left(-{1 \over 4}\right) + 4 \over 4\left(-{1 \over 4}\right) -1} \\ & = 3{1 \over 8} &&& & = -1{5 \over 8} \\ \\ \therefore & \phantom{.} \left({3 \over 4}, 3{1 \over 8}\right) &&& \therefore & \phantom{.} \left(-{1 \over 4}, -1 {5 \over 8}\right) \end{align}

(ii)

\begin{align} {dy \over dx} & = -{19 \over (4x - 1)^2} \\ & = -19(4x - 1)^{-2} \\ \\ {d^2y \over dx^2} & = -19(-2) (4x - 1)^{-3} . (4) \phantom{00000} [\text{Chain rule}] \\ & = 152 (4x - 1)^{-3} \\ & = {152 \over (4x - 1)^3} \\ \\ \text{When } & x = {1 \over 2}, \\ {d^2y \over dx^2} & = {152 \over \left[4\left(1 \over 2\right) - 1\right]^3 } \\ & = 152 \end{align}

\begin{align} y & = x^2 + 2x + 3 \\ \\ {dy \over dx} & = 2x + 2 \\ \\ {d^2y \over dx^2} & = 2 \\ \\ \text{To show } & \left(dy \over dx\right)^2 + \left(d^2y \over dx^2\right)^3 = 4y, \\ \text{L.H.S} & = \left(dy \over dx\right)^2 + \left(d^2y \over dx^2\right)^3 \\ & = (2x + 2)^2 + (2)^3 \\ & = (2x)^2 + 2(2x)(2) + 2^2 + 8 \\ & = 4x^2 + 8x + 4 + 8 \\ & = 4x^2 + 8x + 12 \\ & = 4(x^2 + 2x + 3) \\ & = 4y \\ & = \text{R.H.S} \end{align}

\begin{align} z & = t^2(5t - 1) \\ z & = 5t^3 - t^2 \\ \\ {dz \over dt} & = 5(3)t^2 - 2t \\ & = 15t^2 - 2t \\ \\ {d^2 z \over dt^2} & = 15(2)t - 2 \\ & = 30t - 2 \\ \\ \text{Since } & {dz \over dt} = {d^2z \over dt^2}, \\ 15t^2 - 2t & = 30t - 2 \\ 15t^2 - 32t + 2 & = 0 \\ \\ t & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-32) \pm \sqrt{(-32)^2 - 4(15)(2)} \over 2(15)} \\ & = {32 \pm \sqrt{904} \over 30} \\ & = {32 \pm \sqrt{4} \sqrt{226} \over 30} \\ & = {32 \pm 2\sqrt{226} \over 30} \\ & = {2(16 \pm \sqrt{226}) \over 30} \\ & = {16 \pm \sqrt{226} \over 15} \\ \\ \therefore a & = 16, b = 226 \end{align}

\begin{align} xy - 3 & = 2x^2 \\ xy & = 2x^2 + 3 \\ y & = {2x^2 + 3 \over x} \\ y & = {2x^2 \over x} + {3 \over x} \\ y & = 2x + 3x^{-1} \\ \\ {dy \over dx} & = 2 + 3(-1)x^{-2} \\ & = 2 - 3x^{-2} \\ & = 2 - {3 \over x^2} \\ \\ {d^2y \over dx^2} & = - 3(-2)x^{-3} \\ & = 6x^{-3} \\ & = {6 \over x^3} \\ \\ \\ \text{To show } & x^2 {d^2y \over dx^2} + x{dy \over dx} = y, \\ \text{L.H.S} & = x^2 {d^2y \over dx^2} + x{dy \over dx} \\ & = x^2 \left(6 \over x^3\right) + x \left(2 - {3 \over x^2}\right) \\ & = {6x^2 \over x^3} + 2x - {3x \over x^2} \\ & = {6 \over x} + 2x - {3 \over x} \\ & = {3 \over x} + 2x \\ & = 2x + 3x^{-1} \\ & = y \\ & = \text{R.H.S} \end{align}

(i)

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

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

\begin{align} \left({dy \over dx}\right)^2 & = \left({3 \over 2} \sqrt{x} + {1 \over \sqrt{x}}\right)^2 \\ & = \left({3 \over 2}\sqrt{x}\right)^2 + 2\left({3 \over 2}\sqrt{x}\right)\left(1 \over \sqrt{x}\right) + \left(1 \over \sqrt{x}\right)^2 \\ & = {9 \over 4} x + 3 + {1 \over x} \\ \\ \therefore {d^2y \over dx^2} & \ne \left({dy \over dx}\right)^2 \end{align}

(iii)

\begin{align} {dy \over dx} & = {3 \over 2} \sqrt{x} + {1 \over \sqrt{x}} \\ \\ \text{For } & x >0, \phantom{.} \sqrt{x} > 0 \\ \therefore \text{For } & x > 0, {dy \over dx} > 0 \\ \\ \\ {d^2 y \over dx^2} & = {3 \over 4\sqrt{x}} - {1 \over 2\sqrt{x^3}} \\ \\ \text{Let } & x = 1, \\ {d^2 y \over dx^2} & = {3 \over 4\sqrt{1}} - {1 \over 2\sqrt{1^3}} \\ & = {1 \over 4} \\ \\ \text{Let } & x = 0.25, \\ {d^2 y \over dx^2} & = {3 \over 4 \sqrt{0.25}} - {1 \over 2\sqrt{0.25^3}} \\ & = -2.5 \\ \\ \therefore \text{For } & x > 0, {d^2 y \over dx^2} \text{ is not always positive} \end{align}

\begin{align} \text{Let } f(x) & = -x^3 \\ \\ f'(x) & = -3x^2 \\ \text{For } x > 0, & \phantom{.} x^2 > 0 \text{ and } f'(x) < 0 \\ \\ f''(x) & = -3x \\ \text{For } x > 0, & \phantom{.} f''(x) < 0 \end{align}