(a)

\begin{align} {d \over dx}(x^{1.5}) & = 1.5x^{0.5} \\ & = 1.5 \sqrt{x} \end{align}

(b)

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

(c)

\begin{align} {d \over dx} (x^{101}) & = 101 x^{100} \end{align}

(d)

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

(a)

\begin{align} {d \over dx} \sqrt[3]{x^4 - x^2 + 8} & = {d \over dx} (x^4 - x^2 + 8)^{1 \over 3} \\ & = {1 \over 3} (x^4 - x^2 + 8)^{-{2 \over 3}} . (4x^3 - 2x) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 3}(4x^3 - 2x) (x^4 - x^2 + 8)^{-{2 \over 3}} \\ & = {1 \over 3}(2x)(2x^2 - 1) (x^4 - x^2 + 8)^{-{2 \over 3}} \\ & = {2 \over 3} x (2x^2 - 1) (x^4 - x^2 + 8)^{-{2 \over 3}} \end{align}

(b)

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

(c) Besides the approach below, you can use quotient rule as well.

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

(d)

\begin{align} u & = x &&& v & = \sqrt{x} - 2 \\ & &&& & = x^{1 \over 2} - 2 \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}x^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x}} \end{align} \begin{align} \require{cancel} {d \over dx} \left( x \over \sqrt{x} - 2 \right) & = { (\sqrt{x} - 2)(1) - (x)\left({1 \over 2\sqrt{x}}\right) \over (\sqrt{x} - 2)^2 } \phantom{00000} [\text{Quotient rule}] \\ & = { \sqrt{x} - 2 - {x \over 2\sqrt{x}} \over (\sqrt{x} - 2)^2 } \\ & = { \sqrt{x} - 2 - {x \over 2\sqrt{x}} \over (\sqrt{x} - 2)^2 } \times {2\sqrt{x} \over 2\sqrt{x}} \\ & = { 2\sqrt{x} \left( \sqrt{x} - 2 - {x \over 2\sqrt{x}} \right) \over 2\sqrt{x} (\sqrt{x} - 2)^2 } \\ & = { 2x - 4\sqrt{x} - x \over 2\sqrt{x} (\sqrt{x} - 2)^2 } \\ & = { x - 4\sqrt{x} \over 2\sqrt{x} (\sqrt{x} - 2)^2} \\ & = { \sqrt{x} \sqrt{x} - 4\sqrt{x} \over 2\sqrt{x} (\sqrt{x} - 2)^2} \\ & = { \cancel{\sqrt{x}} (\sqrt{x} - 4) \over 2 \cancel{\sqrt{x}} (\sqrt{x} - 2)^2} \\ & = { \sqrt{x} - 4 \over 2 (\sqrt{x} - 2)^2 } \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} y & = {\pi \sqrt{x} \over 2} + {3 \over \sqrt{x}} \\ & = {\pi \over 2} x^{1 \over 2} + 3x^{-{1 \over 2}} \\ \\ {dy \over dx} & = {\pi \over 2} \left(1 \over 2\right) x^{-{1 \over 2}} + 3 \left(-{1 \over 2}\right) x^{-{3 \over 2}} \\ & = {\pi \over 4} x^{-{1 \over 2}} - {3 \over 2} x^{-{3 \over 2}} \\ & = {\pi \over 4\sqrt{x}} - {3 \over 2\sqrt{x^3}} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {\pi \over 4\sqrt{1}} - {3 \over 2\sqrt{(1)^3}} \\ & = {\pi \over 4(1)} - {3 \over 2(1)} \\ & = {\pi \over 4} - {3 \over 2} \end{align}

(i)

\begin{align} y & = {u \over v} \\ & = u v^{-1} \\ \\ {dy \over dx} & = u {d \over dx} (v^{-1}) + (v^{-1}) {d \over dx} (u) \phantom{00000} [\text{Product rule}] \\ & = u {d \over dx} \left(1 \over v\right) + {1 \over v} . {du \over dx} \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} {d \over dx} \left(1 \over v\right) & = {d \over dx} \left(v^{-1}\right) \\ & = (-1) v^{-2} . {dv \over dx} \phantom{00000} [\text{Chain rule}] \\ & = (-1) \left(1 \over v^2\right) . {dv \over dx} \\ & = -{1 \over v^2} . {dv \over dx} \phantom{0} \text{ (Shown)} \end{align}

Last part

\begin{align} \text{From (i), } {dy \over dx} & = u {d \over dx} \left(1 \over v\right) + {1 \over v} . {du \over dx} \\ \\ \text{From (ii), } & {d \over dx} \left(1 \over v\right) = -{1 \over v^2} . {dv \over dx}, \\ {dy \over dx} & = u \left( -{1 \over v^2} . {dv \over dx}\right) + {1 \over v} . {du \over dx} \\ & = {-u \over v^2} . {dv \over dx} + {1 \over v} . {du \over dx} \\ & = {-u {dv \over dx} \over v^2} + { {du \over dx} \over v} \\ & = {-u {dv \over dx} \over v^2} + { v {du \over dx} \over v^2} \\ & = { v {du \over dx} - u{dv \over dx} \over v^2} \phantom{0} \text{ (Shown)} \end{align}

(a)

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

(b)

\begin{align} y & = (x^2 + 1)(4x - 1) \\ & = 4x^3 - x^2 + 4x - 1 \\ \\ {dy \over dx} & = 12x^2 - 2x + 4 \end{align}

(i)

\begin{align} u & = 1 &&& v & = 3x + 1 \\ {du \over dx} & = 0 &&& {dv \over dx} & = 3 \end{align} \begin{align} {d \over dx} \left(1 \over 3x + 1\right) & = {(3x + 1)(0) - (1)(3) \over (3x + 1)^2} \\ & = {0 - 3 \over (3x + 1)^2} \\ & = {-3 \over (3x + 1)^2} \end{align}

(ii)

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

\begin{align} u & = x + 1 &&& v & = \sqrt{1 - 2x} \\ & &&& & = (1 - 2x)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (1 - 2x)^{-{1 \over 2}} . (-2) \\ & &&& & = (-1) (1 - 2x)^{-{1 \over 2}} \\ & &&& & = {-1 \over \sqrt{1 - 2x}} \end{align} \begin{align} {dy \over dx} & = {(\sqrt{1 - 2x}) (1) - (x + 1)\left(-1 \over \sqrt{1 - 2x}\right) \over (\sqrt{1 - 2x})^2} \\ & = { \sqrt{1 - 2x} + {x + 1 \over \sqrt{1 - 2x}} \over 1 - 2x } \\ & = { \sqrt{1 - 2x} + {x + 1 \over \sqrt{1 - 2x}} \over 1 - 2x } \times {\sqrt{1 - 2x} \over \sqrt{1 - 2x}} \\ & = { 1 - 2x + x + 1 \over (1 - 2x) \sqrt{1 - 2x} } \\ & = { 2 - x \over (1 - 2x) (1 - 2x)^{1 \over 2}} \\ & = { 2 - x \over (1 - 2x)^{3 \over 2}} \\ & = { 2 - x \over \sqrt{(1 - 2x)^3} } \\ & = { -x + 2 \over \sqrt{(1 - 2x)^3} } \\ \\ \therefore a & = -1, b = 2 \end{align}

(i)

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

(ii)

\begin{align} {dy \over dx} & = {3x - 5 \over 2\sqrt{x - 1}} \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = {3x - 5 \over 2\sqrt{x - 1}} \\ 0 & = 3x - 5 \\ 5 & = 3x \\ {5 \over 3} & = x \\ \\ \text{Substitute } & x = {5 \over 3} \text{ into eqn of curve,} \\ y & = \left({5 \over 3} - 3\right) \sqrt{ {5 \over 3} - 1} \\ & = -{4 \over 3} \sqrt{ 2 \over 3} \\ & = -{4 \over 3} \left( \sqrt{2} \over \sqrt{3} \right) \\ & = -{4\sqrt{2} \over 3\sqrt{3}} \times {\sqrt{3} \over \sqrt{3}} \\ & = -{4\sqrt{6} \over 3(3)} \\ & = -{4\sqrt{6} \over 9} \\ \\ \therefore & \phantom{.} \left(1{2 \over 3}, -{4\sqrt{6} \over 9}\right) \end{align}

\begin{align} g(x) & = f(u) \\ \\ g'(x) & = f'(u) . {du \over dx} \phantom{00000} [\text{Chain rule}] \\ & = {1 \over u} . (2x + 1) \\ & = {1 \over x^2 + x} . (2x + 1) \\ & = {2x + 1 \over x^2 + x} \end{align}