(a)

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

(b)

\begin{align} {d \over dx} (3x - 2)^5 & = 5(3x - 2)^4 . (3) \\ & = 15(3x - 2)^4 \end{align}

(c)

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

(d)

\begin{align} {d \over dx} (1 - 6x)^{9} & = 9(1 - 6x)^8 . (-6) \\ & = -54(1 - 6x)^8 \end{align}

(e)

\begin{align} {d \over dx} (5 - 4x^2)^6 & = 6(5 - 4x^2)^5 . (-8x) \\ & = -48x (5 - 4x^2)^5 \end{align}

(f)

\begin{align} {d \over dx} (-2x + x^2)^3 & = 3(-2x + x^2)^2 .(-2 + 2x) \\ & = 3(2x - 2)(-2x + x^2)^2 \\ & = 3(2)(x - 1)(x^2 - 2x)^2 \\ & = 6(x - 1)[x(x - 2)]^2 \\ & = 6(x - 1)(x)^2 (x - 2)^2 \\ & = 6x^2 (x - 1) (x - 2)^2 \end{align}

(a)

\begin{align} {d \over dx} \left( {3 \over 2x + 5} \right) & = {d \over dx} [3(2x + 5)^{-1}] \\ & = 3(-1)(2x + 5)^{-2} . (2) \\ & = -6(2x + 5)^{-2} \\ & = -6 \left[ 1 \over (2x + 5)^2 \right] \\ & = -{6 \over (2x + 5)^2} \end{align}

(b)

\begin{align} {d \over dx} \left( {5 \over 3x^2 + 7} \right) & = {d \over dx} [5(3x^2 + 7)^{-1}] \\ & = 5(-1)(3x^2 + 7)^{-2} . (6x) \\ & = -30x(3x^2 + 7)^{-2} \\ & = -30x \left[ {1 \over (3x^2 + 7)^2 } \right] \\ & = -{30x \over (3x^2 + 7)^2} \end{align}

(c)

\begin{align} {d \over dx} \left[ {6 \over (3 - 5x)^3} \right] & = {d \over dx} [6(3 - 5x)^{-3}] \\ & = 6(-3)(3 - 5x)^{-4} . (-5) \\ & = 90(3 - 5x)^{-4} \\ & = 90 \left[ {1 \over (3 - 5x)^4} \right] \\ & = {90 \over (3 - 5x)^4} \end{align}

(d)

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

(e)

\begin{align} {d \over dx} \sqrt{6 - 3x^2} & = {d \over dx} (6 - 3x^2)^{1 \over 2} \\ & = {1 \over 2}(6 - 3x^2)^{-{1 \over 2}} . (-6x) \\ & = -3x(6 - 3x^2)^{-{1 \over 2}} \\ & = -3x \left( 1 \over \sqrt{6 - 3x^2} \right) \\ & = - {3x \over \sqrt{6 - 3x^2}} \end{align}

(f)

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

(a)

\begin{align} {d \over dx} (x^2 + 5x)^6 & = 6(x^2 + 5x)^5 . (2x + 5) \\ & = 6(2x + 5) (x^2 + 5x)^5 \end{align}

(b)

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

(c)

\begin{align} {d \over dx} \left[ {1 \over 2(x^3 - 1)^6 } \right] & = {d \over dx} \left[ {1 \over 2} (x^3 - 1)^{-6} \right] \\ & = {1 \over 2} (-6) (x^3 - 1)^{-7} . (3x^2) \\ & = -9x^2 (x^3 - 1)^{-7} \\ & = -{9x^2 \over (x^3 - 1)^7} \end{align}

(d)

\begin{align} {d \over dx} (x^4 + 5x - \pi)^5 & = 5(x^4 + 5x - \pi)^4 . (4x^3 + 5) \\ & = 5(x^4 + 5x - \pi)^4(4x^3 + 5) \end{align}

(a)

\begin{align} h & = {1 \over (2 - 3t + 5t^2)^4} \\ & = (2 - 3t + 5t^2)^{-4} \\ \\ {dh \over dt} & = (-4)(2 - 3t + 5t^2)^{-5} . (-3 + 10t) \\ & = (-4) \left[{1 \over (2 - 3t + 5t^2)^5} \right] (-3 + 10t) \\ & = -{4(-3 + 10t) \over (2 - 3t + 5t^2)^5} \\ \end{align}

(b)

\begin{align} y & = {1 \over (\sqrt{r} - 1)^3} \\ & = (\sqrt{r} - 1)^{-3} \\ & = \left( r^{1 \over 2} - 1 \right)^{-3} \\ \\ {dy \over dr} & = (-3) \left( r^{1 \over 2} - 1 \right)^{-4} . \left({1 \over 2}r^{-{1 \over 2}} \right) \\ & = (-3) \left[ {1 \over (\sqrt{r} - 1)^4 } \right] \left[ {1 \over 2} \left(1 \over \sqrt{r}\right) \right] \\ & = -{3 \over 2\sqrt{r} (\sqrt{r} - 1)^4} \end{align}

(a)

\begin{align} {dy \over dx} & = {d \over dx} (3x - 2)^3 \\ & = 3(3x - 2)^2 . (3) \\ & = 9(3x - 2)^2 \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 9[3(2) - 2]^2 \\ & = 144 \end{align}

(b)

\begin{align} y & = \sqrt{7 - 3x} \\ & = (7 - 3x)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(7 - 3x)^{-{1 \over 2}} . (-3) \\ & = -{3 \over 2} (7 - 3x)^{-{1 \over 2}} \\ & = - {3 \over 2\sqrt{7 - 3x}} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = - {3 \over 2\sqrt{7 - 3(1)}} \\ & = -{3 \over 4} \end{align}

(c)

\begin{align} y & = {1 \over 3x - 2} \\ & = (3x - 2)^{-1} \\ \\ {dy \over dx} & = (-1)(3x - 2)^{-2} . (3) \\ & = -3(3x - 2)^{-2} \\ & = -3 \left[ 1 \over (3x - 2)^2 \right] \\ & = -{3 \over (3x - 2)^2} \\ \\ \text{Substitute } & y = 1 \text{ into eqn of curve,} \\ 1 & = {1 \over 3x - 2} \\ 3x - 2 & = 1 \\ 3x & = 3 \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{3 \over [3(1) - 2]^2} \\ & = -3 \end{align}

(d)

\begin{align} y & = {1 \over (2x - 5)^3} \\ & = (2x - 5)^{-3} \\ \\ {dy \over dx} & = (-3)(2x - 5)^{-4} . (2) \\ & = -6(2x - 5)^{-4} \\ & = -6 \left[ {1 \over (2x - 5)^4} \right] \\ & = -{6 \over (2x - 5)^4} \\ \\ \text{Substitute } & y = {1 \over 8} \text{ into eqn of curve,} \\ {1 \over 8} & = {1 \over (2x - 5)^3} \\ (2x - 5)^3 & = 8 \\ \sqrt[3]{(2x - 5)^3} & = \sqrt[3]{8} \\ 2x - 5 & = 2 \\ 2x & = 7 \\ x & = {7 \over 2} \\ \\ \text{Substitute } & x = {7 \over 2} \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{6 \over \left[2 \left({7 \over 2}\right) - 5 \right]^4} \\ & = -{3 \over 8} \end{align}

(a)

\begin{align} y & = (x + 2)^2\sqrt{x + 2} \\ & = (x + 2)^2(x + 2)^{1 \over 2} \\ & = (x + 2)^{5 \over 2} \phantom{00000000} [a^m \times a^n = a^{m + n}] \\ \\ {dy \over dx} & = {5 \over 2}(x + 2)^{3 \over 2} . (1) \\ & = {5 \over 2}(x + 2)^{3 \over 2} \\ & = {5 \over 2} \sqrt{(x + 2)^3} \end{align}

(b)

\begin{align} y & = {(2 - 3x^2)^2 \over \sqrt{2 - 3x^2}} \\ & = {(2 - 3x^2)^2 \over (2 - 3x^2)^{1 \over 2}} \\ & = (2 - 3x^2)^{3 \over 2} \phantom{00000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \\ {dy \over dx} & = {3 \over 2}(2 - 3x^2)^{1 \over 2} . (- 6x) \\ & = {3 \over 2}(-6x)(2 - 3x^2)^{1 \over 2} \\ & = -9x\sqrt{2 - 3x^2} \end{align}

(i)

(ii) Apply the result from (i)

\begin{align} {d \over dx}\left({x \over 1 + x}\right) & = {d \over dx}\left( 1 - {1 \over 1 + x} \right) \\ & = {d \over dx}(1) - {d \over dx}\left[ (1 + x)^{-1} \right] \\ & = 0 - (-1)(1 + x)^{-2} . (1) \\ & = (1 + x)^{-2} \\ & = {1 \over (1 + x)^2} \end{align}

(i)

\begin{align} {2x - 1 \over (x - 1)^2} & = {A \over (x - 1)} + {B \over (x - 1)^2} \\ & = {A(x - 1) \over (x - 1)^2} + {B \over (x - 1)^2} \\ & = {A(x - 1) + B \over (x - 1)^2} \\ \\ 2x - 1 & = A(x - 1) + B \\ \\ \text{Let } & x = 1, \\ 2(1) - 1 & = A(0) + B \\ 1 & = B \\ \\ 2x - 1 & = A(x - 1) + 1 \\ \\ \text{Let } & x = 0, \\ 2(0) - 1 & = A(0 - 1) + 1 \\ -1 & = A(-1) + 1 \\ -1 & = -A + 1 \\ A & = 1 + 1 \\ & = 2 \\ \\ \therefore {2x - 1 \over (x - 1)^2} & = {2 \over (x - 1)} + {1 \over (x - 1)^2} \\ \\ \therefore A & = 2, B = 1 \end{align}

(ii) Apply the results from part (i)

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

\begin{align} y & = (2 - 5x)^3 + 1 \\ \\ {dy \over dx} & = 3(2 - 5x)^2 . (-5) \\ & = -15(2 - 5x)^2 \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = -15(2 - 5x)^2 \\ {0 \over -15} & = (2 - 5x)^2 \\ 0 & = (2 - 5x)^2 \\ 0 & = 2 - 5x \\ 5x & = 2 \\ x & = {2 \over 5} \\ \\ \text{Substitute } & x = {2 \over 5} \text{ into eqn of curve,} \\ y & = \left[2 - 5\left(2 \over 5\right)\right]^3 + 1 \\ & = 1 \\ \\ \therefore & \phantom{.} \left({2 \over 5}, 1 \right) \end{align}

\begin{align} y & = \sqrt{x^2 - 4x + 1} \\ & = (x^2 - 4x + 1)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2} (x^2 - 4x + 1)^{-{1 \over 2}} . (2x - 4) \\ & = {1 \over 2}(2x - 4) (x^2 - 4x + 1)^{-{1 \over 2}} \\ & = (x - 2) (x^2 - 4x + 1)^{-{1 \over 2}} \\ & = {x - 2 \over \sqrt{x^2 - 4x + 1}} \\ \\ \text{When } & {dy \over dx} = 2, \\ 2 & = {x - 2 \over \sqrt{x^2 - 4x + 1}} \\ 2\sqrt{x^2 - 4x + 1} & = x - 2 \\ \left(2 \sqrt{x^2 - 4x + 1} \right)^2 & = (x - 2)^2 \\ (2)^2 (x^2 - 4x + 1) & = x^2 - 2(x)(2) + 2^2 \\ 4(x^2 - 4x + 1) & = x^2 - 4x + 4 \\ 4x^2 - 16x + 4 & = x^2 - 4x + 4 \\ 3x^2 - 12x & = 0 \\ x^2 - 4x & = 0 \\ x(x - 4) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & x - 4 & = 0 \\ &&&& x & = 4 \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = {0 - 2 \over \sqrt{(0)^2 - 4(0) + 1}} &&& {dy \over dx} & = {4 - 2 \over \sqrt{(4)^2 - 4(4) + 1}} \\ & = -2 &&& & = 2 \\ \\ \therefore \text{Reject } & x = 0 \\ \\ &&&& \text{Substitute } & x = 4 \text{ into eqn of curve,} \\ &&&& y & = \sqrt{(4)^2 - 4(4) + 1} \\ &&&& & = 1 \\ \\ &&&& \therefore & \phantom{.} (4, 1) \end{align}

\begin{align*} \text{Circumference} & = 2\pi r \\ \\ \text{Circumference of } A & = 2\pi (3) = 6\pi \\ \text{Circumference of } B & = 2\pi (1) = 2\pi \\ \text{Circumference of } C & = 2\pi (2) = 4\pi \\ \\ {dy \over dx} & = {dy \over du} \times {du \over dx} \\ \\ {dy \over du} & = {2\pi \over 4\pi} \phantom{000000} [\text{How fast } C \text{ changes compared to } B] \\ & = {1 \over 2} \\ \\ {du \over dx} & = {6\pi \over 2\pi} \phantom{000000} [\text{How fast } B \text{ changes compared to } A] \\ & = 3 \\ \\ {dy \over dx} & = {1 \over 2} \times 3 \\ & = {3 \over 2} \end{align*}

(a)

\begin{align} y & = [(x^2 - 1)^2 +4]^5 \\ & = [(x^2)^2 - 2(x^2)(1) + 1^2 + 4 ] ^5 \\ & = (x^4 - 2x^2 + 1 + 4)^5 \\ & = (x^4 - 2x^2 + 5)^5 \\ \\ {dy \over dx} & = 5(x^4 - 2x^2 + 5)^4 . [4x^3 - 2(2)x] \\ & = 5(x^4 - 2x^2 + 5)^4 (4x^3 - 4x) \\ & = 5(4x^3 - 4x) (x^4 - 2x^2 + 5)^4 \\ & = 5(4x)(x^2 - 1) (x^4 - 2x^2 + 5)^4 \\ & = 20x (x^2 - 1) (x^4 - 2x^2 + 5)^4 \end{align}

(b)

\begin{align} {d \over dx} \left[ 4 - \sqrt{16 - x^2} \right] & = {d \over dx} \left[ 4 - (16 - x^2)^{1 \over 2} \right] \\ & = - {1 \over 2} (16 - x^2)^{-{1 \over 2}} . (-2x) \\ & = x (16 - x^2)^{-{1 \over 2}} \\ \\ y & = {1 \over (4 - \sqrt{16 - x^2})^2} \\ & = \left(4 - \sqrt{16 - x^2} \right)^{-2} \\ \\ {dy \over dx} & = (-2) \left(4 - \sqrt{16 - x^2} \right)^{-3} . \left[ x (16 - x^2)^{-{1 \over 2}} \right] \\ & = \left[ - {2 \over \left(4 - \sqrt{16 - x^2} \right)^{3} } \right] \left[ {x \over \sqrt{16 - x^2} } \right] \\ & = -{2x \over \sqrt{16 - x^2} \left(4 - \sqrt{16 - x^2} \right)^{3} } \end{align}

\begin{align} y & = {1 \over \sqrt{x^2 + 3} } \\ & = (x^2 + 3)^{-{1 \over 2}} \\ \\ {dy \over dx} & = -{1 \over 2} (x^2 + 3)^{-{3 \over 2}} . (2x) \\ & = - x (x^2 + 3)^{-{3 \over 2}} \\ \\ \text{To show } & (x^2 + 3){dy \over dx} + xy = 0, \\ \text{L.H.S} & = (x^2 + 3){dy \over dx} + xy \\ & = (x^2 + 3) \left[ -x(x^2 + 3)^{-{3 \over 2}} \right] + x(x^2 + 3)^{-{1 \over 2}} \\ & = -x (x^2 + 3)^{-{1 \over 2}} + x(x^2 + 3)^{-{1 \over 2}} \\ & = 0 \\ & = \text{R.H.S} \end{align}