(a)

\begin{align} {d \over dx}(3x^{-7}) & = 3(-7)(x^{-8}) \\ & = -21x^{-8} \end{align}

(b)

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

(a)

\begin{align} {d \over dx} (2x^3 + 4x - 3) & = 2(3)x^2 + 4 - 0 \\ & = 6x^2 + 4 \end{align}

(b)

\begin{align} {d \over dx} (-4x^3 + 5x^2 - 12) & = -4(3)x^2 + 5(2)x - 0 \\ & = -12x^2 + 10x \end{align}

(c)

\begin{align} {d \over dx} \left( {\pi^2 x^2 \over 3} - {2x^4 \over 5} \right) & = {d \over dx} \left( {\pi^2 \over 3}x^2 - {2 \over 5}x^4 \right) \\ & = {\pi^2 \over 3}(2)x - {2 \over 5}(4)x^3 \\ & = {2\pi^2 \over 3}x - {8 \over 5}x^3 \end{align}

(d)

\begin{align} {d \over dx} \left( {3x^8 \over 20} - {\pi x^9 \over 11} \right) & = {d \over dx} \left( {3 \over 20}x^8 - {\pi \over 11}x^9 \right) \\ & = {3 \over 20}(8)x^7 - {\pi \over 11}(9)x^8 \\ & = {6 \over 5}x^7 - {9\pi \over 11}x^8 \end{align}

(e) Note a and b are constants.

\begin{align} {d \over dx} (3a + bx^2) & = (0) + b(2)x \\ & = 2bx \end{align}

(f) Note a and b are constants.

\begin{align} {d \over dx} \left( {a \over 2} - {b \over 2}x^3 \right) & = (0) - {b \over 2}(3)x^2 \\ & = - {3 \over 2}bx^2 \end{align}

(a)

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

(b)

\begin{align} {d \over dx} \left( 100x^2 + {100 \over x} \right) & = {d \over dx} (100x^2 + 100x^{-1}) \\ & = 100(2)x + 100(-1)x^{-2} \\ & = 200x - 100 \left(1 \over x^2\right) \\ & = 200x - {100 \over x^2} \end{align}

(c)

\begin{align} {d \over dx} \left( 9x^2 - {3 \over x^2} \right) & = {d \over dx} (9x^2 - 3x^{-2}) \\ & = 9(2)x - 3(-2)x^{-3} \\ & = 18x + 6 \left(1 \over x^3\right) \\ & = 18x + {6 \over x^3} \end{align}

(d)

\begin{align} {d \over dx} \left( {6 \over x^3} - {1 \over x} + 3 \right) & = {d \over dx} ( 6x^{-3} - x^{-1} + 3 ) \\ & = 6(-3)x^{-4} - (-1)x^{-2} + 0 \\ & = -18 \left(1 \over x^4\right) + x^{-2} \\ & = -{18 \over x^4} + {1 \over x^2} \end{align}

(e)

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

(f)

\begin{align} {d \over dx} (8x^2 + 3x - \sqrt{x}) & = {d \over dx} (8x^2 + 3x - x^{1 \over 2}) \\ & = 8(2)x + 3 - \left({1 \over 2}\right)x^{-{1 \over 2}} \\ & = 16x + 3 - {1 \over 2} \left(1 \over \sqrt{x}\right) \\ & = 16x + 3 - {1 \over 2\sqrt{x}} \end{align}

(a)

\begin{align} {dy \over dx} & = 6.5(4)x^3 \\ & = 26x^3 \end{align}

(b)

\begin{align} s & = {4 \over t^5} \\ & = 4t^{-5} \\ \\ {ds \over dt} & = 4(-5)t^{-6} \\ & = -20t^{-6} \\ & = -{20 \over t^6} \end{align}

(c)

\begin{align} f'(x) & = -3(-4)x^{-5} \\ & = 12x^{-5} \\ & = {12 \over x^5} \end{align}

(d)

\begin{align} g'(u) & = 0.5 \left(3 \over 5\right) u^{-{2 \over 5}} \\ & = {3 \over 10} u^{-{2 \over 5}} \\ & = {3 \over 10} \left(1 \over \sqrt[5]{u^2}\right) \\ & = {3 \over 10 \sqrt[5]{u^2}} \end{align}

(e)

\begin{align} h'(x) & = -{5 \over 4} \left(-{1 \over 5}\right) x^{-{6 \over 5}} \\ & = {1 \over 4} x^{-{6 \over 5}} \\ & = {1 \over 4} \left(1 \over \sqrt[5]{x^6}\right) \\ & = {1 \over 4 \sqrt[5]{x^6} } \end{align}

(a)

\begin{align} {d \over dx} \left({3x^2 + x - 1 \over \sqrt{x}}\right) & = {d \over dx} \left({3x^2 \over x^{1 \over 2}} + {x \over x^{1 \over 2}} - {1 \over x^{1 \over 2}} \right) \\ & = {d \over dx} \left( 3x^{3 \over 2} + x^{1 \over 2} - x^{- {1 \over 2}} \right) \phantom{000000000000000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = 3\left({3 \over 2}\right)x^{1 \over 2} + \left({1 \over 2}\right)x^{-{1 \over 2}} - \left(-{1 \over 2}\right)x^{-{3 \over 2}} \\ & = {9 \over 2}\sqrt{x} + {1 \over 2} \left(1 \over \sqrt{x}\right) + {1 \over 2}\left(1 \over \sqrt{x^3}\right) \\ & = {9 \over 2} \sqrt{x} + {1 \over 2\sqrt{x}} + {1 \over 2\sqrt{x^3}} \end{align}

(b)

\begin{align} {d \over dx} \left({6x^2 - \sqrt{x} + 2 \over 2x}\right) & = {d \over dx} \left( {6x^2 \over 2x} - {x^{1 \over 2} \over 2x} + {2 \over 2x} \right) \\ & = {d \over dx} \left( 3x - {1 \over 2}x^{-{1 \over 2}} + {1 \over x} \right) \phantom{0000000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = {d \over dx} \left( 3x - {1 \over 2}x^{-{1 \over 2}} + x^{-1} \right) \\ & = 3 -{1 \over 2}\left(-{1 \over 2}\right)x^{-{3 \over 2}} + (-1)x^{-2} \\ & = 3 + {1 \over 4}\left(1 \over \sqrt{x^3}\right) - {1 \over x^2} \\ & = 3 + {1 \over 4\sqrt{x^3} } - {1 \over x^2} \end{align}

(a)

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

(b)

\begin{align} f(t) & = t(\sqrt{t} + 3) \\ & = t(t^{1 \over 2} + 3) \\ & = t^{3 \over 2} + 3t \phantom{000000} [ a^m \times a^n = a^{m + n}] \\ \\ f'(t) & = {3 \over 2}t^{1 \over 2} + 3 \\ & = {3 \over 2} \sqrt{t} + 3 \end{align}

(c)

\begin{align} g(r) & = (1 + \sqrt{r})(1 - \sqrt{r}) \\ & = (1)^2 - (\sqrt{r})^2 \phantom{000000} [ (a + b)(a - b) = a^2 - b^2] \\ & = 1 - r \\ \\ g'(r) & = 0 - 1 \\ & = -1 \end{align}

(d)

\begin{align} t & = 3z^2(2 - \sqrt{z}) \\ & = 3z^2(2 - z^{1 \over 2}) \\ & = 6z^2 - 3z^{5 \over 2} \phantom{000000} [a^m \times a^n = a^{m + n}] \\ \\ {dt \over dz} & = 6(2)z - 3 \left(5 \over 2\right)z^{3 \over 2} \\ & = 12z - {15 \over 2} \sqrt{z^3} \end{align}

(e)

\begin{align} s & = (3p + 2)^2 \\ & = (3p)^2 + 2(3p)(2) + 2^2 \phantom{000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ & = 9p^2 + 12p + 4 \\ \\ {ds \over dp} & = 9(2)p + 12 \\ & = 18p + 12 \\ & = 6(3p + 2) \end{align}

(f)

\begin{align} h(x) & = {(2x + 1)(x - 1) \over x} \\ & = {2x^2 - x - 1 \over x} \\ & = {2x^2 \over x} - {x \over x} - {1 \over x} \\ & = 2x - 1 - x^{-1} \\ \\ h'(x) & = 2 - 0 - (-1)x^{-2} \\ & = 2 + x^{-2} \\ & = 2 + {1 \over x^2} \end{align}

(a)

\begin{align} {dy \over dx} & = 5(2)x - 4 + 0 \\ & = 10x - 4 \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 10(1) - 4 \\ & = 6 \end{align}

(b)

\begin{align} y & = \sqrt{x}(3 - 2x) \\ & = x^{1 \over 2} (3 - 2x) \\ & = 3x^{1 \over 2} - 2x^{3 \over 2} \phantom{00000000} [a^m \times a^n = a^{m + n}] \\ \\ {dy \over dx} & = 3\left( {1 \over 2} \right)x^{-{1 \over 2}} - 2\left({3 \over 2}\right)x^{1 \over 2} \\ & = {3 \over 2}x^{-{1 \over 2}} - 3x^{1 \over 2} \\ \\ \text{When } & x = 4, \\ {dy \over dx} & = {3 \over 2}(4)^{-{1 \over 2}} - 3(4)^{1 \over 2} \\ & = {3 \over 4} - 6 \\ & = -5{1 \over 4} \end{align}

(c)

\begin{align} y & = {(x - 1)(2x + 3) \over x} \\ & = {2x^2 + x - 3 \over x} \\ & = {2x^2 \over x} + {x \over x} - {3 \over x} \\ & = 2x + 1 - 3x^{-1} \\ \\ {dy \over dx} & = 2(1) + 0 - 3(-1)x^{-2} \\ & = 2 + 3 \left(1 \over x^2\right) \\ & = 2 + {3 \over x^2} \\ \\ \text{When } & x = -2, \\ {dy \over dx} & = 2 + {3 \over (-2)^2} \\ & = 2{3 \over 4} \end{align}

The x-coordinates of P and Q can be found by solving simultaneous equations using the equations of the curve and the line.

\begin{align} 3x^2 - 8y & = -5 \\ -8y & = -3x^2 - 5 \\ 8y & = 3x^2 + 5 \\ y & = {3 \over 8}x^2 + {5 \over 8} \\ \\ {dy \over dx} & = {3 \over 8}(2)(x) + 0 \\ & = {3 \over 4}x \\ \\ \text{Eqn of line: } \phantom{00} 4y - 3x & = 7 \\ 4y & = 3x + 7 \\ 8y & = 6x + 14 \phantom{000} \text{ --- (1)} \\ \\ \text{Eqn of curve: } \phantom{00} 3x^2 - 8y & = -5 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute (1)} & \text{ into (2)}, \\ 3x^2 - (6x + 14) & = -5 \\ 3x^2 - 6x - 14 & = -5 \\ 3x^2 - 6x - 9 & = 0 \\ x^2 - 2x - 3 & = 0 \\ (x - 3)(x + 1) & = 0 \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x + 1 & =0 \\ x & = 3 &&& x & = -1 \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = {3 \over 4}(3) &&& {dy \over dx} & = {3 \over 4}(-1) \\ & = 2{1 \over 4} &&& & = -{3 \over 4} \end{align}

\begin{align} y & = {10 \over x} - x \\ & = 10x^{-1} - x \\ \\ {dy \over dx} & = 10(-1)x^{-2} - 1 \\ & = -10x^{-2} - 1 \\ & = -{10 \over x^2} - 1 \\ \\ \text{Let } & {dy \over dx} = -{7 \over 2}, \\ -{7 \over 2} & = -{10 \over x^2} - 1 \\ -{5 \over 2} & = -{10 \over x^2} \\ {5 \over 2} & = {10 \over x^2} \\ 5x^2 & = 20 \\ x^2 & = {20 \over 5} \\ & = 4 \\ x & = \pm \sqrt{4} \\ & = \pm 2 \end{align} \begin{align} \text{Substitute } & x = 2 \text{ into eqn of curve,} &&& \text{Substitute } & x = -2 \text{ into eqn of curve,} \\ y & = {10 \over 2} - 2 &&& y & = {10 \over -2} - (-2) \\ & = 3 &&& & = -3 \\ \\ \therefore & \phantom{.} (2, 3) &&& \therefore & \phantom{.} (-2, -3) \end{align}

The steepness of a curve is the size of its gradient, ignoring the sign. The smallest possible size is 0, so the point on the curve where it is least steep is the point where gradient = 0.

\begin{align} {dy \over dx} & = 2(2)x - 3 \\ & = 4x - 3 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4x - 3 \\ 3 & = 4x \\ {3 \over 4} & = x \\ \\ \text{Substitute } & x = {3 \over 4} \text{ into eqn of curve,} \\ y & = 2\left(3 \over 4\right)^2 - 3\left(3 \over 4\right) + 7 \\ & = 5{7 \over 8} \\ \\ \therefore & \phantom{.} \left({3 \over 4}, 5{7 \over 8}\right) \end{align}

(i) To find the values of two unknown constants, we need to form two equations linking the constants.

\begin{align} y & = x^3 + px + q \\ \\ \text{Using } & (3, 16), \\ 16 & = (3)^3 + p(3) + q \\ 16 & = 27 + 3p + q \\ q & = -3p - 11 \phantom{00} \text{---(1)} \\ \\ {dy \over dx} & = (3)x^2 + p \\ & = 3x^2 + p \\ \\ \text{When } & x = 3 \text{ and } {dy \over dx} = 20, \\ 20 & = 3(3)^2 + p \\ 20 & = 27 + p \\ 20 - 27 & = p \\ -7 & = p \\ \\ \text{Substitute } & p = -7 \text{ into (1),} \\ q & = -3(-7) - 11 \\ & = 21 - 11 \\ & = 10 \\ \\ \therefore p & = -7, q = 10 \end{align}

(ii)

\begin{align} {dy \over dx} & = 3x^2 + p \\ \\ \text{When } & {dy \over dx} = 20 \text{ and } p = -7, \\ 20 & = 3x^2 - 7 \\ 20 + 7 & = 3x^2 \\ 27 & = 3x^2 \\ {27 \over 3} & = x^2 \\ 9 & = x^2 \\ \pm \sqrt{9} & = x \\ \pm 3 & = x \\ \\ \text{Since } x = 3 & \text{ is the given point } (3, 16), \\ \therefore x\text{-coordinate} & \text{ of other point is } -3 \\ \\ \text{Substitute } & x = -3 \text{ into eqn of curve,} \\ y & = (-3)^3 + (-7)(-3) + (10) \\ & = -27 + 21 + 10 \\ & = 4 \\ \\ \therefore & \phantom{.} (-3, 4) \end{align}

Note the powers of \(2\) (i.e. \(2^n\)) are \(2\), \(4\), \(8\), \(16\), \(32\), …, \(1024\), \(2048\) and so on…

\begin{align} y & = (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)...(x^{1024} + 1) \\ & = (x^2 - 1)(x^2 + 1)(x^4 + 1)...(x^{1024} + 1) \\ & = (x^4 - 1) (x^4 + 1) ... (x^{1024} + 1) \\ & = (x^{8} - 1) ... (x^{1024} + 1) \\ & \phantom{= }. \\ & \phantom{= }. \\ & \phantom{= }. \\ & = (x^{1024} - 1)(x^{1024} + 1) \\ & = x^{2048} - 1 \\ \\ {dy \over dx} & = 2048x^{2047} \end{align}

Discussion on this concept from Mathematics Stack Exchange (Thread 1, Thread 2)

$$ {d \over dx} (x + x + \phantom{.} ... \phantom{.} + x) = 1 + 1 + \phantom{.} ... \phantom{.} + 1 $$ The step above is wrong as the number of terms depends on \(x\). Differentiating term by term only works when the number of terms is fixed. Also, \(x + x + ... + x\) (\(x\) times) only makes sense when \(x\) is a positive whole number, so the left side is not a smooth curve that can be differentiated. Hence \(x = 2x\) (and so \(1 = 2\)) does not follow.

Discussion on this concept from Mathematics Stack Exchange (Thread 1, Thread 2)

$$ {d \over dx} \underbrace{(1 + 1 + 1 + \phantom{.} ... \phantom{.} + 1)}_{x \text{ times}} = 0 + 0 + 0 + \phantom{.} ... \phantom{.} + 0 $$ The step above is wrong as the number of 1's depends on \(x\). Differentiating term by term only works when the number of terms is fixed. Also, \(1 + 1 + ... + 1\) (\(x\) times) only makes sense when \(x\) is a positive whole number, so it cannot be differentiated like the line \(y = x\), whose gradient is \(1\).

(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} \\ & = {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{ (gradient is not 2)} \\ \\ &&&& \text{Substitute } & x = 4 \text{ into eqn of curve,} \\ &&&& y & = \sqrt{(4)^2 - 4(4) + 1} \\ &&&& & = 1 \\ \\ &&&& \therefore & \phantom{.} (4, 1) \end{align}

Note: Squaring both sides of an equation can introduce an extra root that does not satisfy the original equation, so always check each root. Here \(x = 0\) gives \(2\sqrt{1} = -2\), which is false.

When two gears are in contact, the rims of both gears move through the same distance. In one turn, the rim of a gear moves through a distance equal to its circumference. So when gear A makes x turns, its rim moves through (circumference of A) × x, and similarly for B (u turns) and C (y turns).

\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 \\ \\ \text{Gears } & A \text{ and } B\text{:} \\ 2\pi u & = 6\pi x \\ u & = 3x \\ {du \over dx} & = 3 \phantom{000000} [\text{How fast } B \text{ turns compared to } A] \\ \\ \text{Gears } & B \text{ and } C\text{:} \\ 4\pi y & = 2\pi u \\ y & = {1 \over 2}u \\ {dy \over du} & = {1 \over 2} \phantom{000000} [\text{How fast } C \text{ turns compared to } B] \\ \\ {dy \over dx} & = {dy \over du} \times {du \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}

(a)

\begin{align} u & = x &&& v & = (3x + 2)^2 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 2(3x + 2) . (3) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 6(3x + 2) \end{align} \begin{align} {d \over dx} \left[ x(3x + 2)^2 \right] & = x[6(3x + 2)] + (3x + 2)^2 (1) \\ & = 6x(3x + 2) + (3x + 2)^2 \\ & = (3x + 2) \left[ 6x + (3x + 2) \right] \\ & = (3x + 2) (6x + 3x + 2) \\ & = (3x + 2) (9x + 2) \end{align}

(b)

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

(c)

\begin{align} u & = 2 - 3x &&& v & = (4x + 1)^4 \\ {du \over dx} & = -3 &&& {dv \over dx} & = 4(4x + 1)^3 . (4) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 16(4x + 1)^3 \end{align} \begin{align} {d \over dx} \left[ (2 - 3x)(4x + 1)^4 \right] & = (2 - 3x)\left[ 16(4x + 1)^3 \right] + (4x + 1)^4 (-3) \\ & = 16(2 - 3x)(4x + 1)^3 - 3(4x + 1)^4 \\ & = (4x + 1)^3 [16(2 - 3x) - 3(4x + 1) ] \\ & = (4x + 1)^3 (32 - 48x - 12x - 3) \\ & = (4x + 1)^3 (29 - 60x) \end{align}

(d)

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

(a)

\begin{align} u & = \sqrt{x} &&& v & = (1 + x)^2 \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} &&& {dv \over dx} & = 2(1 + x) .(1) \phantom{00000} [\text{Chain rule}] \\ & = {1 \over 2\sqrt{x}} &&& & = 2(1 + x) \end{align} \begin{align} {d \over dx} [\sqrt{x} (1 + x)^2] & = (\sqrt{x}) [2(1 + x)] + (1 + x)^2 \left(1 \over 2\sqrt{x}\right) \\ & = 2 \sqrt{x} (1 + x) + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {2 \sqrt{x} (1 + x) \over 1} + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {2 \sqrt{x} (1 + x)(2\sqrt{x}) \over 2\sqrt{x}} + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {4x(1 + x) \over 2\sqrt{x}} + {(1 + x)^2 \over 2\sqrt{x}} \\ & = {4x(1 + x) + (1 + x)^2 \over 2\sqrt{x}} \\ & = {(1 + x) [4x + (1 + x)] \over 2\sqrt{x}} \\ & = {(1 + x) (4x + 1 + x) \over 2\sqrt{x}} \\ & = {(1 + x)(5x + 1) \over 2\sqrt{x}} \end{align}

(b)

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

(c)

\begin{align} u & = x^2 &&& v & = \sqrt{1 + 2x^2} \\ &&&& & = (1 + 2x^2)^{1 \over 2} \\ {du \over dx} & = 2x &&& {dv \over dx} & = {1 \over 2}(1 + 2x^2)^{-{1 \over 2}} . (4x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 2x (1 + 2x^2)^{-{1 \over 2}} \\ &&&& & = {2x \over \sqrt{1 + 2x^2}} \end{align} \begin{align} {d \over dx} \left[ x^2 \sqrt{1 + 2x^2} \right] & = (x^2) \left({2x \over \sqrt{1 + 2x^2}} \right) + \left(\sqrt{1 + 2x^2}\right)(2x) \\ & = {2x^3 \over \sqrt{1 + 2x^2} } + 2x \sqrt{1 + 2x^2} \\ & = {2x^3 \over \sqrt{1 + 2x^2}} + {2x \sqrt{1 + 2x^2} \over 1} \\ & = {2x^3 \over \sqrt{1 + 2x^2}} + {2x \sqrt{1 + 2x^2} \left(\sqrt{1 + 2x^2}\right) \over \sqrt{1 + 2x^2}} \\ & = {2x^3 \over \sqrt{1 + 2x^2}} + {2x(1 + 2x^2) \over \sqrt{1 + 2x^2}} \\ & = {2x^3 + 2x(1 + 2x^2) \over \sqrt{1 + 2x^2}} \\ & = {2x^3 + 2x + 4x^3 \over \sqrt{1 + 2x^2}} \\ & = {6x^3 + 2x \over \sqrt{1 + 2x^2}} \\ & = {2x(3x^2 + 1) \over \sqrt{1 + 2x^2}} \end{align}

(d)

\begin{align} u & = 2x - 1 &&& v & = \sqrt{5x^2 - 1} \\ &&&& & = (5x^2 - 1)^{1 \over 2} \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over 2}(5x^2 - 1)^{-{1 \over 2}} . (10x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 5x(5x^2 - 1)^{-{1 \over 2}} \\ &&&& & = {5x \over \sqrt{5x^2 - 1}} \end{align} \begin{align} {d \over dx} \left[ (2x - 1) \sqrt{5x^2 - 1} \right] & = (2x - 1) \left(5x \over \sqrt{5x^2 - 1}\right) + \left( \sqrt{5x^2 - 1} \right) (2) \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + 2 \sqrt{5x^2 - 1} \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + {2\sqrt{5x^2 - 1} \over 1} \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + {2\sqrt{5x^2 - 1} \left(\sqrt{5x^2 - 1}\right) \over \sqrt{5x^2 - 1}} \\ & = {5x(2x - 1) \over \sqrt{5x^2 - 1}} + {2(5x^2 - 1) \over \sqrt{5x^2 - 1}} \\ & = {5x(2x - 1) + 2(5x^2 - 1) \over \sqrt{5x^2 - 1}} \\ & = {10x^2 - 5x + 10x^2 - 2 \over \sqrt{5x^2 - 1}} \\ & = {20x^2 - 5x - 2 \over \sqrt{5x^2 - 1}} \end{align}

(e)

\begin{align} u & = \sqrt{x + 2} &&& v & = x\sqrt{x} + 1 \\ & = (x + 2)^{1 \over 2} &&& & = x(x)^{1 \over 2} + 1 \\ & &&& & = x^{3 \over 2} + 1 \\ {du \over dx} & = {1 \over 2}(x + 2)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] &&& {dv \over dx} & = {3 \over 2}x^{1 \over 2} \\ & = {1 \over 2} (x + 2)^{-{1 \over 2}} &&& & = {3 \over 2} \sqrt{x} \\ & = {1 \over 2\sqrt{x + 2}} &&& & = {3 \sqrt{x} \over 2} \end{align} \begin{align} {d \over dx} \left[ \sqrt{x + 2} \left(x \sqrt{x} + 1\right) \right] & = \sqrt{x + 2} \left( {3 \sqrt{x} \over 2} \right) + \left(x \sqrt{x} + 1\right) \left( {1 \over 2\sqrt{x + 2}} \right) \\ & = {3\sqrt{x} \sqrt{x + 2} \over 2} + {x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3\sqrt{x} \sqrt{x + 2} \left(\sqrt{x + 2}\right) \over 2\sqrt{x + 2}} + {x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3\sqrt{x} (x + 2) \over 2\sqrt{x + 2}} + {x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3\sqrt{x} (x + 2) + x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {3x \sqrt{x} + 6\sqrt{x} + x\sqrt{x} + 1 \over 2\sqrt{x + 2}} \\ & = {4x \sqrt{x} + 6\sqrt{x} + 1 \over 2\sqrt{x + 2}} \end{align}

(f)

\begin{align} u & = 3\sqrt{x} + 3 &&& v & = \sqrt{2 - 2x} \\ & = 3x^{1 \over 2} + 3 &&& & = (2 - 2x)^{1 \over 2} \\ {du \over dx} & = 3\left({1 \over 2}\right)x^{-{1 \over 2}} &&& {dv \over dx} & = {1 \over 2}(2 - 2x)^{-{1 \over 2}} . (-2) \phantom{00000} [\text{Chain rule}] \\ & = {3 \over 2} x^{-{1 \over 2}} &&& & = -(2 - 2x)^{-{1 \over 2}} \\ & = {3 \over 2\sqrt{x}} &&& & = -{1 \over \sqrt{2 - 2x}} \end{align} \begin{align} {d \over dx} \left[ \left( 3\sqrt{x} + 3 \right) \sqrt{2 - 2x} \right] & = \left( 3\sqrt{x} + 3 \right) \left( -{1 \over \sqrt{2 - 2x}} \right) + \sqrt{2 - 2x} \left( {3 \over 2\sqrt{x}} \right) \\ & = {-(3\sqrt{x} + 3) \over \sqrt{2 - 2x}} + {3\sqrt{2 - 2x} \over 2\sqrt{x}} \\ & = {-(3\sqrt{x} + 3)(2\sqrt{x}) \over 2\sqrt{x} \sqrt{2 - 2x}} + { 3\sqrt{2 - 2x} \left(\sqrt{2 - 2x}\right) \over 2\sqrt{x} \sqrt{2 - 2x}} \\ & = {-2\sqrt{x} (3 \sqrt{x} + 3) \over 2\sqrt{x}\sqrt{2 - 2x}} + {3(2 - 2x) \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {-2\sqrt{x} (3 \sqrt{x} + 3) + 3(2 - 2x) \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {-6x - 6\sqrt{x} + 6 - 6x \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {-12x - 6\sqrt{x} + 6 \over 2\sqrt{x}\sqrt{2 - 2x}} \\ & = {6 \left( -2x - \sqrt{x} + 1 \right) \over 2 \sqrt{x(2 - 2x)}} \\ & = { 3(-2x - \sqrt{x} + 1) \over \sqrt{x(2 - 2x)} } \\ & = { 3(1 - \sqrt{x} - 2x) \over \sqrt{2x(1 - x)}} \end{align}

(a)

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

(b)

\begin{align} u & = x^2 (x - 1) &&& v & = \sqrt{6x + 5} \\ & = x^3 - x^2 &&& & = (6x + 5)^{1 \over 2} \\ {du \over dx} & = 3x^2 - 2x &&& {dv \over dx} & = {1 \over 2}(6x + 5)^{-{1 \over 2}} . (6) \phantom{00000} [\text{Chain rule}] \\ &&&& & = 3(6x + 5)^{-{1 \over 2}} \\ &&&& & = {3 \over \sqrt{6x + 5}} \end{align} \begin{align} {d \over dx} \left[ (x^3 - x^2)\sqrt{6x + 5} \right] & = (x^3 - x^2) \left( {3 \over \sqrt{6x + 5}} \right) + \left( \sqrt{6x + 5} \right)(3x^2 - 2x) \\ & = {3(x^3 - x^2) \over \sqrt{6x + 5}} + (3x^2 - 2x)\sqrt{6x + 5} \\ & = {3(x^3 - x^2) \over \sqrt{6x + 5}} + { (3x^2 - 2x)\sqrt{6x + 5} \over 1} \\ & = {3(x^3 - x^2) \over \sqrt{6x + 5}} + {(3x^2 - 2x)\sqrt{6x + 5}(\sqrt{6x + 5}) \over \sqrt{6x + 5}} \\ & = {3x^3 - 3x^2 \over \sqrt{6x + 5}} + {(3x^2 - 2x)(6x + 5) \over \sqrt{6x + 5}} \\ & = {3x^3 - 3x^2 + (3x^2 - 2x)(6x + 5) \over \sqrt{6x + 5}} \\ & = {3x^3 - 3x^2 + 18x^3 + 15x^2 - 12x^2 - 10x \over \sqrt{6x + 5}} \\ & = {21x^3 - 10x \over \sqrt{6x + 5}} \\ & = {x(21x^2 - 10) \over \sqrt{6x + 5}} \end{align}

\begin{align} u & = x &&& v & = \sqrt{5 - x^2} \\ &&&& & = (5 - x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(5 - x^2)^{-{1 \over 2}} . (-2x) \phantom{00000} [\text{Chain rule}] \\ &&&& & = -x (5 - x^2)^{-{1 \over 2}} \\ &&&& & = {-x \over \sqrt{5 - x^2}} \end{align} \begin{align} {dy \over dx} & = (x)\left({-x \over \sqrt{5 - x^2}}\right) + \left(\sqrt{5 - x^2} \right) (1) \\ & = {-x^2 \over \sqrt{5 - x^2}} + \sqrt{5 - x^2} \\ & = {-x^2 \over \sqrt{5 - x^2}} + {\sqrt{5 - x^2} \over 1} \\ & = {-x^2 \over \sqrt{5 - x^2}} + {\sqrt{5 - x^2}\sqrt{5 - x^2} \over \sqrt{5 - x^2}} \\ & = {-x^2 \over \sqrt{5 - x^2}} + {5 - x^2 \over \sqrt{5 - x^2}} \\ & = {-x^2 + 5 - x^2 \over \sqrt{5 - x^2}} \\ & = {5 - 2x^2 \over \sqrt{5 - x^2}} \phantom{0} \text{ (Shown)} \end{align}

(i)

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

(ii)

\begin{align} \text{Substitute } & x = 5 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {3(5) - 2 \over 2\sqrt{(5) - 1}} \\ & = 3{1 \over 4} \end{align}

The points where the curve meets the x-axis are the x-intercepts, where y = 0.

\begin{align} u & = (x - 1)^3 &&& v & = x + 1 \\ {du \over dx} & = 3(x - 1)^2 . (1) &&& {dv \over dx} & = 1 \\ & = 3(x - 1)^2 \end{align} \begin{align} {dy \over dx} & = (x - 1)^3 (1) + (x + 1)[3(x - 1)^2] \\ & = (x - 1)^3 + 3(x + 1)(x - 1)^2 \\ & = (x - 1)^2 [(x - 1) + 3(x + 1)] \\ & = (x - 1)^2 (x - 1 + 3x + 3) \\ & = (x - 1)^2 (4x + 2) \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = (x - 1)^3 (x + 1) \end{align} \begin{align} (x - 1)^3 & = 0 && \text{ or } & x + 1 & =0 \\ x - 1 & = 0 &&& x & = -1 \\ x & = 1 \\ \\ \text{Substitute} & \text{ into } {dy \over dx}, &&& \text{Substitute} & \text{ into } {dy \over dx}, \\ {dy \over dx} & = (1 - 1)^2 [4(1) + 2] &&& {dy \over dx} & = (-1 - 1)^2[4(-1) + 2] \\ & = 0 &&& & = -8 \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} (x - 1)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x - 1}} \end{align} \begin{align} {dy \over dx} & = (x - 3) \left({1 \over 2\sqrt{x - 1}}\right) + \left( \sqrt{x - 1} \right) (1) \\ & = {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}\left(2 \sqrt{x - 1} \right) \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} \text{Let } & {dy \over dx} = 0, \\ 0 & = {3x - 5 \over 2\sqrt{x - 1} } \\ \therefore 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} \\ & = \left(-{4 \over 3}\right) \sqrt{ 2 \over 3} \\ & = \left(-{4 \over 3}\right) {\sqrt{2} \over \sqrt{3}} \\ & = -{4 \sqrt{2} \over 3\sqrt{3}} \\ & = -{4 \sqrt{2} \over 3\sqrt{3}} \times {\sqrt{3} \over \sqrt{3}} \phantom{00000} [\text{Rationalise denominator}] \\ & = -{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}

(i)

\begin{align} u & = \sqrt{x} &&& v & = (x - 3)^4 \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} &&& {dv \over dx} & = 4(x - 3)^3 . (1) \\ & = {1 \over 2\sqrt{x}} &&& & = 4(x - 3)^3 \end{align} \begin{align} {dy \over dx} & = (\sqrt{x}) [4(x - 3)^3] + (x - 3)^4 \left(1 \over 2\sqrt{x}\right) \\ & = 4\sqrt{x} (x - 3)^3 + {1 \over 2\sqrt{x}}(x - 3)^4 \\ & = (x - 3)^3 \left[ 4\sqrt{x} + {1 \over 2\sqrt{x}}(x - 3) \right] \\ & = (x - 3)^3 \left[ {4\sqrt{x} \over 1} + {x - 3 \over 2\sqrt{x}} \right] \\ & = (x - 3)^3 \left[ {4\sqrt{x}(2\sqrt{x}) \over 2\sqrt{x}} + {x - 3 \over 2\sqrt{x}} \right] \\ & = (x - 3)^3 \left( {8x + x - 3 \over 2\sqrt{x}} \right) \\ & = (x - 3)^3 \left( {9x - 3 \over 2\sqrt{x}} \right) \\ & = {(x - 3)^3 (9x - 3) \over 2\sqrt{x}} \\ & = {3(x - 3)^3 (3x - 1) \over 2\sqrt{x}} \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {3(x - 3)^3 (3x - 1) \over 2\sqrt{x}} \\ \therefore 0 & = 3(x - 3)^3 (3x - 1) \end{align} \begin{align} (x - 3)^3 & = 0 && \text{ or } & 3x - 1 & = 0 \\ x - 3 & = 0 &&& 3x & = 1 \\ x & = 3 &&& x & = {1 \over 3} \end{align}

\begin{align} y & = {x^2 - 1 \over x^2 + 1} \\ & = (x^2 - 1)\left(1 \over x^2 + 1\right) \\ & = (x^2 - 1)(x^2 + 1)^{-1} \end{align}
\begin{align} u & = x^2 - 1 &&& v & = (x^2 + 1)^{-1} \\ {du \over dx} & = 2x &&& {dv \over dx} & = (-1)(x^2 + 1)^{-2} . (2x) \\ & &&& & = -2x (x^2 + 1)^{-2} \\ & &&& & = {-2x \over (x^2 + 1)^2} \end{align} \begin{align} {dy \over dx} & = (x^2 - 1) \left[ {-2x \over (x^2 + 1)^2} \right] + (x^2 + 1)^{-1} (2x) \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + \left(1 \over x^2 + 1\right)(2x) \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + {2x \over x^2 + 1} \\ & = {-2x(x^2 - 1) \over (x^2 + 1)^2} + {2x (x^2 + 1) \over (x^2 + 1)(x^2 + 1)} \\ & = {-2x^3 + 2x \over (x^2 + 1)^2} + {2x^3 + 2x \over (x^2 + 1)^2} \\ & = {-2x^3 + 2x + 2x^3 + 2x \over (x^2 + 1)^2} \\ & = {4x \over (x^2 + 1)^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {4x \over (x^2 + 1)^2} \\ \therefore 0 & = 4x \\ 0 & = x \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = {0^2 - 1 \over 0^2 + 1} \\ & = -1 \end{align}

\begin{align} u & = x &&& v & = \sqrt{9 - x^2} \\ & &&& & = (9 - x^2)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (9 - x^2)^{-{1 \over 2}} . (-2x) \\ & &&& & = -x (9 - x^2)^{-{1 \over 2}} \\ & &&& & = {-x \over \sqrt{9 - x^2}} \end{align} \begin{align} {dy \over dx} & = (x) \left( {-x \over \sqrt{9 - x^2}} \right) + \left( \sqrt{9 - x^2}\right) (1) \\ & = { -x^2 \over \sqrt{9 - x^2}} + \sqrt{9 - x^2} \\ \\ \\ y & = x\phantom{00}\text{--- (1)} \\ \\ y & = x\sqrt{9 - x^2}\phantom{00}\text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x & = x\sqrt{9 - x^2} \\ x - x\sqrt{9 - x^2} & = 0 \\ (x)\left(1 - \sqrt{9 - x^2}\right) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & 1 - \sqrt{9 - x^2} & = 0 \\ & &&& 1 & = \sqrt{9 - x^2} \\ & &&& 1^2 & = 9 - x^2 \\ & &&& 1 & = 9 - x^2 \\ & &&& x^2 & = 9 - 1 \\ & &&& x^2 & = 8 \\ & &&& x & = \pm \sqrt{8} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = { -(0)^2 \over \sqrt{9 - (0)^2} } + \sqrt{9 - (0)^2} &&& {dy \over dx} & = { -(\pm \sqrt{8})^2 \over \sqrt{9 - (\pm \sqrt{8})^2} } + \sqrt{9 - (\pm \sqrt{8})^2} \\ & = 3 &&& & = {-8 \over \sqrt{1}} + \sqrt{1} \\ &&&& & = -7 \end{align}

Recall that parallel lines have the same gradient

\begin{align} 5x - y & = 0 \\ y & = 5x \\ y & = 5x + 0 \\ \\ \text{Comparing with } & y = mx + c, \\ m & = 5 \\ \\ \therefore \text{Gradient of tangent} & = 5 \end{align}
\begin{align} u & = 3x - 1 &&& v & = x - 2 \\ {du \over dx} & = 3 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = (3x - 1)(1) + (x - 2)(3) \\ & = 3x - 1 + 3x - 6 \\ & = 6x - 7 \\ \\ \text{When } & {dy \over dx} = 5, \\ 5 & = 6x - 7 \\ 12 & = 6x \\ {12 \over 6} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = [3(2) - 1](2 - 2) \\ & = 0 \\ \\ \therefore & \phantom{.} (2, 0) \end{align}

Since the curve meets the x-axis at point A, the y-coordinate of A is 0.

\begin{align} u & = x - a &&& v & = \sqrt{x - b} \\ & &&& & = (x - b)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (x - b)^{-{1 \over 2}} . (1) \\ & &&& & = {1 \over 2} (x - b)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x - b}} \end{align} \begin{align} {dy \over dx} & = (x - a) \left(1 \over 2\sqrt{x - b}\right) + \left( \sqrt{x - b} \right) (1) \\ & = {x - a \over 2\sqrt{x - b}} + \sqrt{x - b} \\ & = {x - a \over 2\sqrt{x - b}} + {\sqrt{x - b} \over 1} \\ & = {x - a \over 2\sqrt{x - b}} + {\sqrt{x - b} (2\sqrt{x - b}) \over 2\sqrt{x - b}} \\ & = {x - a \over 2\sqrt{x - b}} + {2(x - b) \over 2\sqrt{x - b}} \\ & = {x - a + 2(x - b) \over 2\sqrt{x - b}} \\ & = {x - a + 2x - 2b \over 2\sqrt{x - b}} \\ & = {3x - a - 2b \over 2\sqrt{x - b}} \\ \\ \\ \text{When } & x = b + 1, \\ {dy \over dx} & = {3(b + 1) - a - 2b \over 2\sqrt{b + 1 - b} } \\ & = {3b + 3 - a - 2b \over 2\sqrt{1}} \\ & = {b + 3 - a \over 2} \\ \\ \text{Substitute } & A(b + 1, 0) \text{ into eqn of curve,} \\ 0 & = (b + 1 - a)\sqrt{b + 1 - b} \\ 0 & = (b + 1 - a) \sqrt{1} \\ 0 & = (b + 1 - a)(1) \\ 0 & = b + 1 - a \\ a & = b + 1 \\ \\ \therefore {dy \over dx} & = {b + 3 - (b + 1) \over 2} \\ & = {b + 3 - b - 1 \over 2} \\ & = {2 \over 2} \\ & = 1 \phantom{0} \text{ (Shown)} \end{align}

(i)

Since \(f(a) = 0\), by the Factor Theorem, \((x - a)\) is a factor of \(f(x)\).

(ii)

\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ f(x) & = (x - a) Q(x) + 0 \\ f(x) & = (x - a) Q(x), \text{ where } Q(x) \text{ denotes a polynomial} \end{align} \begin{align} u & = x - a &&& v & = Q(x) \\ {du \over dx} & = 1 &&& {dv \over dx} & = Q'(x) \end{align} \begin{align} f'(x) & = (x - a)Q'(x) + Q(x) . (1) \\ & = (x - a)Q'(x) + Q(x) \\ \\ f'(a) & = (a - a)Q'(a) + Q(a) \\ & = 0[Q'(a)] + Q(a) \\ & = 0 + Q(a) \\ & = Q(a) \\ \\ \text{If } & f'(a) = 0, \\ Q(a) & = 0 \\ \\ \implies (x - a) & \text{ is a factor of } Q(x) \\ \therefore (x - a) & \text{ is a repeated factor of } f(x)\text{,} \\ & \text{ i.e. } (x - a)^2 \text{ is a factor of } f(x) \end{align}

(a)

\begin{align} u & = x^2 + 2 &&& v & = 3 - x^3 \\ {du \over dx} & = 2x &&& {dv \over dx} & = -3x^2 \end{align} \begin{align} {dy \over dx} & = { (3 - x^3)(2x) - (x^2 + 2)(-3x^2) \over (3 - x^3)^2} \\ & = { 2x(3 - x^3) + 3x^2 (x^2 + 2) \over (3 - x^3)^2} \\ & = { 6x - 2x^4 + 3x^4 + 6x^2 \over (3 - x^3)^2 } \\ & = { x^4 + 6x^2 + 6x \over (3 - x^3)^2} \end{align}

(b)

\begin{align} u & = 1 - \sqrt{t} &&& v & = 1 + \sqrt{t} \\ & = 1 - t^{1 \over 2} &&& & = 1 + t^{1 \over 2} \\ {du \over dt} & = -{1 \over 2}t^{-{1 \over 2}} &&& {dv \over dt} & = {1 \over 2}t^{-{1 \over 2}} \\ & = -{1 \over 2\sqrt{t}} &&& & = {1 \over 2\sqrt{t}} \end{align} \begin{align} g'(t) & = { (1 + \sqrt{t}) \left(-{1 \over 2\sqrt{t}}\right) - (1 - \sqrt{t})\left(1 \over 2\sqrt{t}\right) \over (1 + \sqrt{t})^2} \\ & = { { -(1 + \sqrt{t}) \over 2\sqrt{t}} - {1 - \sqrt{t} \over 2\sqrt{t}} \over (1 + \sqrt{t})^2 } \\ & = { { -(1 + \sqrt{t}) - (1 - \sqrt{t}) \over 2 \sqrt{t}} \over (1 + \sqrt{t})^2} \\ & = { { -1 - \sqrt{t} - 1 + \sqrt{t} \over 2\sqrt{t}} \over (1 + \sqrt{t})^2} \\ & = { { -2 \over 2\sqrt{t}} \over (1 + \sqrt{t})^2 } \\ & = { {-1 \over \sqrt{t}} \over (1 + \sqrt{t})^2 } \\ & = { -1 \over \sqrt{t} (1 + \sqrt{t})^2 } \end{align}

(c)

\begin{align} u & = (x^2 - 1)^3 &&& v & = 3 - x^4 \\ {du \over dx} & = 3(x^2 - 1)^2 . (2x) \phantom{00000} [\text{Chain rule}] &&& {dv \over dx} & = -4x^3 \\ & = 6x(x^2 - 1)^2 \end{align} \begin{align} f'(x) & = {(3 - x^4)[6x(x^2 - 1)^2] - (x^2 - 1)^3 (-4x^3) \over (3 - x^4)^2 } \\ & = { 6x(3 - x^4)(x^2 - 1)^2 + 4x^3(x^2 - 1)^3 \over (3 - x^4)^2} \\ & = { 2x(x^2 - 1)^2 [ 3(3 - x^4) + 2x^2(x^2 - 1)] \over (3 - x^4)^2} \\ & = { 2x(x^2 - 1)^2 (9 - 3x^4 + 2x^4 - 2x^2) \over (3 - x^4)^2} \\ & = { 2x(x^2 - 1)^2 (9 - x^4 - 2x^2) \over (3 - x^4)^2} \end{align}

(d)

\begin{align} u & = a^2 + 2a - 7 &&& v & = \sqrt{a} \\ & &&& & = a^{1 \over 2} \\ {du \over da} & = 2a + 2 &&& {dv \over da} & = {1 \over 2} a^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{a}} \end{align} \begin{align} {dr \over da} & = { (\sqrt{a})(2a + 2) - (a^2 + 2a - 7)\left(1 \over 2\sqrt{a}\right) \over (\sqrt{a})^2} \\ & = { \sqrt{a} (2a + 2) - {a^2 + 2a - 7 \over 2\sqrt{a}} \over a} \\ & = { \sqrt{a} (2a + 2) - {a^2 + 2a - 7 \over 2\sqrt{a}} \over a} \times {2\sqrt{a} \over 2\sqrt{a}} \\ & = { \sqrt{a} (2a + 2)(2\sqrt{a}) - (a^2 + 2a - 7) \over 2a \sqrt{a}} \\ & = { 2a(2a + 2) - a^2 - 2a + 7 \over 2a\sqrt{a}} \\ & = {4a^2 + 4a - a^2 - 2a + 7 \over 2a \sqrt{a}} \\ & = {3a^2 + 2a + 7 \over 2a \sqrt{a}} \\ & = {3a^2 + 2a + 7 \over 2a (a^{1 \over 2})} \\ & = {3a^2 + 2a + 7 \over 2a^{3 \over 2}} \end{align}

(a)

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

(b)

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

(c)

\begin{align} u & = 3x &&& v & = \sqrt{3x + 1} \\ & &&& & = (3x + 1)^{1 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = {1 \over 2} (3x + 1)^{-{1 \over 2}} . (3) \\ & &&& & = {3 \over 2}(3x + 1)^{-{1 \over 2}} \\ & &&& & = {3 \over 2\sqrt{3x + 1}} \end{align} \begin{align} {d \over dx} \left( 3x \over \sqrt{3x + 1} \right) & = {(\sqrt{3x + 1})(3) - (3x) \left({3 \over 2\sqrt{3x + 1}}\right) \over \left(\sqrt{3x + 1}\right)^2} \\ & = {3\sqrt{3x + 1} - {9x \over 2\sqrt{3x + 1}} \over 3x + 1} \\ & = {3\sqrt{3x + 1} - {9x \over 2\sqrt{3x + 1}} \over 3x + 1} \times {2\sqrt{3x + 1} \over 2\sqrt{3x + 1}} \\ & = {3\sqrt{3x + 1} (2\sqrt{3x + 1}) - 9x \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {6(3x + 1) - 9x \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {18x + 6 - 9x \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {9x + 6 \over 2(3x + 1)\sqrt{3x + 1}} \\ & = {3(3x + 2) \over 2(3x + 1)\sqrt{3x + 1}} \end{align}

(d)

\begin{align} u & = 5x^2 &&& v & = \sqrt{3x^2 - 1} \\ & &&& & = (3x^2 - 1)^{1 \over 2} \\ {du \over dx} & = 10x &&& {dv \over dx} & = {1 \over 2} (3x^2 - 1)^{-{1 \over 2}} . (6x) \\ & &&& & = 3x(3x^2 - 1)^{-{1 \over 2}} \\ & &&& & = {3x \over \sqrt{3x^2 - 1}} \end{align} \begin{align} {d \over dx} \left( 5x^2 \over \sqrt{3x^2 - 1} \right) & = {(\sqrt{3x^2 - 1}) (10x) - (5x^2) \left(3x \over \sqrt{3x^2 - 1}\right) \over \left(\sqrt{3x^2 - 1}\right)^2 } \\ & = {10x\sqrt{3x^2 - 1} - {15x^3 \over \sqrt{3x^2 - 1}} \over 3x^2 - 1} \\ & = {10x\sqrt{3x^2 - 1} - {15x^3 \over \sqrt{3x^2 - 1}} \over 3x^2 - 1} \times {\sqrt{3x^2 - 1} \over \sqrt{3x^2 - 1}} \\ & = {10x\sqrt{3x^2 - 1}(\sqrt{3x^2 - 1}) - 15x^3 \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {10x(3x^2 - 1) - 15x^3 \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {30x^3 - 10x - 15x^3 \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {15x^3 - 10x \over (3x^2 - 1)\sqrt{3x^2 - 1}} \\ & = {5x(3x^2 - 2) \over (3x^2 - 1)\sqrt{3x^2 - 1}} \end{align}

\begin{align} u & = x^2 + x + 1 &&& v & = x + 1 \\ {du \over dx} & = 2x + 1 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1) (2x + 1) - (x^2 + x + 1)(1) \over (x + 1)^2} \\ & = {2x^2 + x + 2x + 1 - (x^2 + x + 1) \over (x + 1)^2} \\ & = {2x^2 + 3x + 1 - x^2 - x - 1 \over (x + 1)^2} \\ & = {x^2 + 2x \over (x + 1)^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {(1)^2 + 2(1) \over (1 + 1)^2} \\ & = {3 \over 4} \end{align}

The point where the curve crosses the x-axis is the x-intercept, where y = 0.

\begin{align} u & = x - 3 &&& v & = 2 - x \\ {du \over dx} & = 1 &&& {dv \over dx} & = -1 \end{align} \begin{align} {dy \over dx} & = {(2 - x)(1) - (x - 3)(-1) \over (2 - x)^2} \\ & = {2 - x + (x - 3) \over (2 - x)^2} \\ & = {2 - x + x - 3 \over (2 - x)^2} \\ & = -{1 \over (2 - x)^2} \\ \\ \text{Substitute } & y = 0 \text{ into eqn of curve,} \\ 0 & = {x - 3 \over 2 - x} \\ 0 & = x - 3 \\ 3 & = x \\ \\ \text{Substitute } & x = 3 \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{1 \over [2 - (3)]^2} \\ & = -1 \end{align}

\begin{align} u & = \sqrt{2x - 1} &&& v & = x \\ & = (2x - 1)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(2x - 1)^{-{1 \over 2}} .(2) \phantom{000} [\text{Chain rule}] &&& {dv \over dx} & = 1 \\ & = (2x - 1)^{-{1 \over 2}} \\ & = {1 \over \sqrt{2x - 1}} \end{align} \begin{align} {dy \over dx} & = { (x)\left(1 \over \sqrt{2x - 1}\right) - (\sqrt{2x - 1})(1) \over (x)^2} \\ & = { {x \over \sqrt{2x - 1}} - \sqrt{2x - 1} \over x^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = { {(1) \over \sqrt{2(1) - 1}} - \sqrt{2(1) - 1} \over (1)^2} \\ & = {1 - 1 \over 1} \\ & = 0 \end{align}

(i)

\begin{align} u & = 3x^5 - x^3 &&& v & = x^2 \\ {du \over dx} & = 15x^4 - 3x^2 &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2)(15x^4 - 3x^2) - (3x^5 - x^3)(2x) \over (x^2)^2} \\ & = { x^2(15x^4 - 3x^2) - 2x(3x^5 - x^3) \over x^4} \\ & = { 15x^6 - 3x^4 - 6x^6 + 2x^4 \over x^4 } \\ & = { 9x^6 - x^4 \over x^4 } \\ & = { 9x^6 \over x^4 } - {x^4 \over x^4} \\ & = 9x^2 - 1 \end{align}

(ii)

\begin{align} y & = {3x^5 - x^3 \over x^2} \\ & = {3x^5 \over x^2} - {x^3 \over x^2} \\ & = 3x^3 - x \\ \\ {dy \over dx} & = 9x^2 - 1 \\ \\ \therefore \text{Ans} & \text{wers are the same} \end{align}

\begin{align} u & = \sqrt{x^2 + 1} &&& v & = 3x + 4 \\ & = (x^2 + 1)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(x^2 + 1)^{-{1 \over 2}} . (2x) &&& {dv \over dx} & = 3 \\ & = x(x^2 + 1)^{-{1 \over 2}} \\ & = {x \over \sqrt{x^2 + 1}} \end{align} \begin{align} {dy \over dx} & = {(3x + 4) \left(x \over \sqrt{x^2 + 1}\right) - (\sqrt{x^2 + 1})(3) \over (3x + 4)^2} \\ & = { {x(3x + 4) \over \sqrt{x^2 + 1}} - 3\sqrt{x^2 + 1} \over (3x + 4)^2} \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = { {(0)[3(0) + 4] \over \sqrt{(0)^2 + 1}} - 3\sqrt{(0)^2 + 1} \over [3(0) + 4]^2 } \\ & = {0 - 3 \over 16} \\ & = -{3 \over 16} \end{align}

(i) Use product rule for the differentiation

\begin{align} u & = 1 - x^2 &&& v & = \sqrt{x + 3} \\ & &&& & = (x + 3)^{1 \over 2} \\ {du \over dx} & = -2x &&& {dv \over dx} & = {1 \over 2}(x + 3)^{-{1 \over 2}} . (1) \\ & &&& & = {1 \over 2} (x + 3)^{-{1 \over 2}} \\ & &&& & = {1 \over 2\sqrt{x + 3}} \end{align} \begin{align} {d \over dx} \left[ (1 - x^2) \sqrt{x + 3} \right] & = (1 - x^2) \left(1 \over 2\sqrt{x + 3}\right) + (\sqrt{x + 3})(-2x) \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - 2x \sqrt{x + 3} \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - {2x \sqrt{x + 3} \over 1} \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - {2x \sqrt{x + 3} (2\sqrt{x + 3}) \over 2\sqrt{x + 3}} \\ & = {1 - x^2 \over 2\sqrt{x + 3}} - {4x (x + 3) \over 2\sqrt{x + 3}} \\ & = {1 - x^2 - 4x(x + 3) \over 2\sqrt{x + 3}} \\ & = {1 - x^2 - 4x^2 - 12x \over 2\sqrt{x + 3}} \\ & = {1 - 5x^2 - 12x \over 2\sqrt{x + 3}} \end{align}

(ii) The rate of change of \(y\) with respect to \(x\) is denoted by \({dy \over dx}\)

\begin{align} u & = (1 - x^2)\sqrt{x + 3} &&& v & = x^3 \\ {du \over dx} & = {1 - 5x^2 - 12x \over 2\sqrt{x + 3}} \phantom{00000} [\text{From (i)}] &&& {dv \over dx} & = 3x^2 \end{align} \begin{align} {dy \over dx} & = { (x^3) \left({1 - 5x^2 - 12x \over 2\sqrt{x + 3}}\right) - (1 - x^2)\sqrt{x + 3} (3x^2) \over (x^3)^2} \\ & = { {x^3(1 - 5x^2 - 12x) \over 2\sqrt{x + 3}} - 3x^2 (1 - x^2) \sqrt{x + 3} \over x^6} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = { {(1)^3 [1 - 5(1)^2 - 12(1)] \over 2\sqrt{1 + 3}} - 3(1)^2 [1 - (1)^2] \sqrt{1 + 3} \over (1)^6} \\ & = {-4 - 0 \over 1} \\ & = -4 \end{align}

Method 1: By quotient rule

\begin{align} u & = \sqrt{x + 1} &&& v & = 2x^2 - 1 \\ & = (x + 1)^{1 \over 2} \\ {du \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) &&& {dv \over dx} & = 4x \\ & = {1 \over 2} (x + 1)^{-{1 \over 2}} \\ & = {1 \over 2\sqrt{x + 1}} \end{align} \begin{align} {dy \over dx} & = {(2x^2 - 1) \left(1 \over 2\sqrt{x + 1}\right) - (\sqrt{x + 1})(4x) \over (2x^2 - 1)^2} \\ & = { {2x^2 - 1 \over 2\sqrt{x + 1}} - 4x \sqrt{x + 1} \over (2x^2 - 1)^2} \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = { {2(0)^2 - 1 \over 2\sqrt{0 + 1}} - 4(0) \sqrt{0 + 1} \over [2(0)^2 - 1]^2} \\ & = -{1 \over 2} \end{align}

Method 2: By product rule

$$ y = {\sqrt{x + 1} \over 2x^2 - 1} = (x + 1)^{1 \over 2} (2x^2 - 1)^{-1} $$
\begin{align} u & = (x + 1)^{1 \over 2} &&& v & = (2x^2 - 1)^{-1} \\ {du \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) &&& {dv \over dx} & = (-1)(2x^2 - 1)^{-2} . (4x) \\ & = {1 \over 2} (x + 1)^{-{1 \over 2}} &&& & = -4x(2x^2 - 1)^{-2}\\ & = {1 \over 2\sqrt{x + 1}} &&& & = -{4x \over (2x^2 - 1)^2} \end{align} \begin{align} {dy \over dx} & = (x + 1)^{1 \over 2} \left[-{4x \over (2x^2 - 1)^2}\right] + (2x^2 - 1)^{-1} \left(1 \over 2\sqrt{x + 1}\right) \\ & = -{4x \sqrt{x + 1} \over (2x^2 - 1)^2} + {1 \over 2 (2x^2 - 1) \sqrt{x + 1}} \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = -{4(0) \sqrt{0 + 1} \over [2(0)^2 - 1]^2} + {1 \over 2[2(0)^2 - 1] \sqrt{0 + 1}} \\ & = -{1 \over 2} \end{align}

\begin{align} u & = 5x + 3 &&& v & = 10x - 6 \\ {du \over dx} & = 5 &&& {dv \over dx} & = 10 \end{align} \begin{align} {dy \over dx} & = {d \over dx}\left( {5x + 3 \over 10x - 6} \right)^4 \\ & = (4)\left( {5x + 3 \over 10x - 6} \right)^3 \underbrace{{d \over dx}\left({5x + 3 \over 10x - 6}\right)}_{\text{Quotient rule}} \\ & = 4\left( {5x + 3 \over 10x - 6} \right)^3 \left[ {(10x - 6)(5) - (5x + 3)(10) \over (10x - 6)^2} \right] \\ & = 4\left( {5x + 3 \over 10x - 6} \right)^3 \left[ {50x - 30 - 50x - 30 \over (10x - 6)^2} \right] \\ & = 4 \left[{(5x + 3)^3 \over (10x - 6)^3}\right] \left[{-60 \over (10x - 6)^2} \right] \\ & = 4 \left[ -60(5x + 3)^3 \over (10x - 6)^5 \right] \\ & = {-240(5x + 3)^3 \over (10x - 6)^5} \end{align}
To find the gradient, we need the x-coordinates of the points where y = 16
\begin{align} \text{Substitute } & y = 16 \text{ into eqn of curve,} \\ 16 & = \left( {5x + 3 \over 10x - 6} \right)^4 \\ \pm\sqrt[4]{16} & = {5x + 3 \over 10x - 6} \\ \pm2 & = {5x + 3 \over 10x -6} \end{align} \begin{align} {5x + 3 \over 10x - 6} & = 2 && \text{ or } & {5x + 3 \over 10x - 6} & = -2 \\ 5x + 3 & = 2(10x - 6) &&& 5x + 3 & = -2(10x - 6) \\ 5x + 3 & = 20x - 12 &&& 5x + 3 & = -20x + 12 \\ -15x & = -15 &&& 25x & = 9 \\ x & = 1 &&& x & = {9 \over 25} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = {-240[5(1) + 3]^3 \over [10(1) - 6]^5} &&& {dy \over dx} & = {-240\left[ 5 \left(9 \over 25 \right) + 3\right]^3 \over \left[ 10\left(9 \over 25 \right) - 6 \right]^5} \\ & = -120 &&& & = {1000 \over 3} \end{align}

$$y = \sqrt{x - a \over b - x} = \left( {x - a \over b - x} \right)^{1 \over 2}$$
\begin{align} u & = x - a &&& v & = b - x \\ {du \over dx} & = 1 &&& {dv \over dx} & = -1 \end{align} \begin{align} \require{cancel} {dy \over dx} & = {1 \over 2}\left( {x - a \over b - x} \right)^{-{1 \over 2}} \underbrace{{d \over dx}\left( {x - a \over b - x} \right)}_{\text{Quotient Rule}} \\ & = {1 \over 2}\left( {x - a \over b - x} \right)^{-{1 \over 2}} \left[ { (b - x)(1) - (x - a)(-1) \over (b - x)^2} \right] \\ & = {1 \over 2}\left( {x - a \over b - x} \right)^{-{1 \over 2}} \left[ {b - x + x - a \over (b - x)^2} \right] \\ & = {1 \over 2}\sqrt{b - x \over x - a} \left[ {b - a \over (b - x)^2} \right] \\ \\ \text{When } & x = {a + b \over 2}, \\ {dy \over dx} & = {1 \over 2}\sqrt{b - \left({a + b \over 2}\right) \over \left({a + b \over 2}\right) - a} \left[{b - a \over \left(b - \left({a + b \over 2}\right)\right)^2}\right] \\ & = {1 \over 2}\sqrt{{2b \over 2} - {a + b \over 2} \over {a + b \over 2} - {2a \over 2}} \left[{b - a \over \left({2b \over 2} - {a + b \over 2}\right)^2}\right] \\ & = {1 \over 2}\sqrt{{2b - a - b \over 2} \over {a + b - 2a \over 2}} \left[{b - a \over \left({2b - a - b \over 2}\right)^2}\right] \\ & = {1 \over 2}\sqrt{{b - a \over 2} \over {b - a \over 2}} \left[{b - a \over \left({b - a \over 2}\right)^2}\right] \\ & = {1 \over 2}\sqrt{1} \left[{(b - a) \over \left({(b - a)^2 \over 4}\right)}\right] \\ & = {1 \over 2}(1)\left[(b - a) \div \left( {(b - a)^2 \over 4} \right)\right] \\ & = {1 \over 2}\left[ \cancel{(b - a)} \times {4 \over (b - a)^{\cancel{2}}} \right] \\ & = {1 \over \cancel{2}}\left( \cancel{4}^2 \over b - a \right) \\ & = {2 \over b - a} \text{ (Shown)} \end{align}

(i) \(y = {25 \over 4}\) is a horizontal line with gradient = 0. The gradient of the curve at A is equal to 0.

\begin{align} y & = x\left({k \over \sqrt{x}} - 1 \right) \\ & = {kx \over \sqrt{x}} - x \\ & = {kx \over x^{1 \over 2}} - x \\ & = kx^{1 \over 2} - x \\ \\ {dy \over dx} & = k \left(1 \over 2\right) x^{-{1 \over 2}} - 1 \\ & = {k \over 2\sqrt{x}} - 1 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {k \over 2\sqrt{x}} - 1 \\ 1 & = {k \over 2\sqrt{x}} \\ \sqrt{x} & = {k \over 2} \\ x & = \left(k \over 2\right)^2 \\ & = {k^2 \over 4} \\ \\ \therefore & \phantom{.} A \left({k^2 \over 4}, {25 \over 4}\right) \\ \\ \text{Substitute } & A \left({k^2 \over 4}, {25 \over 4}\right) \text{ into eqn of curve,} \\ {25 \over 4} & = k \left(k^2 \over 4\right)^{1 \over 2} - {k^2 \over 4} \\ {25 \over 4} & = k \left(k \over 2\right) - {k^2 \over 4} \\ {25 \over 4} & = {k^2 \over 2} - {k^2 \over 4} \\ {25 \over 4} & = {2k^2 \over 4} - {k^2 \over 4} \\ {25 \over 4} & = {2k^2 - k^2 \over 4} \\ 25 & = 2k^2 - k^2 \\ 25 & = k^2 \\ \pm \sqrt{25} & = k \\ \\ k & = 5 \text{ or } k = -5 \text{ (Reject, since } k > 0) \end{align}

(ii)

\begin{align} {dy \over dx} & = {k \over 2\sqrt{x}} - 1 \\ \\ \text{As } x & \rightarrow \infty, {k \over 2\sqrt{x}} \rightarrow 0 \\ \therefore \text{As } x & \rightarrow \infty, {dy \over dx} \rightarrow -1 \\ \\ \therefore \text{Gradient of } l & = -1 \\ \\ \\ y & = mx + c \\ y & = (-1)x + c \\ y & = -x + c \\ \\ \text{From the diagram, } & l \text{ passes through } O(0, 0). \\ \text{Using } & O(0, 0), \\ 0 & = -0 + c \\ 0 & = c \\ \\ \therefore \text{Eqn of } l: & \phantom{0} y = -x \end{align}

(i)

\begin{align} u & = x^{n + 1} - 1 &&& v & = x - 1 \\ {du \over dx} & = (n + 1)x^n &&& {dv \over dx} & = 1 \end{align} \begin{align} {d \over dx} \left( {x^{n + 1} - 1 \over x - 1} \right) & = {(x - 1)[(n + 1)x^n] - (x^{n + 1} - 1)(1) \over (x - 1)^2} \\ & = { (n + 1)(x^{n + 1} - x^n) - (x^{n + 1} - 1) \over (x - 1)^2} \\ & = { (n + 1)x^{n + 1} - (n + 1)x^n - x^{n + 1} + 1 \over (x - 1)^2} \\ & = { n x^{n + 1} - (n + 1)x^n + 1 \over (x - 1)^2 } \\ \\ {d \over dx} (1 + x + x^2 + ... + x^{n - 1} + x^n) & = 0 + 1 + 2x + ... + (n - 1)x^{n - 2} + n x^{n - 1} \\ \\ \therefore 1 + 2x + ... +(n - 1)x^{n - 2} + nx^{n - 1} & = { n x^{n + 1} - (n + 1)x^n + 1 \over (x - 1)^2 } \end{align}

(ii)

\begin{align} \text{From } & \text{(i),} \\ 1 + 2x + ... +(n - 1)x^{n - 2} + nx^{n - 1} & = { n x^{n + 1} - (n + 1)x^n + 1 \over (x - 1)^2 } \\ \\ \text{Let } & x = 3 \text{ and } n = 10, \\ 1 + 2(3) + ... + 9(3)^8 + 10 (3)^9 & = {(10)(3)^{11} - (11)(3)^{10} + 1 \over (3 - 1)^2} \\ & = {1\,771\,470 - 649\,539 + 1 \over 4} \\ & = 280 \phantom{.} 483 \end{align}

(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) Note a is a constant, so \({1 \over 3}\sqrt{a}\) is a constant and its derivative is 0.

\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) \\ \\ u & = x^2 + x \\ {du \over dx} & = 2x + 1 \\ \\ 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}