(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 is a constant.

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

(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)

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

(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 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)

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

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 $x$.

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 $x$.