(a)

\begin{align} {d \over dx} (x + 2)^5 & = 5(x + 2)^4 .(1) \\ & = 5(x + 2)^4 \end{align}

(b)

\begin{align} {d \over dx} (2x - 1)^4 & = 4(2x - 1)^3 . (2) \\ & = 8(2x - 1)^3 \end{align}

(c)

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

(d)

\begin{align} {d \over dx} (1 - 4x)^{10} & = 10(1 - 4x)^9 . (-4) \\ & = -40(1 - 4x)^9 \end{align}

(e)

\begin{align} {d \over dx} (2 - 3x^2)^4 & = 4(2 - 3x^2)^3 . (-6x) \\ & = -24x (2 - 3x^2)^3 \end{align}

(f)

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

(a)

\begin{align} {d \over dx} \left( {4 \over 2x + 7} \right) & = 4(-1)(2x + 7)^{-2} . (2) \\ & = -8(2x + 7)^{-2} \\ & = -8 \left[ 1 \over (2x + 7)^2 \right] \\ & = -{8 \over (2x + 7)^2} \end{align}

(b)

\begin{align} {d \over dx} \left( {2 \over 6x^2 + 5} \right) & = {d \over dx} [2(6x^2 + 5)^{-1}] \\ & = 2(-1)(6x^2 + 5)^{-2} . (12x) \\ & = -24x(6x^2 + 5)^{-2} \\ & = -24x \left[ {1 \over (6x^2 + 5)^2 } \right] \\ & = -{24x \over (6x^2 + 5)^2} \end{align}

(c)

\begin{align} {d \over dx} \left[ {3 \over (3 - 4x)^3} \right] & = {d \over dx} [3(3 - 4x)^{-3}] \\ & = 3(-3)(3 - 4x)^{-4} . (-4) \\ & = 36(3 - 4x)^{-4} \\ & = 36 \left[ {1 \over (3 - 4x)^4} \right] \\ & = {36 \over (3 - 4x)^4} \end{align}

(d)

\begin{align} {d \over dx} \left[ {6 \over (2 - x)^2} \right] & = {d \over dx} [6(2 - x)^{-2}] \\ & = 6(-2)(2 - x)^{-3} . (-1) \\ & = 12(2 - x)^{-3} \\ & = 12 \left[ {1 \over (2 - x)^3} \right] \\ & = {12 \over (2 - x)^3} \end{align}

(a)

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

(b)

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

(c)

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

(d)

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

(a)

\begin{align} {dy \over dx} & = {d \over dx} (3x - 1)^4 \\ & = 4(3x - 1)^3 . (3) \\ & = 12(3x - 1)^3 \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 12[3(1) - 1]^3 \\ & = 96 \end{align}

(b)

\begin{align} y & = \sqrt{5 - 2x} \\ & = (5 - 2x)^{1 \over 2} \\ \\ {dy \over dx} & = {d \over dx} (5 - 2x)^{1 \over 2} \\ & = {1 \over 2}(5 - 2x)^{-{1 \over 2}} . (-2) \\ & = -(5 - 2x)^{-{1 \over 2}} \\ & = - {1 \over \sqrt{5 - 2x}} \\ \\ \text{When } & x = {1 \over 2}, \\ {dy \over dx} & = - {1 \over \sqrt{5 - 2\left({1 \over 2}\right)}} \\ & = -{1 \over 2} \end{align}

(c)

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

(d)

\begin{align} {dy \over dx} & = {d \over dx}\left[ {1 \over (4x - 5)^3} \right] \\ & = {d \over dx}(4x - 5)^{-3} \\ & = (-3)(4x - 5)^{-4} . (4) \\ & = -12(4x - 5)^{-4} \\ & = -12 \left[ {1 \over (4x - 5)^4} \right] \\ & = -{12 \over (4x - 5)^4} \\ \\ \text{Substitute } & y = {1 \over 27} \text{ into eqn of curve,} \\ {1 \over 27} & = {1 \over (4x - 5)^3} \\ (4x - 5)^3 & = 27 \\ \sqrt[3]{(4x - 5)^3} & = \sqrt[3]{27} \\ 4x - 5 & = 3 \\ 4x & = 8 \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into } {dy \over dx}, \\ {dy \over dx} & = -{12 \over [4(2) - 5]^4} \\ & = -{12 \over 81} \\ & = -{4 \over 27} \end{align}

(a)

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

(b)

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

(c)

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

(d)

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

(e)

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

(f)

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

(a)

\begin{align} {dy \over dx} & = {d \over dx} (x^2 - 2x - 4)^3 \\ & = (3)(x^2 - 2x - 4)^2 . (2x - 2) \\ & = (3)(2x - 2)(x^2 - 2x - 4)^2 \\ & = (6x - 6)(x^2 - 2x - 4)^2 \\ \\ \text{When } & x = -1, \\ {dy \over dx} & = [6(-1) - 6][(-1)^2 - 2(-1) - 4]^2 \\ & = -12 \end{align}

(b)

\begin{align} y & = {1 \over \sqrt{1 + x}} \\ & = (1 + x)^{-{1 \over 2}} \\ \\ {dy \over dx} & = \left(-{1 \over 2}\right)(1 + x)^{-{3 \over 2}}(1) \\ & = \left(-{1 \over 2}\right)\left[ {1 \over \sqrt{(1 + x)^3}} \right] \\ & = - {1 \over 2\sqrt{(1 + x)^3}} \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = - {1 \over 2\sqrt{(1 + 3)^3}} \\ & = - {1 \over 16} \end{align}

\begin{align} {dy \over dx} & = {d \over dx} (1 - x)^4 \\ & = 4(1 - x)^3 . (-1) \\ & = -4(1 - x)^3 \\ \\ \text{When } & {dy \over dx} = -4, \\ -4 & = -4(1 - x)^3 \\ 1 & = (1 - x)^3 \\ \sqrt[3]{1} & = \sqrt[3]{(1 - x)^3} \\ 1 & = 1 - x \\ x & = 1 - 1 \\ x & = 0 \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = (1 - 0)^4 \\ & = 1 \\ \\ \therefore \text{Coordinates } & \text{of the point is } (0, 1) \end{align}

\begin{align} y & = \sqrt[3]{x^2 - 2x + 5} \\ & = (x^2 - 2x + 5)^{1 \over 3} \\ \\ {dy \over dx} & = {1 \over 3}(x^2 - 2x + 5)^{-{2 \over 3}} . (2x - 2) \\ & = {1 \over 3} \left[ {1 \over (x^2 - 2x + 5)^{2 \over 3} } \right] (2x - 2) \\ & = {1 \over 3}\left[ {2x - 2 \over (x^2 - 2x + 5)^{2 \over 3}} \right] \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = {1 \over 3}\left[ {2x - 2 \over (x^2 - 2x + 5)^{2 \over 3}} \right] \\ 0 & = {2x - 2 \over (x^2 - 2x + 5)^{2 \over 3}} \\ \\ \therefore 2x - 2 & = 0 \\ 2x & = 2 \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = \sqrt[3]{(1)^2 - 2(1) + 5} \\ & = \sqrt[3]{4} \\ \\ \therefore \text{Coordinates } & \text{of the point is } (1, \sqrt[3]{4}) \end{align}

Take note $a$ is a constant, so ${d \over dx}(a) = 0 $

\begin{align} \text{When } & x = 2 \text{ and } {dy \over dx} = -{1 \over 3}, \\ -{1 \over 3} & = -{3 \over 2}\sqrt{a - 2} \\ {-{1 \over 3} \over -{3 \over 2}} & = \sqrt{a - 2} \\ {2 \over 9} & = \sqrt{a - 2} \\ \left({2 \over 9}\right)^2 & = (\sqrt{a - 2})^2 \\ {4 \over 81} & = a - 2 \\ {4 \over 81} + 2 & = a \\ {166 \over 81} & = a \end{align}

\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}

\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}

(i)

\begin{align} y & = (x^2 + 1)^n \\ \\ {dy \over dx} & = n(x^2 + 1)^{n - 1} . (2x) \\ & = 2n x (x^2 + 1)^{n - 1} \phantom{0000} \text{ (Shown)} \end{align}

(ii)

\begin{align} {dy \over dx} & = 2n x (x^2 + 1)^{n - 1} \\ \\ {dy \over dx} |_{x = 1} & = 2n(1)[(1)^2 + 1]^{n - 1} \\ & = 2n (2)^{n - 1} \\ & = n 2^n \\ \\ {d^2y \over dx^2} & = (2nx)[(n - 1)(x^2 + 1)^{n - 2} . (2x)] + (x^2 + 1)^{n - 1} (2n) \\ & = (4nx^2)(n - 1)(x^2 + 1)^{n - 2} + 2n (x^2 + 1)^{n - 1} \\ \\ {d^2y \over dx^2} |_{x = 1} & = [4n(1)^2] (n - 1)[(1)^2 + 1]^{n - 2} + 2n[(1)^2 + 1]^{n - 1} \\ & = (4n)(n - 1)(2)^{n - 2} + 2n (2)^{n - 1} \\ & = n (n - 1)(2^2)(2^{n - 2}) + n (2)(2^{n-1}) \\ & = n (n - 1)(2^n) + n (2^n) \\ & = n(2^n) [ (n - 1) + 1] \\ & = n(2^n) (n) \\ & = n^2 2^n \\ \\ { {dy \over dx} |_{x = 1} \over {d^2y \over dx^2} |_{x = 1}} & = { n 2^n \over n^2 2^n} \\ & = {1 \over n} \phantom{000} \text{ (Shown)} \end{align}

\begin{align} f(x) & = \sqrt{1 + \sqrt{x}} \\ & = \left(1 + x^{1 \over 2} \right)^{1 \over 2} \\ \\ f'(x) & = {1 \over 2} \left(1 + x^{1 \over 2} \right)^{-{1 \over 2}} . \left({1 \over 2} x^{-{1 \over 2}}\right) \\ & = {1 \over 2} (1 + \sqrt{x})^{-{1 \over 2}} \left[ {1 \over 2} \left(1 \over \sqrt{x}\right) \right] \\ & = {1 \over 2} \left(1 \over \sqrt{1 + \sqrt{x}} \right) \left(1 \over 2\sqrt{x}\right) \\ & = {1 \over 4 \sqrt{x} \sqrt{1 + \sqrt{x}}} \\ & = {1 \over 4 \sqrt{x(1 + \sqrt{x})}} \\ & = {1 \over 4 \sqrt{x + x\sqrt{x}}} \phantom{00} \text{ (Shown)} \end{align}

\begin{align} x^2 + y^2 & = 1 \\ y^2 & = 1 - x^2 \\ y & = \pm \sqrt{1 - x^2} \\ y & = \pm (1 - x^2)^{1 \over 2} \\ \\ {dy \over dx} & = \pm {1 \over 2} (1 - x^2)^{-{1 \over 2}} . (-2x) \\ & = \mp x(1 - x^2)^{-{1 \over 2}} \\ & = \mp {x \over \sqrt{1 - x^2}} \\ \\ \text{When } & x = {1 \over 2}, \\ {dy \over dx} & = \mp { {1 \over 2} \over \sqrt{1 - \left(1 \over 2\right)^2} } \\ & = \mp { {1 \over 2} \over \sqrt{3 \over 4}} \\ & = \mp \left( {1 \over 2} \div \sqrt{3 \over 4} \right) \\ & = \mp \left( {1 \over 2} \div {\sqrt{3} \over 2} \right) \\ & = \mp \left( {1 \over 2} \times {2 \over \sqrt{3}} \right) \\ & = \mp {2 \over 2\sqrt{3}} \\ & = \mp {1 \over \sqrt{3}} \end{align}

\begin{align} {d \over dx} \left(1 \over x^2 + 1\right) & = {d \over dx} (x^2 + 1)^{-1} \\ & = (-1)(x^2 + 1)^{-2} . (2x) \\ & = (-2x)(x^2 + 1)^{-2} \\ & = -{2x \over (x^2 + 1)^2} \\ \\ {d \over dx} \left(1 \over x^2 + 2\right) & = {d \over dx} (x^2 + 2)^{-1} \\ & = (-1)(x^2 + 2)^{-2} . (2x) \\ & = (-2x)(x^2 + 2)^{-2} \\ & = -{2x \over (x^2 + 2)^2} \\ \\ {d \over dx} \left(1 \over x^2 + 3\right) & = {d \over dx} (x^2 + 3)^{-1} \\ & = (-1)(x^2 + 3)^{-2} . (2x) \\ & = (-2x)(x^2 + 3)^{-2} \\ & = -{2x \over (x^2 + 3)^2} \\ \\ \\ {dy \over dx} & = {d \over dx} \left( {1 \over x^2 + 1} + {1 \over x^2 + 2} + {1 \over x^2 + 3} \right)^{-1} \\ & = (-1) \left( {1 \over x^2 + 1} + {1 \over x^2 + 2} + {1 \over x^2 + 3} \right)^{-2} \left( -{2x \over (x^2 + 1)^2} -{2x \over (x^2 + 2)^2} -{2x \over (x^2 + 3)^2} \right) \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = (-1) \left({1 \over 1^2 + 1} + {1 \over 1^2 + 2} + {1 \over 1^2 + 3} \right)^{-2} \left[ -{2(1) \over (1^2 + 1)^2} - {2(1) \over (1^2 + 2)^2} - {2(1) \over (1^2 + 3)^2} \right] \\ & = {122 \over 169} \end{align}