1. Write in the form au^n is: a (1)/((2x-1)^2) sqrt (x^2-3x) C (2)/(sqrt (2-x^2)) d sqrt [3](x^3-x^2) (4)/((3-x)^3) f (10)/(x^2)-3 2 Find the gradient function (dy)/(dx) for: a y=(4x-5)^2 b y=(1)/(5-2x) y=sqrt (3x-x^2) d y=(1-3x)^4 e y=6(5-x)^3 f y=sqrt [3](2x^3-x^2) g y=(6)/((5x-4)^2) h y=(4)/(3x-x^2) y=2(x^2-(2)/(x))^3 3 Find the gradient of the tangent to: a y=sqrt (1-x^2) at x=(1)/(2) b y=(3x+2)^6 at x=-1 C y=(1)/((2x-1)^4) at x=1 d y=6times sqrt [3](1-2x) at x=0 e y=(4)/(x+2sqrt (x)) at x=4 f y=(x+(1)/(x))^3 at x=1 7

1. The correct answer is option a: $\frac {1}{(2x-1)^{2}}$.<br /><br />Explanation: The expression $\frac {1}{(2x-1)^{2}}$ can be written in the form $au^{n}$, where $a$ is a constant, $u$ is a variable, and $n$ is a positive integer. In this case, $a = 1$, $u = 2x-1$, and $n = 2$.<br /><br />2. The gradient function $\frac {dy}{dx}$ for each of the given functions is as follows:<br /><br />a) $y=(4x-5)^{2}$<br />$\frac {dy}{dx} = 2(4x-5) \cdot 4$<br /><br />b) $y=\frac {1}{5-2x}$<br />$\frac {dy}{dx} = -\frac {2}{(5-2x)^{2}}$<br /><br />c) $y=\sqrt {3x-x^{2}}$<br />$\frac {dy}{dx} = \frac {3-2x}{2\sqrt {3x-x^{2}}}$<br /><br />d) $y=(1-3x)^{4}$<br />$\frac {dy}{dx} = 4(1-3x)^{3} \cdot (-3)$<br /><br />e) $y=6(5-x)^{3}$<br />$\frac {dy}{dx} = 6 \cdot 3(5-x)^{2} \cdot (-1)$<br /><br />f) $y=\sqrt [3]{2x^{3}-x^{2}}$<br />$\frac {dy}{dx} = \frac {1}{3}(2x^{3}-x^{2})^{-\frac {2}{3}} \cdot (6x^{2}-2x)$<br /><br />g) $y=\frac {6}{(5x-4)^{2}}$<br />$\frac {dy}{dx} = -\frac {12}{(5x-4)^{3}}$<br /><br />h) $y=\frac {4}{3x-x^{2}}$<br />$\frac {dy}{dx} = \frac {4(1-2x)}{(3x-x^{2})^{2}}$<br /><br />i) $y=2(x^{2}-\frac {2}{x})^{3}$<br />$\frac {dy}{dx} = 2 \cdot 3(x^{2}-\frac {2}{x})^{2} \cdot (2x+\frac {2}{x^{2}})$<br /><br />3. The gradient of the tangent to each of the given functions at the specified points is as follows:<br /><br />a) $y=\sqrt {1-x^{2}}$ at $x=\frac {1}{2}$<br />$\frac {dy}{dx} = -\frac {1}{\sqrt {1-x^{2}}}$<br />$\frac {dy}{dx} \bigg|_{x=\frac {1}{2}} = -\frac {1}{\sqrt {1-\frac {1}{4}}} = -\frac {2}{\sqrt {3}}$<br /><br />b) $y=(3x+2)^{6}$ at $x=-1$<br />$\frac {dy}{dx} = 6(3x+2)^{5} \cdot 3$<br />$\frac {dy}{dx} \bigg|_{x=-1} = 6(3(-1)+2)^{5} \cdot 3 = 6(-1)^{5} \cdot 3 = -18$<br /><br />c) $y=\frac {1}{(2x-1)^{4}}$ at $x=1$<br />$\frac {dy}{dx} = -\frac {4}{(2x-1)^{5}} \cdot 2$<br />$\frac {dy}{dx} \bigg|_{x=1} = -\frac {4}{(2(1)-1)^{5}} \cdot 2 = -8$<br /><br />d) $y=6\times \sqrt [3]{1-2x}$ at $x=0$<br />$\frac {dy}{dx} = 6 \cdot \frac {1}{3}(1-2x)^{-\frac {2}{3}} \cdot (-2)$<br />$\frac {dy}{dx} \bigg|_{x=0} = 6 \cdot \frac {1}{3}(1-2(0))^{-\frac {2}{3}} \cdot (-2) = -12$<br /><br />e) $y=\frac {4}{x+2\sqrt {x}}$ at $x=4$<br />$\frac {dy}{dx} = -\frac {4(1+\frac {1}{\sqrt {x}})}{(x+2\sqrt {x})^{2}} \cdot