1

$x = 0$ of $x = 2$	$x = 0$ of $x = ‐ 2$
$x = 0$ of $x = \sqrt[3]{2}$	$x = 0$ of $x = ‐ \sqrt[3]{2}$
$x = 0$ of $x = \sqrt[4]{2}$	Alleen $x = 0$
$x = 0$ of $x = \sqrt[8]{8}$	$x = 0$
$x = 0$ of $x = 4$	$x = 9$ (Subsitueer $\sqrt{x} = t$ .)
$x = 2$	$x = 8$

2

a

$a^{\frac{1}{2}} \cdot b \cdot c^{1 \frac{1}{2}}$	$a^{\frac{1}{3}} \cdot b^{\frac{2}{3}} \cdot c$
$a^{‐ \frac{1}{2}} \cdot b^{0} \cdot c^{\frac{1}{2}}$	$a^{0} \cdot b^{\frac{1}{2}} \cdot c$

b

$\sqrt{a b}$	$\sqrt[6]{a^{3} \cdot b^{2}}$
$\sqrt{\frac{a}{b}}$	$\sqrt[6]{\frac{a^{3}}{b^{2}}}$

c

$2^{\frac{1}{2}} \cdot a^{1 \frac{1}{2}} \cdot b^{‐ \frac{1}{4}} \cdot c^{‐ 1}$

3

$x^{2} - x$	$\frac{1}{\sqrt{x}}$
$\frac{x + a}{a^{2}}$	$\frac{1}{\sqrt{a x} (\sqrt{x} + \sqrt{a})}$

4

a

Dat volgt uit de quotiëntregel.

b

$\frac{\sqrt{x} - \frac{1}{2 \sqrt[]{x}} (x + 1)}{x} \cdot \frac{2 \sqrt{x}}{2 \sqrt{x}} = \frac{2 x - (x + 1)}{2 x \sqrt{x}} = \frac{x - 1}{2 x \sqrt{x}}$

c

$\frac{x - 1}{\sqrt{x}} = \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}} = \sqrt{x} - \frac{1}{\sqrt{x}}$

d

$\frac{x - 1}{2 x \sqrt{x}} = \frac{x}{2 x \sqrt{x}} - \frac{1}{2 x \sqrt{x}} = \frac{1}{2} (\frac{x}{x \sqrt{x}} - \frac{1}{x \sqrt{x}}) = \frac{1}{2} (x^{‐ \frac{1}{2}} - x^{‐ 1 \frac{1}{2}})$

e

$\frac{\sqrt{x} - \frac{1}{\sqrt{x}}}{\sqrt{x} + 1} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{x - 1}{\sqrt{x} (\sqrt{x} + 1)} = \frac{(\sqrt{x} + 1) (\sqrt{x} - 1)}{\sqrt{x} (\sqrt{x} + 1)} = \frac{\sqrt{x} - 1}{\sqrt{x}}$

5

a

$y = \frac{x + 2}{x^{2} - 4} = \frac{x + 2}{(x + 2) (x - 2)} = \frac{1}{x - 2}$

b

Met de kettingregel, levert meteen $\frac{d y}{d x}$ $= ‐ \frac{1}{{(x - 2)}^{2}}$ op.

c

$\frac{d y}{d x} = \frac{x^{2} - 4 - 2 x (x + 2)}{{(x^{2} - 4)}^{2}} = ‐ \frac{x^{2} + 4 x + 4}{{(x^{2} - 4)}^{2}} = ‐ \frac{{(x + 2)}^{2}}{{((x + 2) (x - 2))}^{2}} = ‐ \frac{1}{{(x - 2)}^{2}}$

6

a

$\frac{d y}{d x}$ $= \frac{1}{2 \sqrt{x + 1}} - 1$	$\frac{d y}{d x}$ $= \frac{1}{3 {\sqrt[3]{x + 1}}^{2}} - \frac{1}{3}$
$\frac{d y}{d x}$ $= \frac{1}{\sqrt{2 x + 1}} - 1$	$\frac{d y}{d x}$ $= \frac{2}{3 {\sqrt[3]{2 x + 1}}^{2}} - \frac{1}{3}$
$\frac{d y}{d x}$ $= \frac{x}{\sqrt{x^{2} + 1}} - 1$	$\frac{d y}{d x}$ $= \frac{2 x}{3 {\sqrt[3]{x^{2} + 1}}^{2}} - \frac{2}{3} x$

b

$\frac{d y}{d x}$ $= 5 {(x - 1)}^{4}$	$\frac{d y}{d x}$ $= ‐ \frac{4}{{(x - 1)}^{2}}$
$\frac{d y}{d x}$ $= 5 {(2 x - 1)}^{4} \cdot 2$	$\frac{d y}{d x}$ $= ‐ \frac{4}{{(2 x - 1)}^{2}} \cdot 2$
$\frac{d y}{d x}$ $= 5 {(x^{2} - 1)}^{4} \cdot 2 x$	$\frac{d y}{d x}$ $= ‐ \frac{4}{{(x^{2} - 1)}^{2}} \cdot 2 x$

c

$\frac{d y}{d x}$ $= \frac{‐ 3}{2 \sqrt{2 - x}}$	$\frac{d y}{d x}$ $= 3 {(2 - \sqrt{x})}^{‐ 4} \cdot \frac{1}{2 \sqrt{x}}$
$\frac{d y}{d x}$ $= \sqrt{2 - x} + \frac{‐ x}{2 \sqrt{2 - x}}$	$\frac{d y}{d x}$ $= {(2 - \sqrt{x})}^{‐ 3} + 3 {(2 - \sqrt{x})}^{‐ 4} \cdot \frac{x}{2 \sqrt{x}}$
$\frac{d y}{d x}$ $= 2 x \sqrt{2 - x} + \frac{‐ x^{2}}{2 \sqrt{2 - x}}$	$\frac{d y}{d x}$ $= 2 x {(2 - \sqrt{x})}^{‐ 3} + 3 {(2 - \sqrt{x})}^{‐ 4} \cdot \frac{x^{2}}{2 \sqrt{x}}$

d

$\frac{d y}{d x}$ $= \frac{‐ (3 x^{2} + 2 x - 1)}{{(x^{3} + x^{2} - x)}^{2}}$	$\frac{d y}{d x}$ $= \frac{‐ 8}{{(4 x + 2)}^{2}}$
$\frac{d y}{d x}$ $= \frac{‐ (2 x + 1)}{{(x^{2} + x - 1)}^{2}}$	$\frac{d y}{d x} =$ $0$
$\frac{d y}{d x}$ $= \frac{‐ (x^{2} + 1)}{{(x^{2} + x - 1)}^{2}}$	$\frac{d y}{d x}$ $= \frac{‐ 4 x}{{(x^{2} - 1)}^{2}}$

Opmerking: De functies $y = \frac{x}{x^{3} + x^{2} - x}$ en $y = \frac{x^{2}}{x^{3} + x^{2} - x}$ zijn te vereenvoudigen tot $y = \frac{1}{x^{2} + x - 1}$ en $y = \frac{x}{x^{2} + x - 1}$ ; de functie $y = \frac{2 x + 1}{4 x + 2}$ is te vereenvoudigen tot $y = \frac{1}{2}$ .

7

a

$‐ \frac{3}{4}$	$0$ , $‐ 2$
$0$	$‐ \frac{1}{2} + \sqrt{2}$ , $‐ \frac{1}{2} - \sqrt{2}$
geen $x$	$0$

b

$1$	geen $x$
$\frac{1}{2}$	geen $x$
$‐ 1$ , $0$ , $1$	$0$

c

geen $x$	geen $x$
$1 \frac{1}{3}$	geen $x$
$1 \frac{3}{5}$ , $0$	$0$ of $64$

d

$\frac{1}{3}$ , $‐ 1$	geen $x$
$‐ \frac{1}{2}$	alle $x$ , behalve $‐ \frac{1}{2}$
geen $x$	$0$