1

a

$B = \frac{1}{20} \cdot R$

b

$K = 1,7 \cdot B$

c

$K = 1,7 \cdot 0,05 \cdot R = 0,085 \cdot R$

2

a

$O = 6 r^{2}$ en $V = r^{3}$

b

Dan $r = \sqrt[3]{5}$ en $O = 6 \cdot {\sqrt[3]{5}}^{2} \approx 17,54$

c

$r = V^{ \frac{1}{3}}$ invullen in $O = 6 r^{2}$ geeft: $O = 6 \cdot {(V^{ \frac{1}{3}})}^{2} = 6 \cdot V^{ \frac{2}{3}}$ .

d

$r = \sqrt{\frac{1}{6} \cdot 5}$ , dus $V = {(\sqrt{\frac{1}{6} \cdot 5})}^{3} \approx 0,76$ .

e

$O = 6 \cdot V^{ \frac{2}{3}} \Leftrightarrow \frac{1}{6} O = V^{ \frac{2}{3}} \Leftrightarrow V = {(\frac{1}{6} O)}^{1 \frac{1}{2}}$ , dus $V = 0,07 \cdot O^{1,50}$ .

3

a

$49$ ; $y = {(2 x + 1)}^{2}$ $= 4 x^{2} + 4 x + 1$

b

$‐ 216$ ; $y = ‐ 8 x^{3}$

c

$324$ ; $y = 4 x^{4}$

4

Voor de tweede ketting: $\frac{d u}{d x} = ‐ 2$ , $\frac{d y}{d u} = 3 u^{2}$ en $\frac{d y}{d x} = ‐ 24 x^{2}$ ,
voor de derde ketting: $\frac{d u}{d x} = 4 x$ , $\frac{d y}{d u} = 2 u$ en $\frac{d y}{d x} = 16 x^{3}$ .

5

a

$\frac{d u}{d x}$ $= ‐ x$ en $\frac{d y}{d u}$ $= u$ .

b

$\frac{d u}{d x}$ $= ‐ 1$ als $x = 1$ , dus de raaklijn in $A$ heeft helling $‐ 1$ .
$\frac{d y}{d u}$ $= 2$ als $u = 2$ , dus de raaklijn in $B$ heeft helling $2$ .

c

$‐ 0,001$ en $‐ 0,002$ ; $‐ 0,002$ en $‐ 0,004$ .

d

$Δ u$ uit $Δ x$ door te vermenigvuldigen met $‐ 1$ ; $Δ y$ uit $Δ u$ door te vermenigvuldigen met $2$ ; dus $Δ y$ uit $Δ x$ door te vermenigvuldigen met $‐ 1 \cdot 2 = ‐ 2$ .

6

Klopt.

7

$f$ is de ketting $x \to u \to y$ met $u = x + \frac{1}{x}$ en $y = u^{3}$ ,
dus $f^{'} (x) = 3 u^{2} \cdot (1 - \frac{1}{x^{2}}) = 3 {(x + \frac{1}{x})}^{2} \cdot (1 - \frac{1}{x^{2}})$
$g$ is de ketting $x \to u \to y$ met $u (x) = x^{10}$ en $y (u) = \frac{1}{u}$ ,
dus $g^{'} (x) = ‐ \frac{1}{u^{2}} \cdot 10 x^{9} = \frac{‐10 x^{9}}{{(x^{10})}^{2}}$
$h$ is de ketting $x \to u \to y$ met $u (x) = 2 x + 1$ en $y (u) = \frac{1}{u}$ ,
dus $h^{'} (x) = ‐ \frac{1}{u^{2}} \cdot 2 = ‐ \frac{2}{{(2 x + 1)}^{2}}$
$k$ is de ketting $x \to u \to y$ met $u (x) = 1 - 2 x$ en $y (u) = \frac{2}{u}$ ,
dus $k^{'} (x) = ‐ \frac{2}{u^{2}} \cdot ‐ 2 = \frac{4}{{(1 - 2 x)}^{2}}$

8

$\frac{{(a^{3})}^{\frac{1}{6}}}{{(a^{\frac{1}{6}})}^{2}} = \frac{a^{\frac{3}{6}}}{a^{\frac{2}{6}}} = a^{\frac{3}{6} - \frac{2}{6}} = a^{\frac{1}{6}}$	$b^{2} \cdot \frac{b^{\frac{3}{4}}}{{(b^{‐ 2})}^{\frac{1}{3}}} = b^{2} \cdot \frac{b^{\frac{3}{4}}}{b^{‐ \frac{2}{3}}} = b^{2} \cdot b^{\frac{3}{4} - ‐ \frac{2}{3}} = b^{3 \frac{5}{12}}$
$\frac{{(\frac{1}{c^{2}})}^{3}}{{(\frac{1}{c^{3}})}^{4}} = \frac{{(c^{‐ 2})}^{3}}{{(c^{‐ 3})}^{4}} = \frac{c^{‐ 6}}{c^{‐ 12}} = c^{‐ 6 - ‐ 12} = c^{6}$	$\frac{d^{\frac{1}{2}} \cdot d^{\frac{2}{3}} \cdot d^{\frac{3}{4}}}{{({(d^{\frac{1}{2}})}^{\frac{2}{3}})}^{\frac{3}{4}}} = \frac{d^{\frac{1}{2} + \frac{2}{3} + \frac{3}{4}}}{d^{\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}}} = \frac{d^{\frac{23}{12}}}{d^{\frac{1}{4}}} = d^{\frac{20}{12}} = d^{1 \frac{2}{3}}$

9

$x \sqrt{x^{3}} = x \cdot {(x^{3})}^{\frac{1}{2}} = x \cdot x^{1 \frac{1}{2}} = x^{2 \frac{1}{2}}$	$\frac{1}{x} \sqrt[5]{x^{2}} = x^{‐ 1} \cdot {(x^{2})}^{\frac{1}{5}} = x^{‐ 1} \cdot x^{\frac{2}{5}} = x^{‐ \frac{3}{5}}$
$\sqrt[3]{\sqrt{x}} = {(x^{\frac{1}{2}})}^{\frac{1}{3}} = x^{\frac{1}{6}}$	$\sqrt[3]{\frac{1}{x^{2}}} = {(x^{‐ 2})}^{\frac{1}{3}} = x^{‐ \frac{2}{3}}$

10

a

$x^{1 \frac{1}{2}} = x \cdot x^{\frac{1}{2}} = x \cdot \sqrt{x}$

b

$x^{2 \frac{1}{2}} + \frac{1}{4} x^{1 \frac{1}{2}}$

$2 x^{‐ 1 \frac{1}{2}} + \sqrt{2} \cdot x^{‐ \frac{1}{2}}$

11

$4^{‐ \frac{1}{2}} = {(2^{2})}^{‐ \frac{1}{2}} = 2^{‐ 1} = \frac{1}{2}$	$4^{‐ 1 \frac{1}{2}} = 4^{‐ \frac{1}{2}} \cdot 4^{‐ 1} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$
${(\frac{1}{9})}^{‐ \frac{1}{2}} = {(3^{‐ 2})}^{‐ \frac{1}{2}} = 3$	${(\frac{1}{9})}^{‐ 1 \frac{1}{2}} = {(\frac{1}{9})}^{‐ \frac{1}{2}} \cdot {(\frac{1}{9})}^{‐ 1} = 3 \cdot 9 = 27$
${0,001}^{‐ \frac{2}{3}} = {(10^{‐ 3})}^{‐ \frac{2}{3}} = 10^{2} = 100$	${0,001}^{‐ 1 \frac{2}{3}} = {(10^{‐ 3})}^{‐ 1 \frac{2}{3}} = 10^{5} = 100.000$
${(2 \frac{1}{2})}^{‐ 1} = {(\frac{5}{2})}^{‐ 1} = \frac{2}{5}$	${({(\frac{1}{2})}^{4})}^{‐ \frac{1}{2}} = {(\frac{1}{2})}^{4 \cdot ‐ \frac{1}{2}} = {(\frac{1}{2})}^{‐ 2} = 4$

12

a

$\frac{{(\frac{1}{3} x^{2})}^{3}}{{(3 x)}^{‐ 1}} = \frac{3^{‐ 3} \cdot x^{6}}{3^{‐ 1} \cdot x^{‐ 1}} = 3^{‐ 2} \cdot x^{7} = \frac{1}{9} x^{7}$ ;
${(x \sqrt{x} \sqrt[3]{x})}^{6} = {(x^{1} \cdot x^{\frac{1}{2}} \cdot x^{\frac{1}{3}})}^{6} = {(x^{1 \frac{5}{6}})}^{6} = x^{11}$

b

${(2 x)}^{3} \cdot 2 x^{3} = 8 x^{3} \cdot 2 x^{3} = 16 x^{6}$ ; ${(2 x)}^{‐ 2} \cdot 2 x^{3} = \frac{1}{4} x^{‐ 2} \cdot 2 x^{3} = \frac{1}{2} x$