De kettingregel

Antwoorden bij de opgaven

1. 1. Het functievoorschrift is op te delen in afzonderlijke schakels. Je ziet dat aan het feit dat er maar op één plek een ${\displaystyle x}$ in het voorschrift voor komt.
   2. ${\displaystyle x\to x-2\to {(x-2)}^2\to 3{(x-2)}^2\to 3{(x-2)}^2-2}$
   3. Terugrekenen vanuit ${\displaystyle 3{(x-2)}^2-2=16}$ geeft ${\displaystyle x=-1\vee x=5}$.
2. 1. ${\displaystyle x\to x^2\to x^2-1\to \sqrt{x^2-1}}$
   2. ${\displaystyle x\to x^3\to 3x^3\to 3x^3+1}$
   3. ${\displaystyle x\to x^2\to 3x^2\to 3x^2+2\to {(3x^2+2)}^4}$
3. 1. ${\displaystyle h\left(x\right)=\sqrt{x^2+2}}$
   2. Nee.
   3. ${\displaystyle k\left(x\right)=x+2}$, met afgeleide ${\displaystyle k\text{'}\left(x\right)=1}$.
4. 1. ${\displaystyle f\left(u\right)=u^8}$ en ${\displaystyle u=g\left(x\right)=2x^2+1}$.
   2. ${\displaystyle f\text{'}\left(u\right)=8u^7}$ en ${\displaystyle g\text{'}\left(x\right)=4x}$ geeft ${\displaystyle h\text{'}\left(x\right)=f\text{'}\left(g\left(x\right)\right)\cdot g\text{'}\left(x\right)=8{\left(g\left(x\right)\right)}^7\cdot 4x=32x{(2x^2+1)}^7}$.
5. 1. ${\displaystyle h\left(x\right)={(2x^3+4x)}^4}$
   2. ${\displaystyle h\text{'}\left(x\right)=4(6x^2+4){(2x^3+4x)}^3}$
   3. ${\displaystyle k\left(x\right)=2x^{12}+4x^4}$
   4. ${\displaystyle k\text{'}\left(x\right)=24x^{11}+16x^2}$
6. 1. ${\displaystyle f\left(x\right)=2x^{\frac{1}{2}}}$
      ${\displaystyle f\text{'}\left(x\right)=2+\frac{1}{2}{x}^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}}$
   2. ${\displaystyle f\left(x\right)=x^{1\frac{1}{2}}}$
      ${\displaystyle f\text{'}\left(x\right)=1\frac{1}{2}{x}^{\frac{1}{2}}=1\frac{1}{2}\sqrt{x}}$
   3. ${\displaystyle f\left(x\right)=x^{\frac{1}{3}}}$
      ${\displaystyle f\text{'}\left(x\right)=\frac{1}{3}{x}^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^2}}}$
   4. ${\displaystyle f\left(x\right)=3x^{-\frac{1}{2}}}$
      ${\displaystyle f\text{'}\left(x\right)=-\frac{3}{2}{x}^{-1\frac{1}{2}}=\frac{-3}{2x\sqrt{x}}}$
7. 1. ${\displaystyle {\text{D}}_f=[-\mathrm{5,5}]}$ en ${\displaystyle {\text{B}}_f=\left[\mathrm{0,5}\right]}$
      De grafiek komt niet tot op de ${\displaystyle x}$-as en dat zou wel moeten. (Is een beperking van de grafische rekenmachine.)
   2. ${\displaystyle f\text{'}\left(x\right)=\frac{1}{2}{(25-x^2)}^{-\frac{1}{2}}\cdot (-2x)=\frac{-x}{\sqrt{25-x^2}}}$
   3. Alleen uitspraak D is goed.
   4. ${\displaystyle f\text{'}\left(3\right)=-\frac{3}{4}}$ en ${\displaystyle f\left(3\right)=4}$ geeft voor de raaklijn ${\displaystyle y=-\frac{3}{4}x+6\frac{1}{4}}$.
8. 1. ${\displaystyle f\text{'}\left(x\right)=8x{(x^2-100)}^3}$
   2. ${\displaystyle f\text{'}\left(x\right)=-3{(1-x)}^2}$
   3. ${\displaystyle H\text{'}\left(t\right)=-300{(2-4t)}^2}$
   4. ${\displaystyle 2p^2-4p{(px+3)}^3}$
9. 1. voor elke waarde van ${\displaystyle x}$ behalve ${\displaystyle x=2}$.
   2. ${\displaystyle f\text{'}\left(2\right)=-24}$ en ${\displaystyle f\left(2\right)=12}$, dus ${\displaystyle P=\left(\mathrm{0,60}\right)}$.
10. 1. ${\displaystyle \frac{\text{d}y}{\text{d}x}=\frac{7}{3}{x}^{\frac{4}{3}}=\frac{7}{3}x\sqrt[3]{x}}$
    2. ${\displaystyle f\text{'}\left(x\right)=\frac{-3}{x^4}-\frac{8}{x^3}+\frac{3}{x^2}}$
    3. ${\displaystyle H\text{'}\left(p\right)=-\frac{3}{2\sqrt{p}}\cdot {(1-\sqrt{p})}^3}$
    4. ${\displaystyle g\text{'}\left(x\right)=2-\frac{5}{{(1-x)}^2}}$
11. 1. ${\displaystyle {\text{D}}_f=[-\sqrt{8},\sqrt{8}]}$
    2. Het minimum ligt op de rand van het domein: min.${\displaystyle f(-\sqrt{8})=-\sqrt{8}}$.
       Het maximum bepaal je met behulp van differentiëren. ${\displaystyle f\text{'}\left(x\right)=1-\frac{x}{\sqrt{8-x^2}}=0}$ geeft ${\displaystyle \sqrt{8-x^2}=x}$ en na kwadrateren ${\displaystyle x=2}$ (${\displaystyle x=-2}$ vervalt). Je vindt: max.${\displaystyle f\left(2\right)=4}$. Het bereik wordt ${\displaystyle {\text{B}}_f=[-\sqrt{8}\mathrm{,4}]}$.
    3. ${\displaystyle A=(-\sqrt{8},-\sqrt{8})}$ en ${\displaystyle B=(\sqrt{8},\sqrt{8})}$.
       De helling van lijn ${\displaystyle AB}$ is gelijk aan 1.
       Je moet daarom oplossen ${\displaystyle f\text{'}\left(x\right)=1}$ en dat levert op ${\displaystyle x=0}$.
12. 1. ${\displaystyle 600\cdot 30+500\cdot 70=53000}$ euro.
    2. ${\displaystyle \sqrt{{600}^2+{500}^2}\cdot 70\approx \mathrm{54671,75}}$ euro.
    3. ${\displaystyle K\left(x\right)=30(600-x)+70\sqrt{{500}^2+x^2}}$
    4. De minimale kosten vind je met behulp van ${\displaystyle K\text{'}\left(x\right)=-30+\frac{70x}{\sqrt{{500}^2+x^2}}=0}$.
       Dit geeft ${\displaystyle \sqrt{{500}^2+x^2}=\frac{7}{3}x}$ en na kwadrateren ${\displaystyle \frac{40}{9}{x}^2=250000}$. Je vindt dan ${\displaystyle x\approx 237}$.
       Je kunt dus het beste eerst 363 m langs de straat graven en daarna door het veld recht naar ${\displaystyle C}$ graven.
13. 1. ${\displaystyle f\text{'}\left(x\right)=36x{(1+x^2)}^2}$
    2. ${\displaystyle y\text{'}\left(x\right)=-16{(1-4x)}^3}$
    3. ${\displaystyle R\text{'}\left(t\right)=\frac{\mathrm{7,5}}{\pi \sqrt{\frac{15}{\pi }t}}}$
    4. ${\displaystyle f\text{'}\left(x\right)=\frac{4x}{\sqrt{10+4x^2}}}$
    5. ${\displaystyle K\text{'}\left(p\right)=\frac{-3}{p^2\sqrt{p}}}$
    6. ${\displaystyle f\text{'}\left(x\right)=3x^2+2+\frac{3}{2x\sqrt{x}}-\frac{2}{x^3}}$
14. 1. ${\displaystyle {\text{D}}_f=[-2,\to \rangle }$
    2. ${\displaystyle f\text{'}\left(x\right)=2-\frac{1}{2\sqrt{x+2}}}$
    3. ${\displaystyle f\text{'}\left(x\right)=0}$ geeft ${\displaystyle \sqrt{x+2}=\frac{1}{4}}$ en dus ${\displaystyle x=-1\frac{1}{2}}$ (${\displaystyle x=-2\frac{1}{2}}$ vervalt).
       Je vindt min.${\displaystyle f(-1\frac{1}{2})=-3+\frac{1}{2}\sqrt{2}}$.
    4. ${\displaystyle {\text{B}}_f=[-3+\frac{1}{2}\sqrt{2},\to \rangle }$
    5. ${\displaystyle f\text{'}\left(0\right)=2-\frac{1}{2\sqrt{2}}\approx \mathrm{1,65}=tan\left(\alpha \right)}$ en dus is ${\displaystyle \alpha \approx {\mathrm{58,7}}^o}$.