Differentiëren

Antwoorden bij de opgaven

1. 1. ${\displaystyle \frac{\Delta y}{\Delta x}=\frac{c{(x+h)}^2-c{x}^2}{h}=\frac{2cxh-c{h}^2}{h}=2cx-ch}$
   2. ${\displaystyle f\text{'}\left(x\right)=\underset{h\to 0}{lim}(2cx-ch)=2cx}$
   3. ${\displaystyle f\text{'}\left(x\right)=3\cdot 5x^4=15x^4}$
2. 1. ${\displaystyle f\text{'}\left(x\right)=12\cdot 5x^4=60x^4}$
   2. ${\displaystyle f\text{'}\left(x\right)=12\cdot 5x^4+0=60x^4}$
   3. ${\displaystyle f\text{'}\left(x\right)=12\cdot 5x^4+20\cdot 3x^2=60x^4+60x^2}$
3. 1. ${\displaystyle f\text{'}\left(x\right)=30x^2-60}$ en ${\displaystyle f\text{'}\left(1\right)=-30}$
   2. ${\displaystyle f\text{'}\left(x\right)=2-10x-40x^3}$ en ${\displaystyle f\text{'}\left(1\right)=-48}$
   3. ${\displaystyle f\text{'}\left(x\right)=2x^3-8x}$ en ${\displaystyle f\text{'}\left(1\right)=-6}$
4. 1. ${\displaystyle (\pm 2,0)}$ en ${\displaystyle \left(\mathrm{6,0}\right)}$
      ${\displaystyle y\left(x\right)=x^3-6x^2-4x+24}$ en ${\displaystyle \frac{\text{d}y}{\text{d}x}=y\text{'}\left(x\right)=3x^2-12x-4}$
   2. ${\displaystyle y\text{'}\left(2\right)=-16}$ en ${\displaystyle y\left(2\right)=0}$, dus de raaklijn wordt ${\displaystyle y=-16x+32}$.
5. 1. ${\displaystyle f\text{'}\left(x\right)=\mathrm{1,5}{x}^2-9x+10}$
   2. ${\displaystyle f\text{'}\left(0\right)=10}$
   3. ${\displaystyle f\text{'}\left(x\right)=10}$ geeft ${\displaystyle \mathrm{1,5}{x}^2-9x+10=10}$ en dus ${\displaystyle x=0\vee x=6}$.
      Het zijn dus de punten ${\displaystyle (0,-35)}$ en ${\displaystyle (6,-29)}$.
6. 1. ${\displaystyle f\text{'}\left(x\right)=3x^2-4}$ en ${\displaystyle f\text{'}\left(1\right)=-1}$
   2. ${\displaystyle g\text{'}\left(x\right)=4x^4+6x^2-10x+12}$ en ${\displaystyle g\text{'}\left(1\right)=12}$
   3. ${\displaystyle s\text{'}\left(t\right)=60-\mathrm{9,8}t}$ en ${\displaystyle s\text{'}\left(1\right)=\mathrm{50,2}}$
   4. ${\displaystyle H\text{'}\left(t\right)=4t}$ en ${\displaystyle H\text{'}\left(1\right)=4}$
   5. ${\displaystyle \frac{\text{d}y}{\text{d}x}=-2x+6}$ en ${\displaystyle y\text{'}\left(1\right)=4}$
   6. ${\displaystyle P\text{'}\left(x\right)=3a{x}^2+2bx+c}$ en ${\displaystyle P\text{'}\left(1\right)=3a+2b+c}$
   7. ${\displaystyle \frac{\text{d}TW}{\text{d}q}=\mathrm{1,5}{q}^2-12q-25}$ en ${\displaystyle TW\text{'}\left(1\right)=\mathrm{35,5}}$
   8. ${\displaystyle K\text{'}\left(x\right)=9a{x}^2-6x-3a^2}$ en ${\displaystyle K\text{'}\left(1\right)=9a-6-3a^2}$
7. 1. ${\displaystyle f\text{'}\left(x\right)=2x^3-8x}$ en ${\displaystyle f\text{'}\left(x\right)=0}$ als ${\displaystyle x=0\vee x=\pm 2}$, dus ${\displaystyle \left(\mathrm{0,0}\right),(-2,-8)}$ en ${\displaystyle (2,-8)}$.
   2. ${\displaystyle TW\text{'}\left(q\right)=-3q^2+6q+3}$ en ${\displaystyle TW\text{'}\left(q\right)=0}$ als ${\displaystyle q=1\pm \sqrt{2}}$, dus ${\displaystyle (\mathrm{2,4};\mathrm{16,7})}$ en ${\displaystyle (-\mathrm{0,4};\mathrm{5,3})}$.
   3. ${\displaystyle v\text{'}\left(t\right)=3t^2-4t+1}$ en ${\displaystyle v\text{'}\left(t\right)=0}$ als ${\displaystyle t=\frac{1}{3}\vee t=1}$, dus ${\displaystyle (\frac{1}{3},\frac{4}{27})}$ en ${\displaystyle \left(\mathrm{1,0}\right)}$.
   4. ${\displaystyle TW\text{'}\left(p\right)=40-\mathrm{0,04}p}$ en ${\displaystyle TW\text{'}\left(p\right)=0}$ als ${\displaystyle p=1000}$, dus ${\displaystyle \left(\mathrm{1000,20000}\right)}$.
8. 1. ${\displaystyle f\left(x\right)=(x^2-4)(x^2-9)=0}$ geeft ${\displaystyle x^2=4\vee x^2=9}$ en ${\displaystyle x=\pm 2\vee x=\pm 3}$, dus ${\displaystyle (\pm \mathrm{2,0})}$ en ${\displaystyle (\pm \mathrm{3,0})}$.
   2. ${\displaystyle f\left(x\right)=x^4-13x^2+36}$ en ${\displaystyle f\text{'}\left(x\right)=4x^3-26x}$.
   3. Raaklijn voor ${\displaystyle x=-2}$ is ${\displaystyle y=20x+40}$.
      Raaklijn voor ${\displaystyle x=2}$ is ${\displaystyle y=-20x+40}$.
      Snijpunt ${\displaystyle (0,40)}$.
   4. ${\displaystyle f\text{'}\left(x\right)=0}$ als ${\displaystyle x=0\vee x=\pm \sqrt{\mathrm{8,5}}}$.
   5. Je vindt daarmee de drie extremen: max.${\displaystyle f\left(0\right)=36}$, min.${\displaystyle f(-\sqrt{\mathrm{8,5}})=-\mathrm{6,25}}$ en min.${\displaystyle f\left(\sqrt{\mathrm{8,5}}\right)=-\mathrm{6,25}}$.
9. 1. ${\displaystyle h\left(0\right)=\mathrm{0,5}}$ m
   2. ${\displaystyle h\text{'}\left(x\right)=-\mathrm{0,02}x+\mathrm{0,2}}$ en ${\displaystyle h\text{'}\left(0\right)=\mathrm{0,2}}$ m/s
   3. Het is de veranderingssnelheid van de hoogte van de kogel op ${\displaystyle x=0}$. Het is niet de beginsnelheid!
   4. ${\displaystyle h\text{'}\left(x\right)=0}$ geeft ${\displaystyle x=10}$, daarbij hoort het punt ${\displaystyle (10;\mathrm{1,5})}$.
   5. De snelheid waarmee de hoogte van de baan verandert is er 0. Maar er is ook een voorwaartse snelheidscomponent.
10. 1. ${\displaystyle GTK\left(q\right)=\frac{1200}{q}+\mathrm{0,2}q}$
    2. ${\displaystyle q=0}$; als ${\displaystyle q=0}$ kun je geen gemiddelde kosten bepalen.
    3. min.${\displaystyle GTK\left(77\right)\approx \mathrm{30,98}}$
    4. ${\displaystyle GTK\to \mathrm{0,2}}$ als ${\displaystyle q\to \infty }$ (${\displaystyle \infty }$ is het symbool voor "oneindig groot").
       De productiekosten per eenheid veranderen op den duur met de (vaste) kosten per artikel.
11. 1. ${\displaystyle f\text{'}\left(x\right)=6x^5+8}$
    2. ${\displaystyle f\text{'}\left(x\right)=12x^3-x}$
    3. ${\displaystyle f\text{'}\left(x\right)=-\mathrm{4,5}{x}^2+4}$
    4. ${\displaystyle f\text{'}\left(x\right)=3x^2-4x}$
    5. ${\displaystyle f\text{'}\left(x\right)=-5x^4+16x^3+6x^2-3x+8}$
    6. ${\displaystyle f\text{'}\left(x\right)=8x+4}$
    7. ${\displaystyle f\text{'}\left(x\right)=-3x^2+12x}$
    8. ${\displaystyle f\text{'}\left(x\right)=-3+6x-3x^2}$
12. 1. ${\displaystyle f\text{'}\left(x\right)=9+6x-3x^2}$ en dus is ${\displaystyle f\text{'}\left(0\right)=9}$
    2. Raaklijn door ${\displaystyle \left(\mathrm{0,0}\right)}$ met ${\displaystyle rc=9}$ heeft vergelijking ${\displaystyle y=9x}$.
    3. ${\displaystyle f\text{'}\left(x\right)=0}$ geeft ${\displaystyle x=-1\vee x=3}$, dus de punten ${\displaystyle (-1,-5)}$ en ${\displaystyle \left(\mathrm{3,27}\right)}$.
    4. ${\displaystyle f\text{'}\left(x\right)}$ is maximaal als de afgeleide ervan 0 is: ${\displaystyle 6-6x=0}$.
       Dit geeft ${\displaystyle x=1}$ en dus het punt ${\displaystyle \left(\mathrm{1,11}\right)}$.
13. 1. ${\displaystyle \frac{\text{d}y}{\text{d}x}=3x^2-51x+180}$
    2. Als de afgeleide 0 is heeft de grafiek een raaklijn evenwijdig aan de ${\displaystyle x}$-as.
    3. ${\displaystyle y\text{'}\left(x\right)=3x^2-51x+180=0}$ geeft ${\displaystyle x=5\vee x=12}$.
    4. Bekijk de grafiek. De functie is dalend als .