Differentiation rules > Product rule
123456Product rule

## Solutions to the exercises

Exercise 1
a

Het is niet elke keer nodig om de productregel te gebruiken. Soms kun je bijvoorbeeld gemakkelijk haakjes uitwerken. Bedenk zelf de beste strategie. Soms krijg je daarom een ander antwoord dan hieronder staat. Dat kan dan toch wel goed zijn. $f\prime \left(x\right)=20{x}^{4}-20{x}^{3}+48x-30$

b

$g\prime \left(x\right)=\frac{5}{\sqrt{x}}-\frac{3}{2}\sqrt{x}$

c

$R\prime \left(t\right)=3{\left(t+5\right)}^{4}+12t{\left(t+5\right)}^{3}$

d

$y\prime \left(x\right)=\sqrt{5+{x}^{2}}+\frac{{x}^{2}}{\sqrt{5+{x}^{2}}}$

e

$\frac{\text{d}y}{\text{d}x}=1-\frac{x}{\sqrt{5+{x}^{2}}}$

f

$V\prime \left(r\right)=\frac{5}{{r}^{2}}\cdot {\left(20-r\right)}^{2}+\left(100-\frac{5}{r}\right)\cdot -2\left(20-r\right)=\left(\frac{100}{{r}^{2}}-200+\frac{5}{r}\right)\left(20-r\right)$

Exercise 2
a

$\left(0,0\right)$ and $\left(4,0\right)$.

b

Just do it.

c

$f\prime \left(x\right)=0$ geeft $x=4\vee x=0\vee x=\frac{4}{3}$.
using the graph of $f$ or a sign scheme of $f\prime$ you find min. $f\left(0\right)=0$, max. $f\left(\frac{4}{3}\right)=1438\frac{274}{729}$ en min. $f\left(4\right)=0$.

d

Read out from (a sketch of ) the graph: $0.

Exercise 3
a

$f\prime \left(x\right)=6\sqrt{x}\cdot \left(1-{x}^{3}\right)+4x\sqrt{x}\cdot -3{x}^{2}=6\sqrt{\left(x\right)}\left(1-3{x}^{3}\right)=0$ when $x=0\vee {x}^{3}=\frac{1}{3}$.
There are two values of $x$ where the tangent is parallell to the $x$ -axis, namely $x=0$ and $x=\sqrt[3]{\frac{1}{3}}$.

b

This function has a minimum at the edge of the domein: $f\left(0\right)=0$. And there is a maximum $f\left(\sqrt[3]{\frac{1}{3}}\right)\approx 0,23$.

Exercise 4
a

$\left(0,0\right)$ and $\left(±\sqrt{8},0\right)$.

b

$f\prime \left(x\right)=\sqrt{8-{x}^{2}}-\frac{{x}^{2}}{\sqrt{8-{x}^{2}}}=0$ gives $x=±2$.
You find min. $f\left(-2\right)=-4$ and max. $f\left(2\right)=4$. So ${\text{B}}_{f}=\left[-4,4\right]$.

c

$f\text{'}\left(0\right)=\sqrt{8}\approx 2,83$. So: $y=2,83x$.

Exercise 5
a

$f\prime \left(x\right)=0,5x-1,5\sqrt{x}=0$ gives $x=0\vee x=9$.
There is a max. $f\left(0\right)=0$ and a min. $f\left(9\right)=-6,75$ so (look at the graph) ${\text{B}}_{f}=\left[-6,75;\to ⟩$.

b

$f\prime \prime \left(x\right)=0,5-\frac{0,75}{\sqrt{x}}=0$ gives $x=2,25$, so the inflection point is $\left(2,25;4,640625\right)$.

c

$f\prime \left(x\right)=2$ gives $3\sqrt{x}=x-4$ and so $9x={x}^{2}-8x+16$.
This results in $x=16$(because $x=1$ does not satisfy the equation) and the tangent point $\left(16,0\right)$.
This point should be on the line and that is only possible if $p=-32$.

Exercise 6
a

$A\prime \left(x\right)=6+\sqrt{9-{x}^{2}}-\frac{{x}^{2}}{\sqrt{9-{x}^{2}}}=0$ gives $6\sqrt{9-{x}^{2}}=2{x}^{2}-9$ and this implies ${x}^{4}=\frac{243}{4}$.
The maximum floor area is reached for $x\approx 2,79$ m.
The maximum floor area therefore is approximately $19,8$ m2.