 Differentiation rules > Quotient rule
123456Quotient rule

## Solutions to the exercises

Exercise 1
a

$f\prime \left(x\right)=\frac{-{x}^{2}-2x+16}{{\left({x}^{2}-16x\right)}^{2}}$

b

$y\prime \left(x\right)=\frac{-2x+4}{{\left({x}^{2}-4x+5\right)}^{2}}$

c

$H\prime \left(t\right)=\frac{-3t-48}{3t\sqrt{2t+6}}$

d

$\frac{\text{d}GTK}{\text{d}q}=4q-10-\frac{120}{{q}^{2}}$

e

$f\prime \left(x\right)=\frac{-2{x}^{2}-20}{{\left({x}^{2}-10\right)}^{2}}$

f

$y\prime \left(x\right)=\frac{-24x}{{\left(1-3{x}^{2}\right)}^{2}}$

g

$A\prime \left(r\right)=\frac{4r+16}{\left(4r+8\right)\sqrt{4r+8}}$

h

$GO\prime \left(p\right)=200-\frac{2000}{{p}^{2}}$

Exercise 2
a

${f}^{\prime }\left(x\right)=\frac{8{x}^{2}--24x+32}{{\left({x}^{2}+4\right)}^{2}}$ geeft $x=-4\vee x=1$.
Using the graph you find: min. $f\left(-4\right)=-1$ en max. $f\left(1\right)=4$.

b

$f\left(x\right)=\frac{3}{2}$ gives $x=-\frac{2}{3}\vee x=6$. Using the graph you find: $x<-\frac{2}{3}\vee x>6$.

c

Line through $A\left(-1,5;0\right)$ and $B\left(0,3\right)$ is $y=2x+3$. This line can only be tangent to the graph in $A$ or $B$. Now $f\prime \left(-1,5\right)\ne 2$ and $f\prime \left(0\right)=2$. So line $AB$ is a tangent in $B$.

Exercise 3
a

Assume the width is $x$ cm, then the length is $4x$ cm. And $4{x}^{2}h=1000$ so $h=\frac{250}{{x}^{2}}$ .
from this it follows that the length of the ribbon $L$ is given by: $L\left(x\right)=10x+\frac{1000}{{x}^{2}}$ .

b

$L\prime \left(x\right)=10-\frac{2000}{{x}^{3}}=0$ gives ${x}^{3}=200$ and so $x\approx 5,8$ cm.
Using the graph of $L$ or a sign scheme of $L\prime$ you see that $L$ has a minimum for $x\approx 5,8$. The dimensions of the box are: $5,8×23,4×7,3$(in cm).

Exercise 4
a

Just do it.

b

$f\prime \left(x\right)=0$ when $x=-3\vee x=6$. Using the graph you find: min. $f\left(6\right)=8,75$.

c

$f\prime \left(x\right)$ does not change signs for $x=-3$.

Exercise 5
a

$P\left(R\right)=\frac{144R}{{\left(R+12\right)}^{2}}$

b

$P\prime \left(R\right)=\frac{-144R+1728}{{\left(R+12\right)}^{3}}=0$ gives $R=12$ ohm.
The maximum of the generated power is $P\left(12\right)=3$ Watt.