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

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

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

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

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

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

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

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

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

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

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

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

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

Just do it.

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

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

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

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