$f\text{'}\left(x\right)=8x{(x^2-100)}^3$

$f\text{'}\left(x\right)=-3{(1-x)}^2$

$H\text{'}\left(t\right)=-300{(2-4t)}^2$

$2p^2-4p{(px+3)}^3$

$f\text{'}\left(x\right)=-6{(2x-6)}^2<0$ for every value of $x$ except $x=2$.

$f\text{'}\left(2\right)=-24$ and $f\left(2\right)=12$, so $P=(0,60)$.

$\frac{\text{d}y}{\text{d}x}=\frac{7}{3}{x}^{\frac{4}{3}}=\frac{7}{3}x\sqrt[3]{x}$

$f\text{'}\left(x\right)=\frac{-3}{x^4}-\frac{8}{x^3}+\frac{3}{x^2}$

$H\text{'}\left(p\right)=\mathrm{-}\frac{3}{2\sqrt{p}}\cdot {(1-\sqrt{p})}^2$

$g\text{'}\left(x\right)=2-\frac{5}{{(1-x)}^2}$

${\text{D}}_f=[\mathrm{-}\sqrt{8},\sqrt{8}]$

The minimum lies on the edge of the domein: min. $f\left(\mathrm{-}\sqrt{8}\right)=\mathrm{-}\sqrt{8}$.
You can find the maximum using differentiation. $f\text{'}\left(x\right)=1-\frac{x}{\sqrt{8-x^2}}=0$ gives $\sqrt{8-x^2}=x$ and after squaring $x=2$( $x=-2$ is cancelled). You find: max. $f\left(2\right)=4$. The range becomes ${\text{B}}_f=[\mathrm{-}\sqrt{8},4]$.

$A=(\mathrm{-}\sqrt{8},\mathrm{-}\sqrt{8})$ en $B=(\sqrt{8},\sqrt{8})$.
The slope of line $AB$ is equal to $1$.
You should therefore solve $f\text{'}\left(x\right)=1$ and that gives $x=0$.

$600\cdot 30+500\cdot 70=53000$ euros.

$\sqrt{{600}^2+{500}^2}\cdot 70\approx \mathrm{54671,75}$ euros.

$K\left(x\right)=30(600-x)+70\sqrt{{500}^2+x^2}$

You find the minimal cost using $K\text{'}\left(x\right)=-30+\frac{70x}{\sqrt{{500}^2+x^2}}=0$.
This gives $\sqrt{{500}^2+x^2}=\frac{7}{3}x$ and after squaring $\frac{40}{9}{x}^2=250000$. You find $x\approx 237$.
So the best way is to first dig $363$ m along the street and then through the field straight at $C$.