$f\text{'}\left(x\right)=30x^5-65x^4+10$

$f\text{'}\left(x\right)=2ax+b$

$P\text{'}\left(I\right)=2RI$

$\frac{\text{d}y}{\text{d}x}=4x^3-20x$

$f\text{'}\left(x\right)=-64x^7$

$f\text{'}\left(x\right)=6a{x}^2-3a^2$

$\frac{\text{d}A}{\text{d}r}=2\pi r+2l$

$h\text{'}\left(x\right)=600x-180x^2+12x^3$

$f\text{'}\left(x\right)=\frac{12}{5}{x}^2-6x=0$ geeft $x=0\vee x=\mathrm{2,5}$.
Use the graph or a sign scheme of $f\text{'}$ to find that there are two extrema, namely min. $f\left(\mathrm{2,5}\right)=\mathrm{-6,25}$ and max. $f\left(0\right)=0$.

$f^{\text{'}}(5)=30$

$f\text{'}\text{'}\left(x\right)=\frac{24}{5}x-6=0$ gives $x=\mathrm{1,25}$.
Use the graph or a sign scheme of $f\text{'}\text{'}$ to conclude that there is one inflection point, namely $(\mathrm{1,25};\mathrm{-3,25})$.

$y\text{'}\left(x\right)=3x^2-10x+7$ geeft $y\text{'}\left(2\right)=-1$.

$y\text{'}\left(x\right)=-1$ gives $3x^2-10x+8=0$. This equation has two solutions, so there is one more point on the graph in which the slope of the tangent to the graph equals $-1$.

$I\left(x\right)=x(20-2x)(60-2x)=1200x-160x^2+4x^3$

$I\text{'}\left(x\right)=12x^2-320x+1200=0$ geeft $x=\frac{80\pm \sqrt{2800}}{6}$.
The graph of $I$ shows that the maximum volume occurs when $x=\frac{80-\sqrt{2800}}{6}\approx \mathrm{4,5}$ cm.

$f\left(x\right)=x^5-40x^4+400x^3$ gives $f\text{'}\left(x\right)=5x^4-160x^3+1200x^2$.
And $f\text{'}\left(x\right)=0$ gives: $x=0\vee x=12\vee x=20$.

For $x=0$ the derivative does not have a sign change, both to the left and to the right of $x=0$ the graph of $f$ is increasing.