$f\text{'}\left(x\right)=6cos\left(2x\right)$

$g\text{'}\left(x\right)=\frac{4}{3}\pi sin\left(\frac{2\pi }{30}(x-5)\right)$

$H\text{'}\left(t\right)=880\pi sin\left(440\pi t\right)cos\left(440\pi t\right)$

$y\text{'}\left(x\right)=\frac{1}{2}{(16+{sin}^2\left(x\right))}^{\mathrm{-}\frac{1}{2}}\cdot 2sin\left(x\right)cos\left(x\right)=\frac{sin\left(x\right)cos\left(x\right)}{\sqrt{16+{sin}^2\left(x\right)}}$

$A\text{'}\left(r\right)=-1{\left(sin\left(2r\right)\right)}^{-2}\cdot cos\left(2r\right)\cdot 2=\frac{-2cos\left(2r\right)}{{sin}^2\left(2r\right)}$

$W\text{'}\left(p\right)=2cos\left(p\right)\cdot cos\left(2p\right)+2sin\left(p\right)\cdot \mathrm{-}sin\left(2p\right)\cdot 2=2cos\left(p\right)cos\left(2p\right)-4sin\left(p\right)sin\left(2p\right)$

Well, what would algebraically mean in this case?
working with the period ( $\pi$), the horizontal translation ( $0$), the amplitude ( $3$) and the equilibrium ( $y=1$) of a standard sinuoid gives max. $f\left(0\right)=f\left(\pi \right)=f\left(2\pi \right)=4$ and min. $f\left(\frac{1}{2}\pi \right)=f\left(1\frac{1}{2}\pi \right)=-2$. But you can also use differentiation. Check if you get the same answer if you do that.

$f\left(x\right)=\mathrm{2,5}$ gives $cos\left(2x\right)=\mathrm{0,5}$ and so $x=\frac{1}{6}\pi +k\cdot \pi \vee x=\mathrm{-}\frac{1}{6}\pi +k\cdot \pi$.
On the given domain $f\left(x\right)<\mathrm{2,5}$ when $\frac{1}{6}\pi <x<\frac{5}{6}\pi \vee 1\frac{1}{6}\pi <x<1\frac{5}{6}\pi$.

$f\text{'}\left(x\right)=-6sin\left(2x\right)$.
The slopes are $f\text{'}\left(\frac{1}{6}\pi \right)=f\text{'}\left(1\frac{1}{6}\pi \right)=-3\sqrt{3}$ en $f\text{'}\left(\frac{5}{6}\pi \right)=f\text{'}\left(1\frac{5}{6}\pi \right)=3\sqrt{3}$.

$H\left(0\right)\approx \mathrm{1,06}$ m.

$H\text{'}\left(t\right)=\mathrm{-}\frac{4\pi }{\mathrm{12,25}}sin\left(\frac{2\pi }{\mathrm{12,25}}(t-3)\right)=0$ when $t=3\vee t=\mathrm{9,125}\vee t=\mathrm{15,25}\vee t=\mathrm{21,375}$.
High water occurs at 3:00 am and at 3.15 pm.
You can also reason with the period and the horizontal translation.

$H\text{'}\left(4\right)\approx \mathrm{-0,50}$ m/hour. It is the rate at which the water level (in this case) drops.

When $H\text{'}$ is maximal of minimal, so when the graph of $H$ goes through the equlibrium level.
That is when $t=\mathrm{6,0625}\vee t=\mathrm{12,1875}\vee t=\mathrm{18,3125}$.
You find the approximate times of 6:04 am, 12:11 pm and 6:19 pm.

$f\text{'}\left(x\right)=\mathrm{0,5}+cos\left(x\right)=0$ gives $cos\left(x\right)=\mathrm{-0,5}$ and so $x=\frac{2}{3}\pi \vee x=1\frac{2}{3}\pi$.
Using the graph you find: min. $f\left(1\frac{2}{3}\pi \right)=\frac{5}{6}\pi -\frac{1}{2}\sqrt{3}$ and max. $f\left(\frac{2}{3}\pi \right)=\frac{1}{3}\pi +\frac{1}{2}\sqrt{3}$.

$f\text{'}\left(0\right)=\mathrm{1,5}$, the equation for the tangent to the graph at $O$ is $y=\mathrm{1,5}x$.

The smallest slope is found at $x=\pi$ and $f\text{'}\left(\pi \right)=\mathrm{-0,5}$.

$V\left(t\right)=325sin\left(100\pi t\right)$

$P\left(t\right)=c\cdot {\left(V\left(t\right)\right)}^2=c\cdot 105625{sin}^2\left(100\pi t\right)={sin}^2\left(100\pi t\right)$ when $c=\frac{1}{\left(105625\right)}$.

$P\text{'}\left(t\right)=2sin\left(100\pi t\right)cos\left(100\pi t\right)=0$ gives $100\pi t=k\cdot \mathrm{0,5}\pi$ and so $t=k\cdot \mathrm{0,005}$.
You find max. $f(\mathrm{0,005}+k\cdot \mathrm{0,01})=1$ and min. $f(k\cdot \mathrm{0,01})=0$.

For instance $P\left(t\right)=\mathrm{0,5}-\mathrm{0,5}cos\left(100\pi t\right)$.