Differentiation rules > Derivative of trigonometric functions
123456Derivative of trigonometric functions

## Solutions to the exercises

Exercise 1
a

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

b

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

c

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

d

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

e

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

f

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

Exercise 2
a

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.

b

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

c

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

Exercise 3
a

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

b

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

c

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

d

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

Exercise 4
a

$f\text{'}\left(x\right)=0,5+cos\left(x\right)=0$ gives $cos\left(x\right)=-0,5$ 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. $f\left(\frac{2}{3}\pi \right)=\frac{1}{3}\pi +\frac{1}{2}\sqrt{3}$.

b

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

c

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

Exercise 5

$f\text{'}\left(x\right)=2xcos\left({x}^{2}\right)=0$ when $x=0\vee {x}^{2}=0,5\pi +k\cdot \pi$, so when $x=0\vee x=±\sqrt{0,5\pi +k\cdot \pi }$.
Now check the values of $k$ and see if the corresponding $x$ -values lie between $2\pi \approx 6,28$ and $3\pi \approx 9,42$.
They do for $k=13,14,...,27$. There are $15$ extrema on this interval.

Exercise 6
a

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

b

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

c

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

d

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