By definition, velocity is the first derivative of position, or \(\displaystyle v(t)=p'(t)\).

Given a position 

\(\displaystyle p(t)=t^{2}+2t-5\), we can use the power rule 

\(\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}\) where \(\displaystyle n\neq0\) to determine that \(\displaystyle v(t)=p'(t)=2t+2\).

Therefore, at \(\displaystyle t=4\), 

\(\displaystyle v(4)=p'(4)=2(4)+2=8+2=10\).

In order to find the velocity function from the position function we need to take the first derivative of the position function.

\(\displaystyle v(t)=\frac{d}{dt}[s(t)]\)

When taking the derivative, we will use the power rule which states

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^{n}=nx^{n-1}\)

and by applying this rule to each term we get

\(\displaystyle v(t)=\frac{d}{dt}[s(t)]=\frac{d}{dt}(3t^2+6t)=2\cdot 3t^1+1\cdot 6t^0\).

As such,

\(\displaystyle v(t)=6t+6\).

In order to find the velocity function from the acceleration function we need to take the integral of the acceleration function.

\(\displaystyle v(t)=\int a(t) \, dt\)

When taking the integral, we will use the inverse power rule which states,

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\).

Applying this rule to each term we get, 

 \(\displaystyle v(t)=\int a(t) \, dt=\int t+2 \, dt= \frac{1}{2}t^2+2t+c\).

To find the value of the constant c we will use the initial condition given in the problem.

Setting the initial condition \(\displaystyle t=0\), we get

\(\displaystyle v(0)=5=\frac{1}{2}0^2 +2(0)+c\)

\(\displaystyle c=5\).

As such,

\(\displaystyle v(t)=\frac{1}{2}t^2 +2t+5\)

Finally, we set  \(\displaystyle t=4\), which yields

\(\displaystyle v(4)=21\).

Looking over our problem, we have a composite function \(\displaystyle f\circ g\), where \(\displaystyle f(x)=(g(x))^{3}\) and \(\displaystyle g(x)=6x-7\). We are trying to find the derivative of \(\displaystyle f\circ g\), or \(\displaystyle (f\circ g)'\), which by the Chain Rule can be defined as \(\displaystyle (f\circ g)'=(f'\circ g)\times g'.\)

Therefore:

\(\displaystyle ((6x-7)^{3})'\)

\(\displaystyle =3(6x-7)^{2}(6x-7)'\)

\(\displaystyle =3(6x-7)^{2}(6)\)

\(\displaystyle =18(6x-7)^{2}\)

By definition, velocity is the first derivative of position, or \(\displaystyle v(t)=p'(t)\).

Given \(\displaystyle p(t)=11t^{2}-20t+30\) 

and using the power rule 

\(\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}\) 

for all \(\displaystyle n\neq0\),

we can deduce that 

\(\displaystyle v(t)=p'(t)=22t-20\). 

At time \(\displaystyle t=4\), 

\(\displaystyle v(4)=p'(4)=22(4)-20=88-20=68\).

The velocity of the falling water droplet is given by the first derivative of the position equation: 

\(\displaystyle v(t)=32t+48\).

This derivative was found by using the power rule 

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\). 

At \(\displaystyle t=2\), the speed is given by \(\displaystyle v(2)=112\).

However, we were asked for the droplet's velocity, which is negative when the droplet is falling (moving in the negative direction). So, the final answer is \(\displaystyle -112\).

First, take the derivative of \(\displaystyle s(t)\) by using the Power Rule (\(\displaystyle x^n=nx^{n-1}\)):

\(\displaystyle s'(t)= -24t+36\)

Then, plug in for \(\displaystyle t=2\):

\(\displaystyle s'(t)=-24(2)+36\)

\(\displaystyle s'(t)=-12\)

To find the equation describing the velocity of a particle given the position s(t), utilize the relationship 

\(\displaystyle v(t)=s'(t)\)

to check each of the possible answers.  

The first derivative of 

\(\displaystyle s(t)=\frac{1}{3}t^{3}-\frac{3}{2}t^{2}+\frac{1}{3}\)

is

\(\displaystyle s'(t)=t^{2}-3t\).

Evaluating this at the given time of t=5:

\(\displaystyle s'(5)=5^{2}-(3)(5)=25-15=10\).

Therefore, 

\(\displaystyle v(5)=10\)

which is the condition the question requires.

In order to find \(\displaystyle f'(x)\), we need to remember the definition of the derivative for natural log.

\(\displaystyle \\ g(x)=ln(h(x)) \\ \\ g'(x)=\frac{h'(x)}{h(x)}\)

Now lets find \(\displaystyle f'(x)\).

\(\displaystyle f(x)=ln(\tan(x))\)

\(\displaystyle f'(x)=\frac{\sec^2(x)}{\tan(x)}=\csc(x)\sec(x)\)

In order to find the derivative, we need to use the chain rule and the definition of the derivative of natural log.

Remember that the definition of the derivative of natural log is

\(\displaystyle g(x)=ln(h(x))\)

\(\displaystyle g'(x)=\frac{h'(x)}{h(x)}\).

Remember that the chain rule is

\(\displaystyle [f(g(x))]'=f'(g(x))\cdot g'(x)\).

Lets apply these two rules to our problem.

\(\displaystyle f(x)=ln(ln(x^2))\)

\(\displaystyle f'(x)=\frac{1}{ln(x^2)}\cdot \frac{1}{x^2}\cdot 2x=\frac{2}{xln(x^2)}\)