First we need to find $\displaystyle f'(x)$ before evaluating it at $\displaystyle x=0$.

$\displaystyle f(x)=e^{2x^2+4x}$

$\displaystyle f'(x)=(4x+4)e^{2x^2+4x}$.

This was found by using the definition of the derivative for exponential functions.

The definition is,

$\displaystyle f(x)=e^{g(x)}$

$\displaystyle f'(x)=g'(x)e^{g(x)}$.

Now we can evaluate $\displaystyle f'(x)$ at $\displaystyle x=0$

$\displaystyle f'(0)=(4\cdot0+4)e^{2\cdot (0)^2+4\cdot 0}$

$\displaystyle f'(0)=4\cdot e^{0}=4$

First, set $\displaystyle s(t)$ to equal $\displaystyle 0$. 

$\displaystyle 0=-36(t^2-t-2)=-36(t-2)(t+1)$.

Then solve for $\displaystyle t$:

$\displaystyle t-2=0$ and $\displaystyle t+1=0$.

Therefore, $\displaystyle t=-1, 2$. Since $\displaystyle t$ represents time in seconds, we can rule out $\displaystyle t=-1$.

This means that it takes the cliff diver 2 seconds to reach the water. Next, take the derivative of $\displaystyle s(t)$ by using the Power Rule $\displaystyle x^n=nx^{n-1}$ and plug in for $\displaystyle t$: $\displaystyle \\s'(t)=-72t+36\\ s'(2)=-72(2)+36=-108$.

The cliff diver's velocity was $\displaystyle -108 f/s$ when he/she hit the water.

First, take the first derivative of the equation by using the Power Rule $\displaystyle x^n=nx^{n-1}$ : 

$\displaystyle s'(t)=-8t+2$.

Then, plug in for $\displaystyle t=2$: 

$\displaystyle \\s'(2)=-8(2)+2\\ s'(2)=-16+2=-14$.

Therefore, the velocity is $\displaystyle -14m/s$.

Take the first derivative of the equation by using the Power Rule $\displaystyle x^n=nx^{n-1}$ : 

$\displaystyle s'(t)=2(10)t-16=20t-16$.

Then, plug in $\displaystyle t=3$: 

$\displaystyle 20(3)-16=44$.

Therefore, the velocity at $\displaystyle t=3$ is $\displaystyle 44ft/s$.

Take the first derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle s'(t)=-5(2)t-3=-10t-3$.

Then, plug in $\displaystyle t=1$,

$\displaystyle s'(1)=-10(1)-3=-13$.

Take the first derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle s'(t)=-6(2)t+8=-12t+8$.

Then, plug in for $\displaystyle t=4$,

$\displaystyle \\s'(4)=v(4)\\v(4)=-12(4)+8=-48+8\\ v(4)=-40$.

Take the first derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle s'(t)=-2(2)t+6=-4t+6$.

Then, plug in $\displaystyle t=3$: 

$\displaystyle -4(3)+6=-12+6=-6$.

Take the first derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle s'(t)=8(2)t-6=16t-6$.

Then, plug in for $\displaystyle t=2$,

$\displaystyle \\s'(2)=v(2)\\ v(2)=16(2)-6\\v(2)=32-6=26$.

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

Given a position 

$\displaystyle p(t)=12t^{2}+5(t)-10$, we can use the power rule 

$\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}$ where $\displaystyle n\neq0$ to determine that 

$\displaystyle v(t)=p'(t)=24t+5$.

Therefore, at $\displaystyle t=2$, 

$\displaystyle v(2)=p'(2)=24(2)+5=48+5=53$.

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

Given a position 

$\displaystyle p(t)=15t^{2}+17t-9$, we can use the power rule 

$\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}$ where $\displaystyle n\neq0$ to determine that 

$\displaystyle v(t)=p'(t)=30t+17$.

Therefore, at $\displaystyle t=2$, 

$\displaystyle v(2)=p'(2)=30(2)+17=60+17=77$.