In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

$\displaystyle f'(x)=3x^{4}-6$

Then, you differentiate the velocity function to get the acceleration function:

$\displaystyle f''(x)=12x^3$

Then, find $\displaystyle f''(x)$ when $\displaystyle x=2$.

The answer is: $\displaystyle 96$

In order to find the acceleration of a certain point, you must first find the derivative of the position function which gives us the velocity function and then the derivative of the velocity function which gives us the acceleration function: $\displaystyle p''(t)= v'(t)= a(t)$

In this case, the position function is: $\displaystyle p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

The velocity function is found by taking the derivative of the position function: $\displaystyle v(t)= t^{-\frac{1}{2}} + 3e^t$

The acceleration function is found by taking the derivative of the velocity function: $\displaystyle a(t)= {-\frac{1}{2}}t^{(-\frac{3}{2})} + 3e^t$

Finally, to find the accelaration, substitute $\displaystyle t=4$ into the acceleration function: $\displaystyle a(4)= {-\frac{1}{2}}(4)^{(-\frac{3}{2})} + 3e^{(4)}$

Therefore, the answer is: $\displaystyle 163.7$

 $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0, so

$\displaystyle f (x) = 7x^{2} + 6x - 9$

$\displaystyle f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$\displaystyle f ' (x) = 14\cdot x^{1} + 6$

$\displaystyle f ' (x) = 14x + 6$

First, find the derivative $\displaystyle f'(x)$ of $\displaystyle f(x)= 8x^{3}-7x^{2}+11$.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0, so

$\displaystyle f(x)= 8x^{3}-7x^{2}+11$

$\displaystyle f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$\displaystyle f'(x)= 24x^{2}-14x^{1}$

$\displaystyle f'(x)= 24x^{2}-14x$

Now, differentiate $\displaystyle f'(x)$ to get $\displaystyle f''(x)$.

$\displaystyle f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$\displaystyle f''(x)= 48x^{1}-14x^{0}$

$\displaystyle f''(x)= 48x-14$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$, so

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

First, find the derivative $\displaystyle f'(x)$ of $\displaystyle f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$.

Recall that $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$, and the derivative of a constant is 0.

$\displaystyle f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$\displaystyle f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate $\displaystyle f'(x)$ to get $\displaystyle f''(x)$.

$\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$\displaystyle f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$\displaystyle f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$\displaystyle f''(x) =10.8x^{2} + 4.2x$

To find the second derivative, first we need to find the first derivative. So for the given function, we get the first derivative to be 

$\displaystyle f'(x)=tan(x)sec(x)$. 

Now we have to take the derivative of the derivative. To do this we need to use the product rule as shown below

 Thus, we get

 $\displaystyle \\f''(x)=tan(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}sec(x)+sec(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}tan(x)\\f''(x)=tan(x)\cdot sec(x)\cdot tan(x)+sec(x)\cdot sec^2(x)\\f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$.

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\displaystyle \small lna^{b}=blna$ to solve an easier derivative.

Let $\displaystyle \small y=lnx^{lnx}$.

Now we'll take the natural log of both sides to get 

$\displaystyle \small lny=ln(lnx^{lnx})=lnxln(lnx)$.

Now we can use implicit differentiation to solve for $\displaystyle \small y'$.

The derivative of $\displaystyle \small lny$ is $\displaystyle \small y'/y$, and the derivative of $\displaystyle \small \small \small lnx(ln(lnx))$ can be found using the product rule, which states 

$\displaystyle \small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'$ where $\displaystyle \small u$ and $\displaystyle \small v$ are functions of $\displaystyle \small x$.

Letting $\displaystyle \small u=lnx$ and $\displaystyle \small v=ln(lnx)$ 

(which means $\displaystyle \small u'=1/x$ and $\displaystyle \small v'=1/(xlnx)$) we get our derivative to be $\displaystyle \small ln(lnx)/x+1/x$.

Now we have $\displaystyle \small y'/y=ln(lnx)/x+1/x$, but $\displaystyle \small y=lnx^{lnx}$, so subbing that in we get 

$\displaystyle \small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x$.

Multiplying both sides by $\displaystyle \small lnx^{lnx}$, we get 

$\displaystyle \small y'=lnx^{lnx}(ln(lnx)/x+1/x)$.

That is our derivative.

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$\displaystyle -\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$,  $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

The derivative of the function is equal to

$\displaystyle f'(x)=3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$