First, find the first derivative.

You should get $\displaystyle f'(x)=6x^{2}-14x+8$.

Next, differentiate again. 

You should get $\displaystyle f''(x)=12x-14$.

Finally, plug in x=2 to get $\displaystyle f''(2)=10$.

To find the rate of change of the velocity of the fluid at any point, we must take the derivative of the function:

$\displaystyle v'(x)=-16\sin(x)+2x+16\cos(x)=16(\cos(x)-\sin(x))+2x$

The derivative was found using the following rules:

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

The derivative of the function is equal to

$\displaystyle f'=\frac{1}{2}\tan(4x)^{-\frac{1}{2}}(4\sec^2(4x))=\frac{2\sec^2(4x)}{\sqrt{\tan(4x)}}$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the chain rule was used on the square root, the inner tangent function, and the function inside the secant. 

The derivative of the function is equal to

$\displaystyle f'=2\cos(2x)(-2\sin(2x))=-4\cos(2x)\sin(2x)$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Notice that the chain rule was used on the cosine squared function, the cosine function, and the function inside the sine function.

To find the derivative of the function, we use both the product rule and the chain rule

$\displaystyle f'=x'\cos(x^2)+x\cos(x^2)'=\cos(x^2)-x\sin(x^2)(2x)=\cos(x^2)-2x^2\sin(x^2)$

To find the derivative of the function, we use both the quotient rule and the chain rule

$\displaystyle f'=(x^5'*\sin(2x)-x^5\sin(2x)')/\sin^2(2x))$

$\displaystyle f'=\frac{5x^4\sin(2x)-2x^5\cos(2x)}{\sin^2(2x)}$

To calculate the derivative of this function at the desired point, first recall that, 

$\displaystyle \sin(x)\frac{d}{dx}=\cos(x)$

$\displaystyle f(x)=\sin x$

$\displaystyle f^{'}(x)=\cos x$

Now, substitute the value into the derivative function to solve.

$\displaystyle f'(\frac{\pi }{4}) = cos\frac{\pi }{4} = cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at $\displaystyle x=2$. In order to find the tangent line, we will need to evaluate the derivative of $\displaystyle f(x)$ at $\displaystyle x=2$.

$\displaystyle f(x)=x^2-\frac{1}{1-x^2}=x^2-(1-x^2)^{-1}$

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

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

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

$\displaystyle f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at $\displaystyle x=2$ is $\displaystyle \frac{32}{9}$. Because the tangent line and the normal line are perpendicular, the product of their slopes must equal $\displaystyle -1$.

(slope of tangent)(slope of normal) = $\displaystyle -1$

$\displaystyle \left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$\displaystyle slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at $\displaystyle x=2$, it will pass through the point $\displaystyle \left ( 2,f(2) \right )$. Be careful to use the original equation for $\displaystyle f$, not its derivative.

$\displaystyle f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $\displaystyle -\frac{9}{32}$ and passes through the piont $\displaystyle \left ( 2,\frac{13}{3} \right )$. We can now use point-slope form to find the equation of the normal line.

$\displaystyle y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by $\displaystyle 32(3)=96$.

$\displaystyle 96y-416=-27(x-2)$

$\displaystyle 27x + 96y = 470$

The answer is $\displaystyle 27x + 96y = 470$.

The answer is $\displaystyle -8$.

Since velocity is the first derivative of the position function, take the derivative once. Then, recall that the acceleration function is the second derivative of position thus the derivative needs to be taken one more time.

$\displaystyle p=t^{3}-16t^{2}-7t$

$\displaystyle velocity =p'=3t^{2}-32t-7$

$\displaystyle acceleration =p''=6t-32$

$\displaystyle p''(4)=6(4)-32 = -8$

To find any relative minimum, one first needs to find the critical points by setting the first derivative equal to zero:

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

However, the first derivative is positive for all real values of x, since the exponential function is always positive.  Thus, there are no values for which $\displaystyle f'(x) = 0$, and therefore no critical points and no relative minimum.