Use the power rule to find the derivative.

Remember the power rule is:

\(\displaystyle (x^n)'=nx^{n-1}\)

Now lets apply this to our problem.

\(\displaystyle \frac{d}{dx}3x^3=9x^2\)

\(\displaystyle \frac{d}{dx}2x^2=4x\)

\(\displaystyle \frac{d}{dx}x=1\)

Recall that the derivative of a constant is zero.

\(\displaystyle \frac{d}{dx}1=0\)

Thus, the derivative is \(\displaystyle 9x^2+4x+1\)

Find the derivative using the power rule. 

Remember the power rule is:

\(\displaystyle (x^n)'=nx^{n-1}\)

Now lets apply this to our problem.

\(\displaystyle \frac{d}{dx}x^4=4x^3\)

Now, substitute \(\displaystyle 5\) for \(\displaystyle x\).

\(\displaystyle 4(5)^3=500\)

First, find the derivative using the power rule. 

Remember the power rule is:

\(\displaystyle (x^n)'=nx^{n-1}\)

Now lets apply this to our problem.

\(\displaystyle \frac{d}{dx}3x^2=6x\)

\(\displaystyle \frac{d}{dx}7x=7\)

\(\displaystyle \frac{d}{dx}-7=0\)

Thus, the derivative is \(\displaystyle 6x+7\). Now, substitute \(\displaystyle 2\) for \(\displaystyle x\).

\(\displaystyle 6(2)+7=19\)

Use the quotient rule to find the derivative.

Remember the quotient rule is:

\(\displaystyle (\frac{f(x)}{g(x)})'=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}\)

Now lets apply this to our problem.

\(\displaystyle \frac{x^2(0)-4(2x)}{x^4}\)

\(\displaystyle -\frac{8x}{x^4}\)

\(\displaystyle -\frac{8}{x^3}\) is the derivative.

First, find the derivative using the quotient rule. 

Remember the quotient rule is:

\(\displaystyle (\frac{f(x)}{g(x)})'=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}\)

Now lets apply this to our problem.

\(\displaystyle \frac{d}{dx}\frac{x^5}{3}=\frac{3(5x^4)-x^5(0)}{9}\)

\(\displaystyle =\frac{15x^4}{9}=\frac{5x^4}{3}\)

Now, substitute \(\displaystyle 6\) for \(\displaystyle x\).

\(\displaystyle \frac{5(6^4)}{3}=2160\)

First, find the derivative using the quotient rule. 

Remember the quotient rule is:

\(\displaystyle (\frac{f(x)}{g(x)})'=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}\)

Now lets apply this to our problem.

\(\displaystyle \frac{d}{dx}\frac{-x^3}{2}=\frac{2(-3x^2)-(-x^3)0}{4}\)

\(\displaystyle =\frac{-6x^2}{4}=\frac{-3x^2}{2}\)

Now, substitute \(\displaystyle 4\) for \(\displaystyle x\).

\(\displaystyle \frac{-3(4^2)}{2}=\frac{-48}{2}=-24\)

Find the derivative using the product rule. 

Remember the product rule is:

\(\displaystyle (f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\)

Now lets apply this to our problem.

\(\displaystyle -x\sin (x)+\cos (x)\)

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice.

Note that

\(\displaystyle d[sin(u)]=cos(u)du\)

\(\displaystyle d[cos(u)]=-sin(u)du\)

 \(\displaystyle f(x)=\frac{x^3}{6\pi}-9cos(\frac{1}{3}x)\)

\(\displaystyle f'(x)=\frac{x^2}{2\pi}+3sin(\frac{1}{3}x)\)

\(\displaystyle f''(x)=\frac{x}{\pi}+cos(\frac{1}{3}x)\)

Set the second derivative to zero and find the values that satisfy the equation.

\(\displaystyle \frac{x}{\pi}+cos(\frac{1}{3}x)=0\)

\(\displaystyle x=-2.28\)

To verify this is a point of inflection, notice how \(\displaystyle f''(x)=\frac{x}{\pi}+cos(\frac{1}{3}x)\) crosses the x-axis at this point, indicating a change in signs:

Now, plug this value back in to the original function to find the value of the function that matches:

\(\displaystyle f(-2.28)=\frac{(-2.28)^3}{6\pi}-9cos(\frac{1}{3}(-2.28))=-7.15\)

The point of inflection is \(\displaystyle (-2.28,-7.15)\)

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

To take this derivative, we'll make use of the Product rule: \(\displaystyle d[uv]=udv+vdu\)

Taking the derivative the function \(\displaystyle f(x)=3x^2(x+2)^2\) at \(\displaystyle x=3\)

The slope of the tangent is

 \(\displaystyle f'(x)=6x(x+2)^2+6x^2(x+2)\)

\(\displaystyle f'(3)=6(3)(3+2)^2+6(3)^2(3+2)=720\)

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

\(\displaystyle -\frac{1}{720}\)

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

\(\displaystyle d[a^u]=a^uduln(a)\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function  \(\displaystyle f(x)=2^{x^3}3^{x^2}\) at the point \(\displaystyle x=1\)

The slope of the tangent is

 \(\displaystyle f'(x)=3x^2(2^{x^3})(3^{x^2})ln(2)+2x(2^{x^3})(3^{x^2})ln(3)\)

\(\displaystyle f'(1)=3(1)^2(2^{1^3})(3^{1^2})ln(2)+2(1)(2^{1^3})(3^{1^2})ln(3)\)

\(\displaystyle f'(1)=25.7\)

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value, thus the answer is

\(\displaystyle -0.04\).