To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})\)

Remember that anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}\)

This problem is best solved by using the power rule. For each variable, multiply by the exponent and reduce the exponent by one:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(2*4x^{2-1})+(1*5x^{1-1})-(0*3x^0-1)\)

Treat \(\displaystyle 3\) as \(\displaystyle 3x^0\) since anything to the zero power is one.

Remember, anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(8x^{1})+(5x^{0})-0\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=8x+5\)

The average rate of change of \(\displaystyle f\) on interval \(\displaystyle [a,b]\) is 

\(\displaystyle \frac{f (b ) - f(a)}{b-a}\)

Substitute:

\(\displaystyle \frac{f (4 ) - f(3)}{4-3} = \frac{4^{4} - 4^{3}}{1} = \frac{256 - 64}{1} = 192\)

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2*x^{2-1})+(1*5x^{1-0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0\)

Remember that anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5\)

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat \(\displaystyle 8\) as \(\displaystyle 8x^0\), as anything to the zero power is one.

That means this problem will look like this:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{1-1})+(0*8x^{0-1})\)

Notice that \(\displaystyle (0*8x^{0-1})=0\), as anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{0})\)

Remember, anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5*1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5\)

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat \(\displaystyle 5\) as \(\displaystyle 5x^0\), as anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})+(0*5x^{0-1})\)

Notice that \(\displaystyle (0*5x^{0-1})=0\), as anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2\)

To get \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x)\), we can use the power rule.

Since the exponent of the \(\displaystyle x\) is \(\displaystyle 1\), as \(\displaystyle 2x=2x^1\), we lower the exponent by one and then multiply the coefficient by that original exponent:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x)=1*2x^{1-1}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x)=2x^{0}\)

Anything to the \(\displaystyle 0\) power is \(\displaystyle 1\).

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x)=2*1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x)=2\)

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat \(\displaystyle 8\) as \(\displaystyle 8x^0\) since anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{2-1})+(0*8x^{0-1})\)

Notice that \(\displaystyle 0*8x^{0-1}=0\) since anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{2-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x\)

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat \(\displaystyle 1\) as \(\displaystyle 1x^0\) since anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})+(0*x^{0-1})\)

Notice that \(\displaystyle (0*x^{0-1})=0\) since anything times zero is zero.

That leaves us with \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})\).

Simplify.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})\)

As stated earlier, anything to the zero power is one, leaving us with:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1\)

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat \(\displaystyle 4\) as \(\displaystyle 4x^0\) since anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})+(0*4x^{0-1})\)

Notice that \(\displaystyle (0*4x^{0-1})=0\) since anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{1})+(1*13x^{0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})\)

Just like it was mentioned earlier, anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13\)