To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

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

Simplify.

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

Remember that anything to the zero power is equal to one.

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

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

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat \(\displaystyle 12\) as \(\displaystyle 12x^0\) since anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(0*12x^{0-1})\)

Notice that \(\displaystyle (0*12x^{0-1})=0\) since anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})\)

Simplify.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})\)

As stated before, anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3\)

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0\)

Anything to the zero power is one.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12\)

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat \(\displaystyle 15\) as \(\displaystyle 15x^0\) since anything to the zero power is one.

For this problem that would look like this:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})\)

Notice that \(\displaystyle (0*15x^{0-1})=0\) since anything times zero is zero.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x\)

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

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

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

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

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

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

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

To find the derivative of the problem, we can use the power rule. The power rule says to multiply the coefficient of the variable by the exponent of the variable and then lower the exponent value by one.

To make that work, we're going to treat \(\displaystyle 3\) as \(\displaystyle 3x^0\), since anything to the zero power is one.

This means that \(\displaystyle 4x+3\) is the same as \(\displaystyle 4x^1+3x^0\).

Now use the power rule:

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

Anything times zero is zero.

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

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

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

To find the derivative of \(\displaystyle x^4-3x^2+8\), we can use the power rule.

The power rule states that we multiply each variable by its current exponent and then lower the exponent of each variable by one.

Since \(\displaystyle x^0=1\), we're going to treat \(\displaystyle 8\) as \(\displaystyle 8x^0\).

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

Anything times zero is zero, so our final term \(\displaystyle (0*8x^{0-1})=0\), regardless of the power of the exponent.

Simplify what we have.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^4-3x^2+8x^0=(4x^{3})-(6x^{1})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^4-3x^2+8x^0=4x^{3}-6x\)

Our final solution, then, is \(\displaystyle 4x^3-6x\).

For this problem, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one. 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(x)=(3*\frac{2}{3}x^{3-1})+(2*7x^{2-1})-(1*12x^{1-1})\)

Simplify.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(x)=(2x^{2})+(14x^{1})-(12x^{0})\)

Anything to the zero power is one, so \(\displaystyle 12x^0=12(1)=12\).

Therefore, \(\displaystyle f'(x)=2x^{2}+14x-12\).

We use the power rule on each term of the function.

The first term

\(\displaystyle 6x^2\)

becomes

\(\displaystyle 2 \times 6 x^2 = 12 x\).

The second term

\(\displaystyle 4x\)

becomes

\(\displaystyle 4\).

The final term, 7, is a constant, so its derivative is simply zero.