To take the second derivative, we need to start with the first derivative.

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.

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

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

Now we repeat the process, but using \(\displaystyle 8x^3+\frac{1}{2}\) as our expression.

We're going to treat \(\displaystyle \frac{1}{2}\) as being \(\displaystyle \frac{1}{2}x^0\) since anything to the zero power is equal to one.

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

Notice that \(\displaystyle (0*\frac{1}{2}x^{0-1})=0\) since anything times zero is zero.

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

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}\)

To take the second derivative, we need to start with the first derivative.

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.

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

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

Now we repeat the process but use \(\displaystyle 3\), or \(\displaystyle 3x^0\), as our expression.

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

As stated before, anything times zero is zero.

Therefore, \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(3x^0)=0\).

To find the second derivative, first we need to start with the first derivative.

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.

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

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

Now we repeat the process using \(\displaystyle 10x+12\) as our expression.

We're going to treat \(\displaystyle 12\) as \(\displaystyle 12x^0\).

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

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

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

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

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

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

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

To find the second derivative, we start by taking the first derivative.

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.

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

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

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

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

Now we repeat the process, but using \(\displaystyle -16x\) as our expression.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16\)

To find the second derivative, we need to start by finding the first derivative.

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.

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

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

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

Now we repeat the process but using \(\displaystyle 3x^2+4x\) as our expression.

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

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

Remember, anything to the zero power is one.

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

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

To find the second derivative, we need to start by finding the first derivative.

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.

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

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

Now we repeat the process, but we use \(\displaystyle 8x+\frac{2}{3}\) as our expression.

For this problem, we're going to say that \(\displaystyle \frac{2}{3}=\frac{2}{3}x^0\) since, as stated before, anything to the zero power is one.

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

Notice that \(\displaystyle (0*\frac{2}{3}x^{0-1})=0\) as anything times zero is zero.

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

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

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

To find the second derivative, we need to start with the first derivative.

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))*g'(x)\)

Plug in our equations.

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

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

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

From here, we can use our normal power rule to find the second derivative.

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

Anything times zero is zero.

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

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

Anything to the zero power is one.

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

To find the second derivative, we need to find the first derivative first. To find the first derivative, we can use 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.

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

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

Now follow the same process but for \(\displaystyle 8x+5\).

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

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

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

Therefore the second derivative will be the line \(\displaystyle y=8\).

The second derivative is just the derivative of the first derivative. So first we find the first derivative of \(\displaystyle f(x)\). Remember the derivative of \(\displaystyle \sin x\) is \(\displaystyle \cos x\), and the derivative for \(\displaystyle \cos x\) is \(\displaystyle -\sin x\).

 \(\displaystyle f'(x)= cos(x)+sin(x)\)

Then to get the second derivative, we just derive this function again. So

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

Take the derivative \(\displaystyle f'\) of \(\displaystyle f\), then take the derivative of \(\displaystyle f'\).

\(\displaystyle f (x) = 6x^{3} - 12x^{2} + 4x -8\)

\(\displaystyle f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0\)

\(\displaystyle f '(x) = 18x^{2} - 24x+ 4\)

\(\displaystyle f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0\)

\(\displaystyle f ' '(x) = 36x - 24\)