Finding Second Derivative of a Function - Math

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Question

Let .

Find the second derivative of .

Answer

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

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

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Question

Define $f (x) = 6x^{3} - 12x^{2} + 4x -8$ .

What is ?

Answer

Take the derivative of , then take the derivative of .

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

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

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

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

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Question

Define $g(x) = 7x^{5} - 6x^{3} - 17x$ .

What is ?

Answer

Take the derivative of , then take the derivative of .

$g(x) = 7x^{5} - 6x^{3} - 17x$

$g' (x) = 5\cdot 7x^{5-1} - 3\cdot 6x^{3-1} - 17$

$g' (x) = 35x^{4} - 18x^{2} - 17$

$g'' (x) = 4\cdot 35x^{4-1} - 2\cdot 18x^{2-1} - 0$

$g'' (x) = 140x^{3} - 36x$

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Question

Define $f(x) = \frac{1}{e^{4x}}$ .

What is ?

Answer

Rewrite:

$f(x) = \frac{1}{e^{4x}}$

$f(x) = e^{-4x}$

Take the derivative of , then take the derivative of .

$f'(x) = -4 e^{-4x}$

$f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}$

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Question

Define $g(x) = \cos \left (x^{2} \right )$ .

What is ?

Answer

Take the derivative of , then take the derivative of .

$g(x) = \cos \left (x^{2} \right )$

$g'(x) = 2x \cdot [-\sin (x^2)]$

$g'(x) = - 2x \sin (x^2)$

$g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]$

$g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

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Question

What is the second derivative of ?

Answer

To get the second derivative, first we need to find the first derivative.

To do that, 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.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2x^{2-1})+(15x^{1-0})$

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

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+51$

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

Now we do the same process again, but using as our expression:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero will be zero.

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

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

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(21)$

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

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Question

What is the second derivative of ?

Answer

To get the second derivative, first we need to find the first derivative.

To do that, 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 as , as anything to the zero power is one.

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})+(08x^{0-1})$

Notice that $(08x^{0-1})=0$ as anything times zero will be zero.

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

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

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=51$

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

Now to get the second derivative we repeat those steps, but instead of using , we use .

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=(05x^{0-1})$

Notice that $(05x^{0-1})=0$ as anything times zero will be zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=0$

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Question

What is the second derivative of ?

Answer

To get the second derivative, first we need to find the first derivative.

To do that, 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 as , as anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})$

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

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

Now we repeat the process using as the expression.

Just like before, we're going to treat as .

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=(23x^{2-1})+(02x^{0-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}3x^2+2=(23x^{2-1})$

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

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

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Question

If , what is ?

Answer

The question is asking us for the second derivative of the equation. First, we need to find the first derivative.

To do that, 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 as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})+(08x^{0-1})$

Notice that $08x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{1})$

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

Now we do the exact same process but using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=112x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12x^{0}$

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

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12$

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Question

What is the second derivative of ?

Answer

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

To do that, 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 as since anything to the zero power is one.

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

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

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

Simplify.

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

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

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

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

Now we can repeat the process using or as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0x^{0-1}$

As pointed out before, anything times zero is zero, meaning that $\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0$ .

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Question

What is the second derivative of ?

Answer

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

To do that, 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 as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})+(04x^{0-1})$

Notice that $(04x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{1})+(113x^{0})$

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(131)$

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

Now we repeat the process using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(124x^{1-1})+(013x^{0-1})$

Like before, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+10)=(124x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=24$

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Question

What is the second derivative of $2x^4+\frac{1}{2}x$ ?

Answer

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.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{4-1})+(1\frac{1}{2}x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{3})+(\frac{1}{2}x^{0})$

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

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

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(38x^{3-1})+(0\frac{1}{2}x^{0-1})$

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

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(38x^{2})$

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

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Question

What is the second derivative of ?

Answer

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 as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

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

Simplify.

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=31$

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

Now we repeat the process but use , or , as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^0)=03x^{0-1}$

As stated before, anything times zero is zero.

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

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Question

What is the second derivative of ?

Answer

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.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{2-1})+(112x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{1})+(112x^{0})$

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

Anything to the zero power is one.

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

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

Now we repeat the process using as our expression.

We're going to treat as .

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(110x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(110x^{1-1})$

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

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

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

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

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Question

What is the second derivative of ?

Answer

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 as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{2-1})+(015x^{0-1})$

Notice that $(015x^{0-1})=0$ since anything times zero is zero.

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

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

Now we repeat the process, but using as our expression.

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

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

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

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Question

What is the second derivative of ?

Answer

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 as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{3-1})+(22x^{2-1})+(05x^{0-1})$

Notice that $(05x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{3-1})+(22x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{2})+(22x^{1})$

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

Now we repeat the process but using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(23x^{2-1})+(14x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(23x^{1})+(14x^{0})$

Remember, anything to the zero power is one.

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

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

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Question

What is the second derivative of $4x^2+\frac{2}{3}x$ ?

Answer

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.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(24x^{2-1})+(1\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(24x^{1})+(1\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

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

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(18x^{1-1})+(0\frac{2}{3}x^{0-1})$

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

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

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

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

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Question

What is the second derivative of ?

Answer

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: $\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)$

Plug in our equations.

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

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(12x^{1-1})+(012x^{0-1})$

Anything times zero is zero.

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

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

Anything to the zero power is one.

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

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Question

What is the second derivative of ?

Answer

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:

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(24x^{2-1})+(15x^{1-1})-(03x^0-1)$

Treat as since anything to the zero power is one.

Remember, anything times zero is zero.

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

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

Now follow the same process but for .

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(18x^{1-1})+(05x^{0-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(18x^{0})$

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

Therefore the second derivative will be the line .

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Question

What is the second derivative of ?

Answer

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

To find the first 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 value of the exponent by one.

To make that work, we're going to treat as , since anything to the zero power is one.

This means that is the same as .

Now use the power rule:

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^1+3x^0)=(14x^{1-1})+(03x^{0-1})$

Anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^1+3x^0)=(14x^{0})+0$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^1+3x^0)=(141)$

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

Now we repeat the process, but using .

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^0)=04x^{0-1}$

Since anything times zero is zero, $\frac{\mathrm{d} }{\mathrm{d} x}(4x^0)=0.$

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