Computation of the Derivative - AP Calculus AB

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Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

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Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

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Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

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Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

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Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

← Didn't Know|Knew It →

Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

← Didn't Know|Knew It →

Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

← Didn't Know|Knew It →

Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

← Didn't Know|Knew It →

Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

← Didn't Know|Knew It →

Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

← Didn't Know|Knew It →

Question

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .

is in .

and is the anti derivative of

then we must have,

(2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Tap to reveal answer

Answer

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

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Question

Find the derivative.

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Answer

Use the power rule to find the derivative.

$\frac{d}{dx}12x^3=36x^2$

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Question

Let .

Find the second derivative of .

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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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