Derivatives of functions - 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,

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

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

← 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,

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

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

← 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

Find the derivative of the function

$f=x\cos(x^2)$

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Answer

To find the derivative of the function, we use both the product rule and the chain rule

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

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Question

Find the limit of the function below using L'Hopital's Rule

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}$

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Answer

*********************************

STEPS

*

Given:

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}$

Try the limit. Plug in two for y and check the result:

$\binom{lim}{y\rightarrow 2}=\frac{(2)^3-7(2)^2+10(2)}{(2)^2+(2)-6}$

$\binom{lim}{y\rightarrow 2}=\frac{8-28+20}{4+2-6}$

$\binom{lim}{y\rightarrow 2}=\frac{0}{0}$

Thus, we realize me must use L'Hopital's Rule on the original quotient, deriving the expressions in the numerator and denominator independently

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}$

$\binom{lim}{y\rightarrow 2}\frac{d(y^3-7y^2+10y)}{d(y^2+y-6)}$

$\binom{lim}{y\rightarrow 2}\frac{3y^2-14y+10}{2y+1}$

Try the limit once more:

$\binom{lim}{y\rightarrow 2}=\frac{3(2)^2-14(2)+10}{2(2)+1}$

$\binom{lim}{y\rightarrow 2}=\frac{12-28+10}{4+1}$

$\binom{lim}{y\rightarrow 2}=\frac{-16+10}{5}$

Simplifying the numerator, we arrive at the correct answer:

*

CORRECT ANSWER

*********************************

${\color{Red} \binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{-6}{5}}$

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Question

Find :

$\small f(x)=sin(x)tan(x)$

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Answer

This is a product rule using trigonometric functions:

$\small cos(x)(tan(x))+sec^2(x)sin(x)$

This can be simplified further:

$\small {\color{Red} cos(x)}(\frac{sin(x)}{{\color{Red} cos(x)}})+sin(x)sec^2(x)$

What is in red cancels and you get:

$\small {sin(x)}+sin(x)sec^2(x)$

But you can take this one step further and pull out a sin(x) to get:

$\small f'(x)=\small f(x)=sin(x)(1+sec^2(x))$

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Question

If $f(x)=\sin x$ then $f'\left({\frac{\pi }{4}}\right)=$

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Answer

To calculate the derivative of this function at the desired point, first recall that,

$\sin(x)\frac{d}{dx}=\cos(x)$

$f(x)=\sin x$

$f^{}(x)=\cos x$

Now, substitute the value into the derivative function to solve.

$f'(\frac{\pi }{4}) = cos\frac{\pi }{4} = cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

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Question

If the position of a particle over time is represented by $p=t^{3}-16t^{2}-7t$ then what is the particle's instantaneous acceleration at ?

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Answer

The answer is .

Since velocity is the first derivative of the position function, take the derivative once. Then, recall that the acceleration function is the second derivative of position thus the derivative needs to be taken one more time.

$p=t^{3}-16t^{2}-7t$

$velocity =p'=3t^{2}-32t-7$

$acceleration =p''=6t-32$

$p''(4)=6(4)-32 = -8$

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Question

Let $f(x)=x^2-\frac{1}{1-x^2}$ . Which of the following gives the equation of the line normal to $f(x)$ when ?

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Answer

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at . In order to find the tangent line, we will need to evaluate the derivative of at .

$f(x)=x^2-\frac{1}{1-x^2}=x^2-(1-x^2)^{-1}$

$f'(x)=2x-(-1)(1-x^2)^{-2}(-2x)$

$f'(x)=2x-2x(1-x^2)^{-2}$

$f'(2)=2(2)-2(2)(1-2^2)^{-2}$

$f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at is $\frac{32}{9}$ . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

$\left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at , it will pass through the point $\left ( 2,f(2) \right )$ . Be careful to use the original equation for , not its derivative.

$f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $-\frac{9}{32}$ and passes through the piont $\left ( 2,\frac{13}{3} \right )$ . We can now use point-slope form to find the equation of the normal line.

$y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by .

$96y-416=-27(x-2)$

$27x + 96y = 470$

The answer is $27x + 96y = 470$ .

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Question

Consider the function:

$f(x) = -6e^{-x} + x$

The relative minimum for this function is at:

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Answer

To find any relative minimum, one first needs to find the critical points by setting the first derivative equal to zero:

$f'(x) = 6e^{-x} + 1 = 0$

However, the first derivative is positive for all real values of x, since the exponential function is always positive. Thus, there are no values for which , and therefore no critical points and no relative minimum.

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Question

Calculate the derivative of the following:

$x^{2}+12x-36$

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Answer

Using the power rule which states,

$\frac{d}{dx}x^n=nx^{n-1}$

you can move the from $x^{2}$ to the front and decrease the exponent by which makes it .

For , any term that has an exponent of , the coefficient is its derivative.

Thus, the derivative of is .

Since does not have a variable attached, the derivative will be .

Add your derivatives to get .

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