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Calculus 1 : Rate of Change

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Example Questions

Example Question #401 : Rate Of Change

What is the slope at x=1 given the following function: $f(x) = \sin (ln(2x))$

Possible Answers:

-0.87

2.85

0.77

0.67

Correct answer:

0.77

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: $f(x) = \sin (ln(2x))$

That is done by doing the following: $f'(x) = \cos (ln(2x)) (\frac{1}{2x}) ((1)2x^{(1-1)})$

Therefore, the derivative is: $f'(x) = \frac{\cos (ln(2x))}{x}$

Then, you plug x=1 into the derivative: $f'(1) = \frac{\cos (ln(2(1)))}{(1)}$

Therefore, the answer is: 0.77

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Example Question #402 : Rate Of Change

What is the slope at $x=\frac{1}{2}$ given the following function: $f(x) = (6x+3)^3 + \cos(4x)$

Possible Answers:

156.9

320.4

-293.8

263.7

132.6

Correct answer:

320.4

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: $f(x) = (6x+3)^3 + \cos(4x)$

That is done by doing the following: $f'(x) = {\color{Blue} (3)}(6x+3)^{\color{Blue} {(3-1)}}({\color{Blue} (1)}6x^{\color{Blue} {(1-1)}}+0) + ({\color{Blue} -\sin}(4x))({\color{Blue} (1)}4x^{\color{Blue} {(1-1)}})$

Therefore, the derivative is: $f'(x) = 18x(6x+3)^2 -4\sin(4x)$

Then, you plug $x=\frac{1}{2}$ into the derivative: $f'(\frac{1}{2}) = 18(\frac{1}{2})(6(\frac{1}{2})+3)^2 -4\sin(4(\frac{1}{2}))$

Therefore, the answer is: 320.4

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Example Question #403 : Rate Of Change

Find the slope at x=2 :

f(x)= (ln(2x))^2

Possible Answers:

Answer not listed

1.4

3.5

1.2

6.7

Correct answer:

1.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: f(x)= (ln(2x))^2

That is done by doing the following: $f'(x)= ln(2x) (\frac{2}{x})$

Then plug x=2 into the derivative: $f'(2)= ln(2({\color{Blue} 2})) (\frac{2}{{\color{Blue} 2}})$

Therefore, the answer is: 1.4

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Example Question #404 : Rate Of Change

Find the slope at x=1 :

f(x)= 5 ln (5x^2-3x+1)

Possible Answers:

10.2

13.5

Answer not listed.

7.4

9.3

Correct answer:

9.3

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: f(x)= 5 ln (5x^2-3x+1)

That is done by doing the following: $f'(x)=\frac{50x+ 15}{5x^2+3x-1}$

Then plug x=1 into the derivative: $f'(1)=\frac{50({\color{Blue}1 })+ 15}{5({\color{Blue} 1})^2+3({\color{Blue} 1})-1}$

Therefore, the answer is: 9.3

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Example Question #405 : Rate Of Change

Find the slope at x=1 :

$f(x)= e^{3x^2+2x+1}$

Possible Answers:

10.2

5.4

6.5

23.5

Answer not listed.

Correct answer:

5.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: $f(x)= e^{3x^2+2x+1}$

That is done by doing the following: $f'(x)= e^{3x^2+2x+1} (6x + 2)$

Then plug x=1 into the derivative: $f'(0)= e^{3({\color{Blue} 0})^2+2({\color{Blue} 0})+1} (6({\color{Blue} 0}) + 2)$

Therefore, the answer is: 5.4

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Example Question #406 : Rate Of Change

Find the slope at x=0 :

$f(x)= \cos(\frac{1}{4}x^4 +3x)$

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: $f(x)= \cos(\frac{1}{4}x^4 +3x)$

That is done by doing the following: $f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

Then plug x=0 into the derivative: $f'(0)= -\sin(\frac{1}{4}({\color{Blue} 0})^4 +3({\color{Blue} 0})) ({\color{Blue} 0}^{3} +3)$

Therefore, the answer is:

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Example Question #407 : Rate Of Change

Find the slope at x=1 :

$f(x)= 2x^2 + \tan x - (2x-4)^2$

Possible Answers:

Answer not listed.

9.7

2.4

15.4

16.7

Correct answer:

15.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: $f(x)= 2x^2 + \tan x - (2x-4)^2$

That is done by doing the following: $f'(x)= \sec^2x - 4x+16$

Then plug x=1 into the derivative: $f'(x)= \sec^2({\color{Blue} 1}) - 4({\color{Blue} 1})+16$

Therefore, the answer is: 15.4

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Example Question #408 : Rate Of Change

Find the slope at x=3 :

f(x)= 2(x+2)^2 + 3x

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: f(x)= 2(x+2)^2 + 3x

That is done by doing the following: f'(x)= 4x+11

Then plug x=3 into the derivative: $f'(3)= 4({\color{Blue} 3})+11$

Therefore, the answer is:

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Example Question #409 : Rate Of Change

Find the derivative of the following function:
f(x)=e^xSin(x)

Possible Answers:

e^x(-Cos(x)+Sin(x))

$e^{Cosx}$

e^xCos(x)

-e^x(Cos(x)+Sin(x))

e^x(Cos(x)+Sin(x))

Correct answer:

e^x(Cos(x)+Sin(x))

Explanation:

We notice that this function,

f(x)=e^xSin(x) , is of the form f(x)=g(x)h(x)

such that we must use the Product Rule to find the derivative.

f'(x)=g(x)h'(x)+h(x)g'(x) .

Doing so we find the derivative to be:

f'(x)=e^xCos(x)+e^xSin(x)=e^x(Cos(x)+Sin(x))

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Example Question #410 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameter of the sphere at the instance the rate of shrinkage of the volume is 18.5 times the rate of shrinkage of the surface area?

Possible Answers:

$\frac{37}{\pi}$

$37\pi$

$\frac{74}{\pi}$

Correct answer:

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$A=4\pi r^2$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of shrinkage of the volume is 18.5 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(18.5)8\pi r \frac{dr}{dt}$

r =(18.5)2

r=37

The diameter is then:

d=2r

d=74

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #401 : Rate Of Change

Example Question #402 : Rate Of Change

Example Question #403 : Rate Of Change

Example Question #404 : Rate Of Change

Example Question #405 : Rate Of Change

Example Question #406 : Rate Of Change

Example Question #407 : Rate Of Change

Example Question #408 : Rate Of Change

Example Question #409 : Rate Of Change

Example Question #410 : Rate Of Change

All Calculus 1 Resources

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