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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1321 : Calculus Ii

$f(x)= 8x^{3}-7x^{2}+11$

Give f''(x) .

Possible Answers:

f ''(x) = 24x

f ''(x) = 24x - 11

f ''(x) = 24x -14

f ''(x) = 48x

f ''(x) = 48x -14

Correct answer:

f ''(x) = 48x -14

Explanation:

First, find the derivative f'(x) of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate f'(x) to get f''(x) .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

f''(x)= 48x-14

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Example Question #23 : Calculus Review

Differentiate $x^{999}$ .

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{999}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 1,000x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Example Question #1322 : Calculus Ii

Give the second derivative of $x^{777}$ .

Possible Answers:

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{775}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 603,729 x^{775}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{779}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 604,506 x^{779}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 603,729 x^{777}$

Correct answer:

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{775}$

Explanation:

Find the derivative of $x^{777}$ , then find the derivative of that expression.

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}$

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Example Question #1323 : Calculus Ii

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

Give f'(x) .

Possible Answers:

$f'(x) = 2.4x^{2}-x$

$f'(x) = 2.4x^{2}-1$

$f'(x) = 2.4x^{2}-x + 0.9$

$f'(x) = 2.4x^{3}-x^{2}$

$f'(x) = 2.4x^{3}-x^{2}+ 0.9$

Correct answer:

$f'(x) = 2.4x^{2}-x$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$f'(x)= 2.4x^{2}-1x^{1}$

$f'(x) = 2.4x^{2}-x$

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Example Question #24 : Calculus Review

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give f''(x) .

Possible Answers:

$f''(x) =10.8x^{2} + 2.1x$

$f''(x) =10.8x^{2} + 8.4x - 0.5$

$f''(x) =10.8x^{2} + 4.2x$

$f''(x) =10.8x^{2} + 4.2x - 0.5$

$f''(x) =10.8x^{2} + 8.4x$

Correct answer:

$f''(x) =10.8x^{2} + 4.2x$

Explanation:

First, find the derivative f'(x) of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate f'(x) to get f''(x) .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Example Question #1 : First And Second Derivatives Of Functions

Find the second derivative of the following equation:

f(x)=ln(2x-1)

Possible Answers:

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

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

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

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

$f''(x)=-\frac{4}{(2x-1)^2}$

Correct answer:

$f''(x)=-\frac{4}{(2x-1)^2}$

Explanation:

To find the second derivative, first we need to find the first derivative. The derivative of a natural log is the derivative of operand times the inverse of the operand. So for the given function, we get the first derivative to be

$f'(x)=2\cdot \frac{1}{2x-1}=\frac{2}{2x-1}$ .

Now, we have to take the derivative of the first derivative. To simplify this, we can rewrite the function to be f'(x)=2(2x-1)^-^1 . From here we can use the chain rule to solve for the derivative. First, multiply by the exponent and find the new exponent by subtracting the old one by one. Next multiply by the derivative of (2x-1) and then simplify. Thus, we get

$f''(x)=2\cdot-1\cdot2(2x-1)^-^2=-\frac{4}{(2x-1)^2}$ .

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Example Question #1 : First And Second Derivatives Of Functions

Find the second derivative of the following function.

f(x)=sec(x)

Possible Answers:

f''(x)=sec(x)tan(x)

f''(x)=cot(x)csc(x)

f''(x)=-csc(x)

f''(x)=sec^3(x)tan(x)

$f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$

Correct answer:

$f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$

Explanation:

To find the second derivative, first we need to find the first derivative. So for the given function, we get the first derivative to be

f'(x)=tan(x)sec(x) .

Now we have to take the derivative of the derivative. To do this we need to use the product rule as shown below

Thus, we get

$\\f''(x)=tan(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}sec(x)+sec(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}tan(x)\\f''(x)=tan(x)\cdot sec(x)\cdot tan(x)+sec(x)\cdot sec^2(x)\\f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$ .

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Example Question #1 : First And Second Derivatives Of Functions

Find the second derivative of the given function:

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

Possible Answers:

$\frac{x^2-1}{x^2}$

x^2+2x+1

$\frac{2}{x^3}$

$\frac{2x+2}{x}$

Correct answer:

$\frac{2}{x^3}$

Explanation:

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be
$\\f'(x)=\frac{x\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2x+1)-(x^2+2x+1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x)}{x^2}\\f'(x)=\frac{x\cdot (2x+2)-(x^2+2x+1)\cdot 1}{x^2}\\f'(x)=\frac{x^2-1}{x^2}$ .

Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.

Thus, we get

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

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Example Question #1 : First And Second Derivatives Of Functions

f(x)= 2sin(2x)+e^x

Calculate f''(x).

Possible Answers:

f''(x)= 8sin(2x)+e^x

f''(x)= -8sin(2x)+e^x

f''(x)=-8cos(2x)+e^x

$f''(x)=-4cos(2x)+e^{2x}$

$f''(x)=4sin(x)+e^{2x}$

Correct answer:

f''(x)= -8sin(2x)+e^x

Explanation:

There are two seprate functions that make up f(x) . There is 2sin(2x) and e^x .

On a general note,

$\frac{d}{dx}sin(ax)=acos(ax)$ and

$\frac{d}{dx}cos(ax)=-asin(ax)$ .

Also,

$\frac{d}{dx}e^u=e^u \cdot u'$

With that said, let's calculate f'(x) :

f'(x)= 4cos(2x)+e^x . Notice the e^x term is still unchanged.

And now let's calculate f''(x) .

f'(x)= -8sin(2x)+e^x

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Example Question #2 : First And Second Derivatives Of Functions

$f(x)= \frac{1}{2}e^{2x}+\frac{1}{8}x^2$

$f'(x)=ae^{bx} + \frac{1}{4}x$

Find and .

Possible Answers:

$a=\frac{1}{2}$ , b=2

a=0, b=1

a=2, b=2

a=1, b=2

$a=\frac{1}{2}$ , b=1

Correct answer:

a=1, b=2

Explanation:

To find a and b, first let's calculate f'(x) .

$f'(x)=\frac{1}{2}(2)e^{2x}+\frac{1}4{x}$

Remember that $\frac{d}{dx}e^{ax}=ae^{ax}$ , where a is real number.

is simply the coefficient in front of the exponential, which simplifies to 1, and is the power of the exponent, which is 2.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1321 : Calculus Ii

Example Question #23 : Calculus Review

Example Question #1322 : Calculus Ii

Example Question #1323 : Calculus Ii

Example Question #24 : Calculus Review

Example Question #1 : First And Second Derivatives Of Functions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #2 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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