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Calculus 2 : Calculus II

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

Example Questions

Example Question #136 : First And Second Derivatives Of Functions

What is the second derivative of f(x)=4x^3-2x+6 ?

Possible Answers:

f{}''(x)=24x

f{}''(x)=24x+2

f{}''(x)=4x

f{}''(x)=24

f{}''(x)=24x+C

Correct answer:

f{}''(x)=24x

Explanation:

Before you can take the second derivative, you must take the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=12x^2-2$

Now, take the derivative of this to find the second derivative:

f{}''(x)=24x

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Example Question #1461 : Calculus Ii

What is the derivative of f(x)=5x^3-2x^2+4x?

Possible Answers:

$f{}'(x)=15x^2-4x+4$

$f{}'(x)=15x^2+2x+4$

$f{}'(x)=15x^2-4x-4$

$f{}'(x)=15x^2-4x$

$f{}'(x)=-15x^2-4x+4$

Correct answer:

$f{}'(x)=15x^2-4x+4$

Explanation:

To take the derivative, remember to multiply the exponent by the coefficient and then subtract one from the exponent:

$f{}'(x)=15x^2-4x+4$

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Example Question #335 : Derivatives

What is the derivative of $f(x)=9x^{\frac{2}{3}}-4x^2+x$ ?

Possible Answers:

$f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-8x+1$

$f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-8x-1$

$f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-2x+1$

$f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-8x$

$f{}'(x)=\frac{6}{x^{\frac{2}{3}}}-8x+1$

Correct answer:

$f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-8x+1$

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient and then subtract one from the exponent:

$f{}'(x)=6x^{\frac{-1}{3}}-8x+1$

Simplify so you don't have a negative exponent:

$f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-8x+1$

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Example Question #1462 : Calculus Ii

Find the derivative of the given function

f(x)=x^x

Possible Answers:

$x^{x-1}\ln(x-1)$

$x^x(1+\ln(\ln x))$

$x^x(1+\ln x)$

$x^x\ln(1+\ln x)$

Correct answer:

$x^x(1+\ln x)$

Explanation:

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Example Question #140 : First And Second Derivatives Of Functions

Find the first derivative of the following function: $y=lnxcos(e^{x})$

Possible Answers:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{sin(e^{x})}{x}-e^{x}lnxsin(e^{x})$

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{cos(e^{x})}{x}-e^{x}lnxsin(e^{x})$

$\frac{\mathrm{dy} }{\mathrm{d} x}=lnxsinxe^{x}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=lnxcos(e^{x})+\frac{cosx}{x}$

Correct answer:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{cos(e^{x})}{x}-e^{x}lnxsin(e^{x})$

Explanation:

To solve this problem, we use a combination of the product and chain rules. First, we use the product rule, which looks like this:

$\frac{\mathrm{dy} }{\mathrm{d} x}=lnx\frac{\mathrm{d} }{\mathrm{d} x}cos(e^{x})+\frac{\mathrm{d} }{\mathrm{d} x}lnxcos(e^{x})$ , which simplifies to:

$\frac{dy}{dx}=lnx\frac{d}{dx}cos(e^{x})+\frac{cos(e^{x})}{x}$ . We need to use the chain rule to find the derivative of $cos(e^{x})$ , which looks like this:

$\frac{d}{dx}cos(e^{x})=-sin(e^{x})e^{x}$ . Plugging this back into our equation above, we get:

$\frac{dy}{dx}=-lnxsin(e^{x})e^{x}+\frac{cos(e^{x})}{x}$ , which simplifies to:

$\frac{dy}{dx}=\frac{cos(e^{x})}{x}-e^{x}lnxsin(e^{x})$

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Example Question #141 : First And Second Derivatives Of Functions

What is the first derivative of the following equation?

$y=(x^3+2x)^{cosx}$

Possible Answers:

$\frac{dy}{dx}=(x^{3}+2x)^{sinx}(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x))$

$\frac{dy}{dx}=(x^{3}+2x)^{cosx}(\frac{cosx(3x^{2}+2)}{x^{3}+2x})$

$\frac{dy}{dx}=(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x))$

$\frac{dy}{dx}=(x^{3}+2x)^{cosx}[(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x)]$

Correct answer:

$\frac{dy}{dx}=(x^{3}+2x)^{cosx}[(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x)]$

Explanation:

To solve this problem, we have to use several tricks. First, we take the natural log of each side, so we can bring down the exponent on the right side of the equation, which looks like this:

$lny=cosxln(x^{3}+2x)$ . Now, we differentiate each side of the equation, keeping in mind that we need to use implicit differentiation for the left side:

$\frac{dy}{dx}\frac{1}{y}=\frac{d}{dx}cosxln(x^{3}+2x)$ . To differentiate the right side, we use a combination of the chain and product rules, which looks like this:

$\frac{dy}{dx}\frac{1}{y}=\frac{d}{dx}cosxln(x^{3}+2x)+cosx\frac{d}{dx}ln(x^{3}+2x)$ , which becomes:

$\frac{dy}{dx}\frac{1}{y}=-sinxln(x^{3}+2x)+cosx(\frac{1}{x^{3}+2x})(3x^{2}+2)$ . This simplifies further to: $\frac{dy}{dx}\frac{1}{y}=(\frac{(cosx)(3x^{2}+2)}{x^{3}+2x})-sinxln(x^{3}+2x)$ . Lastly, we multiple both sides of the equation by . This gives us:

$\frac{dy}{dx}=y(\frac{(cosx)(3x^{2}+2)}{x^{3}+2x})-sinxln(x^{3}+2x)$ . Looking back at our original equation gives our value for . Plugging in this value, we get our answer of:

$\frac{dy}{dx}=(x^3+2x)^{cosx}[(\frac{(cosx)(3x^{2}+2)}{x^{3}+2x})-sinxln(x^{3}+2x)]$

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Example Question #341 : Derivative Review

What is the second derivative of f(x)=4x^4-3x^3+2x^2-x ?

Possible Answers:

$f{}''(x)=48x^2-18x-5$

$f{}''(x)=48x^3-18x+4$

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

$f{}''(x)=48x^2+18x+4$

$f{}''(x)=48x^2-18x+4$

Correct answer:

$f{}''(x)=48x^2-18x+4$

Explanation:

Before you can take the second derivative, find the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=16x^3-9x^2+4x-1$

Now, take the second derivative from the first derivative:

$f{}''(x)=48x^2-18x+4$

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Example Question #1463 : Calculus Ii

Find the derivative of f(x)=14x^2-3x+1 .

Possible Answers:

f{}'(x)=24x-3

$f{}'(x)=28x-3x^2$

f{}'(x)=28x-3

$f{}'(x)=28x^2-3$

f{}'(x)=28x+3

Correct answer:

f{}'(x)=28x-3

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

f{}'(x)=28x-3

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Example Question #343 : Derivative Review

What is the second derivative of f(x)=9x^3+4x^2-2x+9 ?

Possible Answers:

f{}''(x)=54x

f{}''(x)=54x+8

f{}''(x)=54x-8

f{}''(x)=24x+8

f{}''(x)=54x+18

Correct answer:

f{}''(x)=54x+8

Explanation:

To take the derivative, multiply the exponent by the coefficient in front of the x term and then also subtract one from the exponent. Take the first derivative:

$f{}'(x)=27x^2+8x-2$

Now, take the second derivative from the first derivative:

f{}''(x)=54x+8

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Example Question #344 : Derivative Review

If f(x)=3x^3 - 6x^2 + 2x - 1 , what is $\frac{d^2 u}{dx^2}$ ?

Possible Answers:

$\frac{x^4}{4}-2x^3+x^2-x$

18x-12

x^2-6x+2

9x^2-12x+2

Correct answer:

18x-12

Explanation:

To find the second derivative, first one has to find the first derivative, then take the derivative of this result.

The derivative of f(x)=3x^3 - 6x^2 + 2x - 1 is 9x^2-12x+2 .

The derivative of 9x^2-12x+2 is 18x-12 , and this is our final answer.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #136 : First And Second Derivatives Of Functions

Example Question #1461 : Calculus Ii

Example Question #335 : Derivatives

Example Question #1462 : Calculus Ii

Example Question #140 : First And Second Derivatives Of Functions

Example Question #141 : First And Second Derivatives Of Functions

Example Question #341 : Derivative Review

Example Question #1463 : Calculus Ii

Example Question #343 : Derivative Review

Example Question #344 : Derivative Review

All Calculus 2 Resources

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