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Calculus 2 : Derivatives

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Example Questions

Example Question #331 : Derivatives

Find the derivative of the function $f(x)=2x\cos(x)$

Possible Answers:

$4x\cos(x)$

$2\cos(x)-2x\sin(x)$

$2x\cos(x)+1$

Correct answer:

$2\cos(x)-2x\sin(x)$

Explanation:

To solve, we use the product rule $\frac{\mathrm{d} }{\mathrm{d} x}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ . Applying, we get $f'(x)= (2x)'\cos(x)+2x(\cos(x))'$ . Taking the derivatives we get the correct solution $2\cos(x)-2x\sin(x)$

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Example Question #332 : Derivative Review

What is the second derivative of f(x)=5x^2-12x+4 ?

Possible Answers:

f{}''(x)=10

f{}''(x)=10x-1

f{}''(x)=x

$f{}''(x)=10x^2$

f{}''(x)=10x

Correct answer:

f{}''(x)=10

Explanation:

To find the second derivative, you must first 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)=10x-12

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

f{}''(x)=10 .

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Example Question #333 : Derivative Review

What is the second derivative of f(x)=3x^3-8x^2+x-4?

Possible Answers:

$f{}''(x)=\frac{18}{x}$

f{}''(x)=18x-16

f{}''(x)=18

f{}''(x)=18x

$f{}''(x)=18x^2-16x$

Correct answer:

f{}''(x)=18x-16

Explanation:

First, 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:

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

Now, take the second derivative:

f{}''(x)=18x-16

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Example Question #332 : Derivatives

What is the derivative of $f(x)=2x^{\frac{5}{4}}$ ?

Possible Answers:

$f{}'(x)=\frac{5}{2}x^{\frac{1}{4}}$

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

$f{}'(x)=\frac{5}{2}x^{\frac{3}{4}}$

$f{}'(x)=\frac{5}{2}x^{\frac{1}{4}}+C$

$f{}'(x)=\frac{3}{2}x^{\frac{1}{4}}$

Correct answer:

$f{}'(x)=\frac{5}{2}x^{\frac{1}{4}}$

Explanation:

Remember that when differentiating, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent as well.

$f{}'(x)=2(\frac{5}{4})x^{\frac{1}{4}}=\frac{10}{4}x^{\frac{1}{4}}$

Simplify to get your answer.

$f{}'(x)=\frac{5}{2}x^{\frac{1}{4}}$

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Example Question #332 : Derivatives

What is the derivative of f(x)=-3t^2+4t-1?

Possible Answers:

f{}'(x)=6t+4

f{}'(x)=-6t-4

f{}'(x)=-6t

f{}'(x)=-6t+4

$f{}'(x)=-6t+4t^2$

Correct answer:

f{}'(x)=-6t+4

Explanation:

To take the derivative, remember to multiply the exponent by the coefficient in front of the x term:

f{}'(x)=-6t+4

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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 #333 : Derivatives

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

Possible Answers:

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

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

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

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

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

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 #334 : Derivatives

Find the derivative of the given function

f(x)=x^x

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Untitled

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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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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #331 : Derivatives

Example Question #332 : Derivative Review

Example Question #333 : Derivative Review

Example Question #332 : Derivatives

Example Question #332 : Derivatives

Example Question #136 : First And Second Derivatives Of Functions

Example Question #333 : Derivatives

Example Question #335 : Derivatives

Example Question #334 : Derivatives

Example Question #140 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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