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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 #1451 : Calculus Ii

Find the derivative of the function:

$f(x)=(x^2+3x)^{5/2}$

Possible Answers:

$\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(2x+3)$

$\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(5x)$

$\frac{5}{2}(2x+3)\sqrt{x^2+3x}$

$\frac{5}{2}(x^2+3x)(2x+3)$

Correct answer:

$\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(2x+3)$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Note that the exponent is the acting as the outer function in the chain rule.

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Example Question #1452 : Calculus Ii

Find the derivative of the function $f(x)= e^\sin(x)$

Possible Answers:

$-e^\sin(x)$

$-\cos(x)e^\cos(x)$

$\cos(x)e^\sin(x)$

$e^\sin(x)$

Correct answer:

$\cos(x)e^\sin(x)$

Explanation:

To find the derivative, you must first use the chain rule. The derivative of $e^\sin(x)$ is $e^\sin(x)$ , and you multiply that by the derivative of the exponent, which is $\cos(x)$ . Putting that all together, you get the final answer as $\cos(x)e^\sin(x)$ .

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Example Question #1453 : Calculus Ii

Find the second derivative of the function $f(x)= 3x^3+\sin(x)$ .

Possible Answers:

$9x^2+\cos(x)$

$18x+\sin(x)$

9x+cos(x)

$18x-\sin(x)$

Correct answer:

$18x-\sin(x)$

Explanation:

Taking the first derivative of the function, using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ and the fact that $\sin'(x)=\cos(x)$ , we get $f'(x)= 9x^2+\cos(x)$ . Using the same rule and the fact that $\cos'(x)=-\sin(x)$ , we get $f''(x)=18x-\sin(x)$

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Example Question #1454 : Calculus Ii

Find the derivative of the function $y(t)= \sin(t)+\frac{5}{t^2}$

Possible Answers:

$\cos(t)+8$

$\cos(t)-\frac{10}{t^3}$

$\sin(t)+1$

$-\sin(t)+10t$

Correct answer:

$\cos(t)-\frac{10}{t^3}$

Explanation:

To find the derivative of the function, we first take the derivative of $\sin(t)$ which is just $\cos(t)$ . Taking the derivative of the second term gets us $\frac{-10}{t^3}$ , using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , where in this case, n=-2 . Using all of the information above and applying, $y'(t)= \cos(t)-\frac{10}{t^3}$

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Example Question #1455 : Calculus Ii

Find the second derivative of the function $f(x)=-\cos(x)$ . Evaluate at $x=\pi/2$ .

Possible Answers:

$-\frac{1}{2}$

Correct answer:

Explanation:

We first take the first derivative of the function which is $f'(x)=\sin(x)$ . We then take the derivative of this new function, which is $f''(x)=\cos(x)$ . Evaluating at $x=\pi/2$ , we get $\cos(\pi/2)=0$ .

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Example Question #1456 : Calculus Ii

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

Possible Answers:

$4x\cos(x)$

$2x\cos(x)+1$

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

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

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

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

f{}''(x)=x

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:

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

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

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

f{}''(x)=18

f{}''(x)=18x

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 #1457 : Calculus Ii

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

Possible Answers:

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

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

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

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

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

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 #1458 : Calculus Ii

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

Possible Answers:

f{}'(x)=-6t

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

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

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

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1451 : Calculus Ii

Example Question #1452 : Calculus Ii

Example Question #1453 : Calculus Ii

Example Question #1454 : Calculus Ii

Example Question #1455 : Calculus Ii

Example Question #1456 : Calculus Ii

Example Question #332 : Derivative Review

Example Question #333 : Derivative Review

Example Question #1457 : Calculus Ii

Example Question #1458 : Calculus Ii

All Calculus 2 Resources

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