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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 #1621 : Calculus Ii

What is v(0) when x(t)=-4t^2+3t-1 ?

Possible Answers:

Correct answer:

Explanation:

To find the value of the velocity at 3, find the derivative of the position function. Remember, when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent.

Therefore,

$v(t)=x'(t)=2\cdot -4t^{2-1}+1\cdot 3t^{1-1}$

v(t)=-8t+3 .

Then, plug in 0 for t.

Therefore,

v(0)=-8(0)+3=3 .

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Example Question #1621 : Calculus Ii

Find $\frac{dy}{dx}$ by implicit differentiation

$y\sqrt{x}-x\sqrt{y}=16$

Possible Answers:

$\frac{2x\sqrt{y}-y}{2y\sqrt{x}-x}$

$\frac{2x\sqrt{x}-y}{2y\sqrt{y}-x}$

$\frac{2y\sqrt{x}-y}{2x\sqrt{y}-x}$

$\frac{2y\sqrt{y}-x}{2x\sqrt{x}-x}$

Correct answer:

$\frac{2y\sqrt{x}-y}{2x\sqrt{y}-x}$

Explanation:

Untitled

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Example Question #1621 : Calculus Ii

What is a possible function for f(x) if

$f^{\left( 4\right )}(x)=120(2x+3)$

Possible Answers:

f(x)=6x^5+120x^4

f(x)=2x^5+15x^4

f(x)=x^5+x^4

f(x)=2x^4+15x^5

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

Correct answer:

f(x)=2x^5+15x^4

Explanation:

Let f(x)=2x^5+15x^4

Step 1: Take the derivative of f(x) four times by using the power rule. The results are below:

f'(x)=10x^4+60x^3

f''(x)=40x^3+180x^2

$f^{(3)}=120x^2+360x$

$f^{(4)}=240x+360=120(2x+3)$

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Example Question #1621 : Calculus Ii

$f(x)=\frac{5^{x^{2}}}{2^{x}}$ . Find f'(x) .

Possible Answers:

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(\frac{5}{2}))}{2^{x}}$

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

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

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}$

Correct answer:

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}$

Explanation:

Since both the numerator and denominator contain a variable, we must use the quotient rule.

$f'(x)=\frac{(2^{x}){\color{Green} \frac{\mathrm{d} }{\mathrm{d} x}(5^{x^{2}})}-(5^{x^{2}}){\color{Blue} \frac{\mathrm{d} }{\mathrm{d} x}(2^{x})}}{(2^{x})^{2}}$

Remember that the derivative of a constant raised to a variable power follows the pattern, $\frac{\mathrm{d} }{\mathrm{d} u}(a^{u})=a^{u}\ln\left | a \right | \mathrm{d} u$ , where a is a constant and u is a function of x.

Applying this we get,

$f'(x)= \frac{(2^{x}){\color{Green} (5^{x^{2}}(\ln5)(2x))}-(5^{x^{2}}){\color{Blue} (2^{x}(\ln2)(1))}}{(2^{x})^{2}}$

Now we simplify by factoring the greatest common factor out of the numerator.

$f'(x)=\frac{2^{x}\cdot 5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{(2^{x})^{2}}$

Then we can cancel a single $2^{x}$ from the numerator and denominator.

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}$

This is the correct answer.

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Example Question #1621 : Calculus Ii

Differentiate $f(t)=\frac{t}{1-e^t}$ .

Possible Answers:

e^t

$-\frac{1}{e^t}$

$\frac{-te^t-1+e^t}{1-2e^t+e^{2t}}$

$\frac{1-e^t+te^t}{1-2e^t+e^{2t}}$

Undefined

Correct answer:

$\frac{1-e^t+te^t}{1-2e^t+e^{2t}}$

Explanation:

Using the Quotient rule for derivatives, we know that the derivative will equal $\frac{(1-e^t)(1)-t(-e^t)}{(1-e^t)^2}$ .

When you simplify this, you get $\frac{1-e^t+te^t}{1-2e^t+e^{2t}}$ .

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Example Question #1621 : Calculus Ii

Differentiate $f(y)=\frac{e^y}{y}$ .

Possible Answers:

e^y

e^y(ln(y))

$\frac{(y-1)e^y}{y^2}$

$\frac{e^y-ye^y}{y^2}$

$\frac{e^y-ye^y}{y}$

Correct answer:

$\frac{(y-1)e^y}{y^2}$

Explanation:

Using the Quotient rule for derivatives, we know that the derivative will equal : $\frac{y(e^y)-e^y(1)}{y^2}$ .

If you simplify this, you will get:

$\frac{(y-1)e^y}{y^2}$

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Example Question #1621 : Calculus Ii

Differentiate $\frac{2}{x}+\frac{1}{4x^3}-\frac{6}{x^5}$ .

Possible Answers:

2lnx-3ln(4x)-30lnx

$-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$

$-2x^{-2}+\frac{1}{4}x^{-4}+30x^{-6}$

$2+\frac{1}{8}x^{-2}-\frac{6}{4}x^{-4}$

$2-\frac{1}{8}x^{-2}+\frac{6}{4}x^{-4}$

Correct answer:

$-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$

Explanation:

Rewriting the original equation gives us $f(x)=2(x^{-1})+\frac{1}{4}x^{-3}-6(x^{-5})$ .

Then, one can just use the power rule to get: $-2x^{-2}-\frac{3}{12}x^{-4}-30x^{-6}$ .

Simplifying this gives us: $-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$ .

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Example Question #1621 : Calculus Ii

Express the derivative of $y(x)=\sqrt {7x^3-6x^2+9x}$ in simplest terms.

Possible Answers:

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt {(7x^3-6x^2+9x)^3}}$

$\large y(x)=\bigg[\frac {7x^2-4x+3}{\sqrt{(7x^3-6x^2+9x)}}\bigg]^{\frac {2}{3}}$

$\large y'(x)=-\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

Correct answer:

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

Explanation:

Screen shot 2016 01 01 at 10.25.54 pm

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Example Question #502 : Derivative Review

Find the derivative of $\large y(x)=(2x+x^{-3})^\frac {1}{4}$ in simplest form.

Possible Answers:

$\large \large y'(x)=\frac {x^\frac {-7}{4}(2x^4-3)}{4(2x^4+1)^\frac {3}{4}}$

$\large \large y'(x)=\frac {(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

$\large \large y'(x)=\frac {x^{\frac {1}{4}}(2x^4-3)}{4(2x^4+1)^\frac {-3}{4}}$

$\large \large y'(x)=\frac {5(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

Correct answer:

$\large \large y'(x)=\frac {(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

Explanation:

Screen shot 2016 01 02 at 8.27.30 am

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Example Question #1622 : Calculus Ii

What is the first derivative of $\large y(x)=tan^4\sqrt {cot(3x)}$ ?

Possible Answers:

$\large - {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2)}$

$\large \frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

$\large -\frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

$\large \large -\frac {6csc^2(3x)\big (tan \sqrt {cot(3x))}\big )^3 \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

Correct answer:

$\large -\frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

Explanation:

Screen shot 2016 01 02 at 9.36.43 am

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #502 : Derivative Review

Example Question #1622 : Calculus Ii

All Calculus 2 Resources

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