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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 #31 : Derivatives

Find the second derivative of the following function:

$f(x)=x^3\cos(x)+e^{2x}$

Possible Answers:

$\cos(x)(6x-x^3)-3x^2\sin(x)+4e^{2x}\:$

$\cos(x)(6x-x^3)-6x^2\sin(x)+2e^{2x}$

$\cos(x)(6x-x^3)+6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x+x^3)-6x^2\sin(x)+4e^{2x}$

Correct answer:

$\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

Explanation:

The first derivative of the function is equal to

$f'(x)=3x^2\cos(x)-x^3\sin(x)+2e^{2x}$

The second derivative - the derivative of the function above - of the original function is equal to

$f''(x)=\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

Both derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$

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Example Question #1341 : Calculus Ii

Find the derivative of the following function:

$f(x)=\sqrt[3]{\cos^2(x)+e^{4x}}$

Possible Answers:

$\frac{-2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{2}{3}}}$

$\frac{-2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{1}{3}}}$

$\frac{-\cos(x)\sin(x)+4e^{4x}}{(\cos^2(x)+e^{4x})^{\frac{2}{3}}}$

$\frac{2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{2}{3}}}$

$\frac{-2\cos(x)\sin(x)+e^{4x}}{3(\cos(x)+e^{4x})^{\frac{2}{3}}}$

Correct answer:

$\frac{-2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{2}{3}}}$

Explanation:

The derivative was found using the following rules:

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

$f(x)=\sqrt[3]{\cos^2(x)+e^{4x}}$

$f'(x)=\frac{1}{3}(\cos^2(x)+e^{4x})^{-\frac{2}{3}}(-2\cos(x)\sin(x)+4e^{4x})$

$f'(x)=\frac{-2\cos(x)\sin(x)+4e^{4x}}{3(\cos^2(x)+e^{4x})^{\frac{2}{3}}}$

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Example Question #1342 : Calculus Ii

Find the derivative of the function:

$f(x)=\frac{e^{2x}+\cos(5x^2)}{e^x}$

Possible Answers:

$\frac{e^{3x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

$\frac{e^{2x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

$\frac{e^{2x}-10xe^x\sin(5x^2)+\cos(5x^2)}{e^x}$

$\frac{e^{2x}+10x\sin(5x^2)+\cos(5x^2)}{e^x}$

Correct answer:

$\frac{e^{2x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

Explanation:

The derivative of the function is equal to

$f(x)=\frac{e^{2x}+\cos(5x^2)}{e^x}$

$f'(x)=\frac{(2e^{2x}-10x\sin(5x^2))e^x-e^x(e^{2x}+\cos(5x^2))}{e^{2x}}$

$f'(x)=\frac{e^{2x}-10x\sin(5x^2)+\cos(5x^2)}{e^x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #212 : Derivatives

Find the derivative of the following function:

$f(x)=\sqrt{e^{\cos(x)}}$

Possible Answers:

$\frac{-\sin(x)e^{\cos(x)}}{\sqrt{e^\cos(x)}}$

$\frac{e^{\cos(x)}}{2\sqrt{e^\cos(x)}}$

$\frac{-\sin(x)e^{\cos(x)}}{2\sqrt{e^\cos(x)}}$

$\frac{\sin(x)e^{\cos(x)}}{\sqrt{e^\cos(x)}}$

Correct answer:

$\frac{-\sin(x)e^{\cos(x)}}{2\sqrt{e^\cos(x)}}$

Explanation:

$f(x)=\sqrt{e^{\cos(x)}}$

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

$f'(x)=\frac{-\sin(x)e^{\cos(x)}}{2\sqrt{e^\cos(x)}}$

The derivative of the function was found using the following rules:

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

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Example Question #1344 : Calculus Ii

Solve for the derivative: y=2cot(2x)

Possible Answers:

-4csc({2x})

$-4sec^2({2x})$

-4sec({2x})tan({2x})

$-2sec^2({2x})$

$-4csc^2({2x})$

Correct answer:

$-4csc^2({2x})$

Explanation:

Write the derivative of cotangent .

cot(x) = -csc^2(x)

The problem requires chain rule since the inner function is . The chain rule is the derivative of the inner function, and the derivative of is .

Take the derivative to obtain $\frac{dy}{dx}$ . Multiply the value obtained by the use of chain rule.

$\frac{dy}{dx} = 2 (-csc^2({2x})) \cdot 2 = -4csc^2({2x})$

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Example Question #21 : First And Second Derivatives Of Functions

Use the chain rule to find the derivative of the function $f(x) = e^{-6x^2}$

Possible Answers:

$-12xe^{-6x^2}$

-6x^2e

$-6x^2e^{-6x^2}$

$e^{-6x^2}$

None of these answers

Correct answer:

$-12xe^{-6x^2}$

Explanation:

The formula for the chain rule works as follows;

$[f(g(x))]' = f'(g(x))\times g'(x)$

Setting f(x) = e^x, g(x)=-6x^2 , we have

$f(g(x)) = e^{-6x^2}$

Hence

$[f(g(x))]' = f'(g(x))\times g'(x) = e^{-6x^2} \times -12x = -12xe^{-6x^2}$ .

(Keep in mind that the derivative of e^x is e^x )

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Example Question #22 : First And Second Derivatives Of Functions

Find the second derivative of: y= sec(2x)

Possible Answers:

8sec(2x)tan^2(2x)+ 8sec^3(2x)

2sec^3(2x)+2sec(2x)tan^2(2x)

2sec^2(2x)+2sec(2x)tan(2x)

4sec^3(2x)+4sec(2x)tan^2(2x)

8sec(4x)tan(4x) +4sec^3(2x)

Correct answer:

4sec^3(2x)+4sec(2x)tan^2(2x)

Explanation:

The derivative of secant is:

$\frac{d}{dx}sec(x) = sec(x)tan(x)$

Since the inner function of secant is not , we will need to use chain rule, which is to multiply the derivative of the inner function .

Solve for the first derivative.

$\frac{d}{dx}sec(2x) = sec(2x)tan(2x) \cdot 2 = 2sec(2x)tan(2x)$

To take the second derivative, we will need to use the product rule and chain rule.

The product rule is: f(x) g'(x) + g(x)f'(x)

To avoid confusion, let f(x) =sec(2x) and g(x)= tan(2x) . Their derivatives are then:

f'(x) = 2sec(2x)tan(2x)

g'(x) = 2sec^2(2x)

The derivative of 2sec(2x)tan(2x) in terms of f(x) and g(x) by product rule is:

$\frac{d^2}{dx^2}:\:2[f(x) g'(x) + g(x)f'(x)]$

Resubstitute the functions and derivative functions into the equation.

$2[sec(2x)\cdot 2sec^2(2x)+ tan(2x)\cdot 2sec(2x)tan(2x)]$

Simplify. The answer is:

4sec^3(2x)+4sec(2x)tan^2(2x)

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Example Question #22 : First And Second Derivatives Of Functions

Find the derivative of: $y=sec(\frac{1}{x})$

Possible Answers:

$\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

$-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

$-ln(x)sec(\frac{1}{x})tan(\frac{1}{x})$

$ln(x)sec(\frac{1}{x})tan(\frac{1}{x})$

$sec(\frac{1}{x})tan(\frac{1}{x})$

Correct answer:

$-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

Explanation:

Write the derivative of secant.

$\frac{d}{dx} sec(x) = sec(x)tan(x)$

Since the inner function is $\frac{1}{x}$ , we will need to apply the chain rule to solve this derivative.

Take the derivative of $\frac{1}{x}$ . Do not mix this with the integration, which is ln(x) .

$\frac{d}{dx}\:\frac{1}{x} = \frac{d}{dx}\:x^{-1} = -1\cdot x^{-2} = -\frac{1}{x^2}$

The derivative of sec(x) is then:

$= sec(\frac{1}{x})tan(\frac{1}{x}) \cdot -\frac{1}{x^2}$

The answer is: $-\frac{sec(\frac{1}{x})tan(\frac{1}{x})}{x^2}$

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Example Question #21 : First And Second Derivatives Of Functions

What is the second derivative of $\sin(x^2)$ ?

Possible Answers:

$-\sin(x^2)$

$2\cos(x^2)-4x^2\cdot \sin(x^2)$

$2x\cdot \cos(x^2)$

$2x\cdot \cos(x^2) - x^4\cdot \sin(x^2)$

Correct answer:

$2\cos(x^2)-4x^2\cdot \sin(x^2)$

Explanation:

We use the chain rule to get the first derivative:

$(f \circ g)' = (f'\circ g)\cdot g'$

This gives us $2x\cdot \cos(x^2)$ . Now we must take the derivative again, which means we have to use the product rule *and* the chain rule. As a reminder, the product rule says that

$(f\cdot g)' = f'\cdot g + f\cdot g'$

Which means the second derivative of the original function is

$2\cos(x^2)+2x\cdot 2x (-\sin(x^2))$ . With some simplifying we can see that this is equal to

$2\cos(x^2)-4x^2\sin(x^2)$

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Example Question #23 : First And Second Derivatives Of Functions

Evaluate the first and second derivative of the function

$f(x)=\pi x$

when x=4

Possible Answers:

$f'(4)=4\pi$

f''(4)=4

$f'(4)=4x\pi$

$f''(4)=4\pi$

$f'(4)=\pi$

f''(4)=0

f'(4)=4x

f''(4)=x

Correct answer:

$f'(4)=\pi$

f''(4)=0

Explanation:

We must find the first and second derivatives.

To do so we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[x^n]=nx^{n-1}$

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]=1*\pi x^{1-1}=\pi x^0 = \pi$

To find the second derivative we apply the power rule again

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x^0]=0*\pi x^{0-1}=0$

We find that

$f'(x)=\pi$ and f''(x)=0

We then evaluate each for x=4 and get

$f'(4)=\pi$

f''(4)=0

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #31 : Derivatives

Example Question #1341 : Calculus Ii

Example Question #1342 : Calculus Ii

Example Question #212 : Derivatives

Example Question #1344 : Calculus Ii

Example Question #21 : First And Second Derivatives Of Functions

Example Question #22 : First And Second Derivatives Of Functions

Example Question #22 : First And Second Derivatives Of Functions

Example Question #21 : First And Second Derivatives Of Functions

Example Question #23 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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