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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 #266 : Derivatives

What is the first derivative of the following equation: y=ln(cos(sinx))

Possible Answers:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{-cosxsin(sinx)}{cosx}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=-cosxtan(sinx)$

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{1}{cosx}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{1}{cos(sinx)}$

Correct answer:

$\frac{\mathrm{dy} }{\mathrm{d} x}=-cosxtan(sinx)$

Explanation:

To solve this problem we apply the chain rule. Applying the chain rule once gives us: $\frac{1}{cos(sinx)}*cos(sinx)$ . Because we still have a composite function, we apply the chain rule to , which gives us -sin(sinx)cosx .

Multiplying everything together gives us $\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{-cosxsin(sinx)}{cos(sinx)}$

This can be further simplified to: $\frac{\mathrm{dy} }{\mathrm{d} x}=-cosxtan(sinx)$

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Example Question #267 : Derivatives

Find the first and second derivatives of the following function:

f(x)=4sinxcosx

Possible Answers:

f'(x)=4sin^3x, f''(x)=-12cos^2xsin^3x

f'(x)=4cos(2x), f''(x)=-8sin(2x)

f'(x)=4cos^2x, f''(x)=-8cosxsinx

f'(x)=4sin(2x), f''(x)=8cos(2x)

Correct answer:

f'(x)=4cos(2x), f''(x)=-8sin(2x)

Explanation:

The first derivative of f(x) can be found through a straightforward application of the product rule:

f(x)=4sinxcosx, f'(x)=4((sinx)(-sinx)+(cosx)(cosx))

=4(cos^2x-sin^2x)

=4cos(2x)

by the double angle formula for the cosine function:

cos(2x)=cos^2x-sin^2x .

To solve for the second derivative of f(x) , simply differentiate f'(x) using the chain rule:

f'(x)=4cos(2x), f''(x)=(-4sin(2x))(2)

=-8sin(2x)

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Example Question #268 : Derivatives

Find the derivative of g(x)=(5x+10+sin(x))^5 .

Possible Answers:

5(5x+10+sin(x))^4(5-cos(x))

5(5+cos(x))^4

5(5x+10+sin(x))^4(5+cos(x))

5(5-cos(x))^4

5(5x+10+sin(x))^4(15+cos(x))

Correct answer:

5(5x+10+sin(x))^4(5+cos(x))

Explanation:

This is a chain rule derivative. We first must take the derivative of the outermost function. Then, we multiply that by the derivative of the inside. Our outside function is something raised to the fifth power. This is simply a power rule. Leaving the inside the same, we then multiply that by the derivative of the inside:

g'(x)=5(5x+10+sin(x))^5(5+cos(x)).

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Example Question #269 : Derivatives

Find the derivative of $\sqrt{3x^2+1}$ .

Possible Answers:

$\frac{3}{2\sqrt{3x^2+1}}$

$\frac{3}{2\sqrt{3x}}$

$\frac{1}{2\sqrt{3x}}$

$\frac{3x^2}{2\sqrt{3x^2+1}}$

$\frac{3x}{\sqrt{3x^2+1}}$

Correct answer:

$\frac{3x}{\sqrt{3x^2+1}}$

Explanation:

This is a chain rule derivative. First, we must take the derivative of a square root function, leaving the inside the same. Then, we multiply that by the derivative of what is inside the square root.

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

$=\frac{3x}{\sqrt{3x^2+1}}$ .

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Example Question #270 : Derivatives

Find the derivative of f(x)=-csc(2x^8-4x^4) .

Possible Answers:

$csc(2x^8-4x^4)cot(2x^8-4x^4)*(\frac{x^9}{18}-\frac{4x^5}{5})$

-csc(2x^8-4x^4)cot(2x^8-4x^4)*16(x^7-x^3)

csc(2x^8-4x^4)cot(2x^8-4x^4)*16(x^7-x^3)

cot^2(2x^8-4x^4)*16(x^7-x^3)

csc^2(2x^8-4x^4)16(x^7-x^3)

Correct answer:

csc(2x^8-4x^4)cot(2x^8-4x^4)*16(x^7-x^3)

Explanation:

This is a chain rule problem involve trigonometric functions. The derivative of cscx=-cscx*cotx. Once you rewrite that derivative (keeping the inside of the functions the same), you just need to multiply by the derivative of the inside. The negative sign will disappear since two negatives make a positive.

f'(x)=csc(2x^8-4x^4)cot(2x^8-4x^4)*16(x^7-x^3).

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Example Question #71 : First And Second Derivatives Of Functions

Find the first derivative of the following function:

$f(x)=e^{\sin^2(x)}$

Possible Answers:

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

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

$e^{\sin^2(x)}$

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

Correct answer:

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

Explanation:

The first derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #271 : Derivative Review

Find the first derivative of the function:

$g(x)=\frac{x^2}{e^{3x}}$

Possible Answers:

$\frac{2xe^{3x}-3x^2e^{3x}}{e^{6x}}$

$\frac{2xe^{3x}-3x^2e^{3x}}{e^{3x^2}}$

$\frac{2xe^{3x}-x^2e^{3x}}{e^{6x}}$

$\frac{2xe^{3x}+3x^2e^{3x}}{e^{6x}}$

Correct answer:

$\frac{2xe^{3x}-3x^2e^{3x}}{e^{6x}}$

Explanation:

The first derivative of the function is equal to

$g'(x)=\frac{2xe^{3x}-3x^2e^{3x}}{e^{6x}}$

and was found using the following rules:

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

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Example Question #272 : Derivative Review

Find the first derivative of the function:

$f(x)=e^{3x^4}\cos(x)$

Possible Answers:

$16x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

$12x^3e^{3x^4}\cos(x)+e^{3x^4}\sin(x)$

$12x^3e^{3x^4}\sin(x)-e^{3x^4}\cos(x)$

$12x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

Correct answer:

$12x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

Explanation:

The derivative of the function is equal to

$f'(x)=12x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

and was found using the following rules:

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

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Example Question #273 : Derivative Review

Find the derivative of the function:

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

Possible Answers:

$\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{(xe^x)^2}$

$\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{xe^x}$

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

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

$\frac{(2x+2)-(x^2+2x+3)(e^x)}{xe^x}$

Correct answer:

$\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{(xe^x)^2}$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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Example Question #1391 : Calculus Ii

What is the derivative of $f(x)=\frac{1}{2}x^2+6x-1?$

Possible Answers:

f{}'(x)=x-6

$f{}'(x)=x^2+6$

f{}'(x)=x+6

f{}'(x)=2x+6

Correct answer:

f{}'(x)=x+6

Explanation:

To take the derivative, remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent. Therefore, your answer should be: $f{}'(x)=(\frac{1}{2}(2))x+6=x+6$ .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #266 : Derivatives

Example Question #267 : Derivatives

Example Question #268 : Derivatives

Example Question #269 : Derivatives

Example Question #270 : Derivatives

Example Question #71 : First And Second Derivatives Of Functions

Example Question #271 : Derivative Review

Example Question #272 : Derivative Review

Example Question #273 : Derivative Review

Example Question #1391 : Calculus Ii

All Calculus 2 Resources

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