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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 #256 : Derivative Review

If y=(3cos(x)sin(x))^2 , find $\frac{dy}{dx}$

Possible Answers:

$\frac{dy}{dx}=18sin(x)cos(2x)$

$\frac{dy}{dx}=18$

$\frac{dy}{dx}=18cos(x)sin(x)cos(2x)$

$\frac{dy}{dx}=18sin(2x)cos(2x)$

Correct answer:

$\frac{dy}{dx}=18cos(x)sin(x)cos(2x)$

Explanation:

By the chain rule:

$\frac{dy}{dx}=2(3cos(x)sin(x))*\frac{d}{dx}(3cos(x)sin(x))$

By the product rule:

$\frac{d}{dx}(3cos(x)sin(x))=3(cos(x)*cos(x)+sin(x)*-sin(x))$

=3cos^2(x)-3sin^2(x)=3cos(2x)

Therefore:

$\frac{dy}{dx}=18cos(x)sin(x)cos(2x)$

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Example Question #257 : Derivative Review

Find the derivative of f(x)=9x^2(4-3x^3) .

Possible Answers:

36x^2-27x^5

18x^2-27x^4

81x^4(4-3x^2)

-162x^3(4-3x^2)

72x-135x^4

Correct answer:

72x-135x^4

Explanation:

First, we should simplify the problem by distributing through the parenthesis.

9x^2(4-3x^3)=36x^2-27x^5 .

Now, since we have a polynomial, we use the power rule to take the derivative. Multiply the coefficient by the exponent, and reduce the power by 1.

f'(x)=36*2x^1-27*5x^4=72x-135x^4 .

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Example Question #258 : Derivative Review

Find $\frac{dy}{dx}$ using implicit differentiation of x^2 +y^2=25 .

Possible Answers:

$\frac{dy}{dx}=\frac{x}{y}$

$\frac{dy}{dx}=\frac{x}{2y}$

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{dy}{dx}=\frac{2x}{y}$

$\frac{dy}{dx}=\frac{-2x}{y}$

Correct answer:

$\frac{dy}{dx}=\frac{-x}{y}$

Explanation:

For implicit differentiation, you take a derivative of both the and components, and add a $\frac{dy}{dx}$ after every derivative. Then, using algebra, solve for the $\frac{dy}{dx}$ in the equation.

The derivative is: $2x+2y\frac{dy}{dx}=0$ , noting that the derivative of a constant equals zero. Now, we simply rearrange the equation.

$\frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}.$

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Example Question #259 : Derivative Review

Find the derivative of ln(cos(3x^2+2)) .

Possible Answers:

-6xtan(3x^2+2)

6xcot(3x^2+2)

6xtan(3x^2+2)

-6xcot(3x^2+2)

Correct answer:

-6xtan(3x^2+2)

Explanation:

This derivative has multiple layers of the chain rule. Whenever woking with a chain rule derivative, always take the derivative of the outside function, leaving the inside function alone. Then, multiply that by the inside derivative. Here, our first outside function is ln(x) . The derivative of that function is $\frac{1}{x}$ . Then, we multiply that by the derivatives of the inside (which is another chain rule. The whole chain looks like this:

$ln(cos(3x^2+2))'=\frac{1}{cos(3x^2+2)}*-sin(3x^2+2)*6x$

$=\frac{-6xsin(3x^2+2)}{cos(3x^2+2)}=-6xtan(3x^2+2)$ .

In the last step, we used the definition $tanx=\frac{sinx}{cosx}$ to simplify the answer.

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Example Question #1381 : Calculus Ii

Find f'(x) .

$f(x)=\frac{1}{3x^3-5x^2}$

Possible Answers:

$\frac{1}{(3x^3-5x^2)^2}$

$\frac{10x-9x^2}{(3x^3-5x^2)^{2}}$

$\frac{-10x+9x^2}{(3x^3-5x^2)^{2}}$

Correct answer:

$\frac{10x-9x^2}{(3x^3-5x^2)^{2}}$

Explanation:

To simplify the problem, it is easiest if we transform the function from in the denominator into the numinator.

$1/{(3x^3-5x^2)}=(3x^3-5x^2)^{-1}$ .

Now, we just take the derivative using the chain rule:

$(3x^3-5x^2)^{-1}'=-1(3x^3-5x^2)^{-2}*(9x^2-10x)$

$=\frac{10x-9x^2}{(3x^3-5x^2)^{2}}$

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Example Question #61 : First And Second Derivatives Of Functions

Find the derivative of cosxsinx .

Possible Answers:

$cos^2{x}-sin^2{x}.$

$-cos^2{x}+sin^2{x}.$

$-cos^2{x}-sin^2{x}.$

$cos^2{x}+sin^2{x}.$

Correct answer:

$cos^2{x}-sin^2{x}.$

Explanation:

To find this derivative, we need to use the product rule.

f(x)=g(x)*h(x)=g'(x)*h(x)+g(x)*h'(x)

$cosxsinx=cosx*cosx+-sinxsinx=cos^2{x}-sin^2{x}.$

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Example Question #262 : Derivatives

What is the second derivative of the following equation?:

$y=x^{x}$

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}=x^{x-1}+x^{x}(1+lnx)^{2}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^{x}(1+lnx)$

$\frac{\mathrm{dy} }{\mathrm{d} x}=x(x^{x-1})$

$\frac{\mathrm{dy} }{\mathrm{d} x}=xlnx$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}=x^{x-1}+x^{x}(1+lnx)^{2}$

Explanation:

This problem requires several tricks. First, we take the natural log of both sides of the equation to bring the exponent down, on the right side of the equation:

$y=x^{x}$

lny=xlnx

Next, differentiate both sides, keeping in mind we will need to use implicit differentiation:

$\frac{1}{y}\frac{\mathrm{dy} }{\mathrm{d} x}=x(\frac{1}{x})+(1)lnx$

This simplifies to: $\frac{1}{y}\frac{\mathrm{dy} }{\mathrm{d} x}=1+lnx$

Multiplying both sides of the equation by y, we get: $\frac{\mathrm{dy} }{\mathrm{d} x}=y+ylnx$ , which factors out as: $\frac{\mathrm{dy} }{\mathrm{d} x}=y(1+lnx)$ , plugging our value for y back into the equation, we get our solution: $\frac{\mathrm{dy} }{\mathrm{d} x}=x^{x}(1+lnx)$

Now that we've found the first derivative, finding the second derivative is comparbly easier. To do this we use the product rule, which looks like this:

$\frac{d^{2}y}{dx^{2}}=x^{x}\frac{\mathrm{d} }{\mathrm{d} x}(1+lnx)+\frac{\mathrm{d} }{\mathrm{d} x}x^{x}(1+lnx)$

This simplifies to: $\frac{x^{x}}{x}+x^{x}(1+lnx)(1+lnx)$ which further simplifies to:

$\frac{d^{2}y}{dx^{2}}=x^{x-1}+x^{x}(1+lnx)^{2}$

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Example Question #263 : Derivatives

Determine: $\frac{\mathrm{d} }{\mathrm{d} x}(x^2y+ln(xy))$

Possible Answers:

$2xy+\frac{1}{x}$

$2(x^2+2xy)+\frac{1}{x}$

$2xy+\frac{1}{x}y$

$2xy+\frac{1}{xy}$

Correct answer:

$2xy+\frac{1}{xy}$

Explanation:

Since the derivative is with respect to , we can treat as a constant and use the power rule and the definition of $\frac{\mathrm{d} }{\mathrm{d} x} ln(x)= \frac{1}{x}$ :

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2y+ln(xy))= 2xy+ \frac{1}{xy}$

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Example Question #264 : Derivatives

What is the first derivative of the following equation:

$y=\frac{sin(cosx)}{cos^{3}x}$

Possible Answers:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3sinxsin(cosx)}{cos^{4}x}-\frac{sinxcos(cosx)}{cos^{3}x}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{cos(cosx)}{cos^{6}x}$

$\frac{\mathrm{dy} }{\mathrm{d} x}=cos^{2}xcos(cosx)$

$\frac{\mathrm{dy} }{\mathrm{d} x}=tan^{3}x$

Correct answer:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3sinxsin(cosx)}{cos^{4}x}-\frac{sinxcos(cosx)}{cos^{3}x}$

Explanation:

To solve this problem, we use the quotient and the chain rule. First we apply the quotient rule, which looks this:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{cos^{3}x\frac{\mathrm{d} }{\mathrm{d} x}sin(cosx)-sin(cosx)\frac{\mathrm{d} }{\mathrm{d} x}cos^{3}x}{cos^{4}x}$

Next, we apply the chain rule, which looks like this:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{cos^{3}xcos(cosx)(-sinx)-sin(cosx)3cos^{2}x(-sinx)}{cos^{6}x}$

Now we simplify the derivative, which looks like this:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3cos^{2}xsinxsin(cosx)-cos^{3}xsinxcos(cosx)}{cos^{6}x}$

We break this into two fractions, to make simplifying easier, which looks like this:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3cos^{2}xsinxsin(cosx)}{cos^{6}x}-\frac{cos^{3}xsinxcos(cosx)}{cos^{6}x}$

This simplifies further, to give us our answer:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3sinxsin(cosx)}{cos^{4}x}-\frac{sinxcos(cosx)}{cos^{3}x}$

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Example Question #265 : Derivatives

What is the first derivative of the following function?

f(x)=lnxcos(sinx)x

Possible Answers:

$f'(x)=lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

$f'(x)=-lnxcos^{2}xcos(sinx)$

$f'(x)=-lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

$\frac{-cos(sinx)}{x^{2}}$

$\frac{-sin(sinx)cosx)}{x}$

Correct answer:

$f'(x)=-lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

Explanation:

We use the product rule to differentiate this function. Applying it looks like this:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}lnx(cos(sinx))+lnx\frac{\mathrm{d} }{\mathrm{d} x}cos(sinx)$

This simplifies to:

$\frac{cos(sinx)}{x}+\frac{\mathrm{d} }{\mathrm{d} x}cos(sinx)lnx$

We apply the chain rule to differentiate cos(sinx) , which becomes -sin(sinx)cosx . Plugging this into the above equation gives us:

$\frac{cos(sinx)}{x}-lnxsin(sinx)cosx$ or $-lnsin(sinx)cosx+\frac{cos(sinx)}{x}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #256 : Derivative Review

Example Question #257 : Derivative Review

Example Question #258 : Derivative Review

Example Question #259 : Derivative Review

Example Question #1381 : Calculus Ii

Example Question #61 : First And Second Derivatives Of Functions

Example Question #262 : Derivatives

Example Question #263 : Derivatives

Example Question #264 : Derivatives

Example Question #265 : Derivatives

All Calculus 2 Resources

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