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Calculus AB : Calculus AB

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45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #7 : Apply The Product Rule And Quotient Rule

Find the derivative of the function $f(x) = \frac{2x^3 - 9x}{x^4}$ .

Possible Answers:

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{28x^3 + 6x^2 - 9}{x^4}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{7x^3 + x^2 - 3}{x^4}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{28x^3 + 6x^2 - 9}{x^8}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{x^3 + x^2}{x^4}$

Correct answer:

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{28x^3 + 6x^2 - 9}{x^4}$

Explanation:

We will use the quotient rule to find the derivative. Let g(x) = 2x^3 - 9x and h(x) = x^4 .

$\frac{d}{dx}[\frac{g(x)}{h(x)}] = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{(6x^2 - 9)x^4 - (2x^3 - 9x)4x^3}{(x^4)^2}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{6x^6 - 9x^4 - 8x^7 + 36x^7}{x^8}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] =\frac{28x^7 + 6x^6 - 9x^4}{x^8}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{x^4(28x^3 + 6x^2 - 9)}{x^8}$

$\frac{d}{dx}[\frac{2x^3 - 9x}{x^4}] = \frac{28x^3 + 6x^2 - 9}{x^4}$

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Example Question #8 : Apply The Product Rule And Quotient Rule

True or False: You may have to use the quotient and power rules for the same function.

Possible Answers:

True

False

Correct answer:

True

Explanation:

Take the function $f(x) = \frac{xe^x}{3 - x}$ for example. We would have to use the product rule at first for the numerator and once that is simplified we would need to use the quotient rule to find the derivative of the entire function.

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Example Question #1 : Apply The Product Rule And Quotient Rule

True or False: For a function in the form of f(x) = g(x)h(x) , if a derivative does not exist for one of g(x) or h(x) but exists for the other, then you are still able to use the product rule to find the derivative of the entire function.

Possible Answers:

False

True

Correct answer:

False

Explanation:

This is not true. In order to use the product rule you need to be able to derive each of the functions and then sum them. So if the derivative does not exist for one of these functions then you would not be able to use the product rule.

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Example Question #191 : Calculus Ab

Which of the following is the derivative of f(x) = tan(x) ?

Possible Answers:

cot^2(x)

sec^2(x)

csc^2(x)

cos^2(x)

Correct answer:

sec^2(x)

Explanation:

Recall that the function $tan(x) = \frac{sin(x)}{cos(x)}$ . Knowing this, we can use the quotient rule to find the derivative.

f(x) = tan(x)

$f(x) = \frac{sin(x)}{cos(x)}$

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

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

$f'(x) = \frac{1}{cos^2(x)}$ (Pythagorean Identity)

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

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Example Question #192 : Calculus Ab

Which of the following is the derivative of f(x) = sec(x) ?

Possible Answers:

tan^2(x)

cos(x)sin(x)

csc^2(x)

tan(x)sec(x)

Correct answer:

tan(x)sec(x)

Explanation:

Recall that $sec(x) = \frac{1}{cos(x)}$

f(x) = sec(x)

$f(x) = \frac{1}{cos(x)}$

$f(x) = cos(x)^{-1}$

$f'(x) = -cos(x)^{-2} * \frac{d}{dx}cos(x)$

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

$f'(x) = \frac{sin(x)}{cos(x)}*\frac{1}{cos(x)}$

f'(x) = tan(x)sec(x) $(tan(x) = \frac{sin(x)}{cos(x)}, sec(x) = \frac{1}{cos(x)})$

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Example Question #62 : Basic Differentiation Rules

Which of the following is the derivative of f(x) = csc(x) ?

Possible Answers:

tancsc(x)

sec^2(x)

cot^2(x)

-cot(x)csc(x)

Correct answer:

-cot(x)csc(x)

Explanation:

Recall that $csc(x) = \frac{1}{sin(x)}$

f(x) = csc(x)

$f(x) = \frac{1}{sin(x)}$

$f(x) = sin(x)^{-1}$

$f'(x) = -sin(x)^{-2}*\frac{d}{dx}sin(x)$

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

$f'(x) = -\frac{cos(x)}{sin(x)}*\frac{1}{sin(x)}$

f'(x) = -cot(x)csc(x) $(\csc(x) = \frac{1}{sin(x)}, cot(x) = \frac{cos(x)}{sin(x)})$

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Example Question #191 : Calculus Ab

What is the derivative for f(x) = cot(x) ?

Possible Answers:

csc^2(x)

sec^2(x)

cot^2(x)

tan^2(x)

Correct answer:

csc^2(x)

Explanation:

Recall that $cot(x) = \frac{1}{tan(x)} = \frac{cos(x)}{sin(x)}$ . We are able to use the quotient rule to find the derivative now.

f(x) = cot(x)

$f(x) = \frac{cos(x)}{sin(x)}$

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

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

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

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

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

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Example Question #193 : Calculus Ab

True or False: We use the Chain Rule to find the derivatives of trigonometric functions.

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is true. The chain rule is what allows us to differentiate composite functions. Composite functions are functions that are within one another. Take tan(2x) for example. tan(x) in itself is a function, but so is 2(x) . We can let f(x) = tan(x) and g(x) = 2x . So the composite function would be, f(g(x)) = tan(2x) .

Now for the Chain Rule we multiply the derivative of the inside function by the coefficient of the outside function while differentiating the outside function as well. F'(x) will be the composite function f(g(x)) . The formula for the Chain Rule is:

F'(x) = g'(x)f'(g(x)) .

Going back to our example of tan(2x) , we know that the derivative of tan(x) is sec^2(x) and the derivative of is . So the derivative of our composite function is 2sec^2(2x) . Thus, every time we are differentiating trigonometric functions we are utilizing the Chain Rule whether we realize it or not.

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Example Question #2 : Use Derivatives Of Trig Functions

What is the derivative of f(x) = tan(x^2 + 3x) ?

Possible Answers:

f'(x) = sec^2(x^2 + 3x)

f'(x) = cot^2(2x + 3)

f'(x) = sec^2(2x + 3)

f'(x) = (2x + 3)sec^2(x^2 + 3x)

Correct answer:

f'(x) = (2x + 3)sec^2(x^2 + 3x)

Explanation:

We must take the derivative of both the quantity of the tangent function and tangent itself to solve for this derivative. Recall that the derivative of tan(x) = sec^2(x)

f(x) = tan(x^2 + 3x)

$f'(x) = \frac{d}{dx}[x^2+3x]*\frac{d}{dx}[tan(x^2 + 3x)]$

f'(x) = (2x + 3)sec^2(x^2 + 3x)

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Example Question #192 : Calculus Ab

What is the derivative of f(x) = 2x + csc(4x) at $x = \pi$ ?

Possible Answers:

7.25

This derivative is undefined at this point

$\pi$

Correct answer:

This derivative is undefined at this point

Explanation:

We must start this problem by finding the derivative. Recall that the derivative of the cosecant function is csc(x) = -cot(x)csc(x)

f(x) = 2x + csc(4x)

$f'(x) = 2 + 4*\frac{d}{dx}[csc(4x)]$

f'(x) = 2 - 4cot(4x)csc(4x)

Now we must find the derivative of $x = \pi$ . So we will plug $\pi$ in for .

$f'(\pi) = 2 - 4cot(4\pi)csc(4\pi)$

$f'(\pi) = 2 - 4(0)(undefined)$

$f'(\pi) = undefined$

And so at the point of the graph of the derivative, the function is undefined.

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #7 : Apply The Product Rule And Quotient Rule

Example Question #8 : Apply The Product Rule And Quotient Rule

Example Question #1 : Apply The Product Rule And Quotient Rule

Example Question #191 : Calculus Ab

Example Question #192 : Calculus Ab

Example Question #62 : Basic Differentiation Rules

Example Question #191 : Calculus Ab

Example Question #193 : Calculus Ab

Example Question #2 : Use Derivatives Of Trig Functions

Example Question #192 : Calculus Ab

All Calculus AB Resources

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