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Calculus AB : Basic Differentiation Rules

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Example Questions

Example Question #2 : Apply The Product Rule And Quotient Rule

If we want to find the derivative of the function f(x) = (x^2 - 2)(4x+3) , which of the following would we use?

Possible Answers:

Chain Rule

None of the above

Quotient Rule

Product Rule

Correct answer:

Product Rule

Explanation:

We would use the product rule in this case. Let g(x) = x^2 - 2 and h(x) = 4x + 3 .

We will use the formula $\frac{d}{dx}[g(x)h(x) = g'(x)h(x) + g(x)h'(x)$

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

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

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

$\frac{d}{dx}[(x^2 -2)(4x+3)] = 12x^2 + 6x - 8$

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Example Question #3 : Apply The Product Rule And Quotient Rule

Find the derivative (do not solve) of the following function, f(x) = 4xsin(x) .

Possible Answers:

$\frac{d}{dx}[4xsin(x)] = 4[xcos(x) + sin(x)]$

$\frac{d}{dx}[4xsin(x)] = 4[xsin(x) + sin(x)]$

$\frac{d}{dx}[4xsin(x)] = 4[xcos(x) + cos(x)]$

$\frac{d}{dx}[4xsin(x)] = 4[xsin(x) + cos(x)]$

Correct answer:

$\frac{d}{dx}[4xsin(x)] = 4[xcos(x) + sin(x)]$

Explanation:

We will let g(x) = 4x and h(x) = sin(x) .

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

$\frac{d}{dx}[4xsin(x)] = 4sin(x) + 4xcos(x)$

$\frac{d}{dx}[4xsin(x)] = 4[xcos(x) + sin(x)]$

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Example Question #4 : Apply The Product Rule And Quotient Rule

Find the derivative (do not solve) of the following function, f(x) =(5x^2 + 8x)(7x - 2) .

Possible Answers:

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 105x^2 + 112x - 36$

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 105x^2 + 112x + 36$

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 56x^2 + 34x - 16$

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 112x^2 + 105x - 36$

Correct answer:

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 105x^2 + 112x - 36$

Explanation:

We will let g(x) = 5x^2 + 8x and h(x) = 7x - 2 .

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

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = (10x+8)(7x-2) + (5x^2+8x)(7)$

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 70x^2 - 20 + 56x - 16 + 35x^2 + 56x$

$\frac{d}{dx}[(5x^2+8x)(7x-2)] = 105x^2 + 112x - 36$

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Example Question #41 : Basic Differentiation Rules

Which of the following would we use to find the derivative of the function $f(x) = \frac{x^2}{x-2}$

Possible Answers:

Quotient Rule

Product Rule

None of the above

Chain Rule

Correct answer:

Quotient Rule

Explanation:

When trying to find the derivative of the quotient of two functions, we use what is called the quotient rule. The quotient rule is similar to the product rule except you are taking the difference of two products and then dividing them further.

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Example Question #161 : Calculus Ab

Which of the following is the correct formula for the quotient rule?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

If two functions are differentiable, then so is their quotient. So we use the quotient rule to find the derivative of the entire function.

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Example Question #6 : Apply The Product Rule And Quotient Rule

Find the derivative of the function $f(x) = \frac{e^x}{x^2}$

Possible Answers:

$\frac{d}{dx}[\frac{e^x}{x^2}] = \frac{x^3 - 2x}{e^2x}$

$\frac{d}{dx}[\frac{e^x}{x^2}] = \frac{xe^x - 2e^x}{x}$

$\frac{d}{dx}[\frac{e^x}{x^2}] = \frac{x^3 - 2x}{e^x}$

$\frac{d}{dx}[\frac{e^x}{x^2}] = \frac{xe^x - 2e^x}{x^3}$

Correct answer:

$\frac{d}{dx}[\frac{e^x}{x^2}] = \frac{xe^x - 2e^x}{x^3}$

Explanation:

We will need to use the quotient rule here. Let g(x) = e^x and h(x) = x^2 .

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

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

$\frac{d}{dx} [\frac{e^x}{x^2}] = \frac{x(xe^x - 2e^x)}{x^4}$

$\frac{d}{dx}[\frac{e^x}{x^2}] = \frac{xe^x - 2e^x}{x^3}$

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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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Calculus AB : Basic Differentiation Rules

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #2 : Apply The Product Rule And Quotient Rule

Example Question #3 : Apply The Product Rule And Quotient Rule

Example Question #4 : Apply The Product Rule And Quotient Rule

Example Question #41 : Basic Differentiation Rules

Example Question #161 : Calculus Ab

Example Question #6 : Apply The Product Rule And Quotient Rule

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

All Calculus AB Resources

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