AP Calculus AB · Lesson 2

Power, product and quotient rules

Know Know it

Power rule: ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1} for any real nn. Rewrite first: x=x12\sqrt{x} = x^{\frac{1}{2}} and 1x2=x−2\frac{1}{x^2} = x^{-2}.

Constant, constant multiple and sum rules: a constant has derivative 0, a constant factor stays, and a sum is differentiated term by term.

Two more to know by heart: ddxex=ex\frac{d}{dx}e^x = e^x, and ddxln⁡x=1x\frac{d}{dx}\ln x = \frac{1}{x} for x>0x > 0.

Worked example Easy

If f(x)=4x3−5x+7f(x) = 4x^3 - 5x + 7, what is f′(2)f'(2)?

  1. 2929
  2. 4343Answer
  3. 4848
  4. 5050

How to solve it

  1. Differentiate term by term: 4x3→12x24x^3 \to 12x^2, −5x→−5-5x \to -5, 7→07 \to 0.
  2. So f′(x)=12x2−5f'(x) = 12x^2 - 5.
  3. Then f′(2)=48−5=43f'(2) = 48 - 5 = 43.

Why each choice is right or wrong

  • A. Incorrect. 29 is f(2)=32−10+7f(2) = 32 - 10 + 7, the value of ff, not the value of its derivative.
  • B. Correct. f′(x)=12x2−5f'(x) = 12x^2 - 5, so f′(2)=12⋅4−5=43f'(2) = 12 \cdot 4 - 5 = 43.
  • C. Incorrect. 48 is 12x212x^2 at 2 alone; it drops the −5-5 that comes from differentiating −5x-5x.
  • D. Incorrect. 50 keeps the constant 7, but the derivative of a constant is 0, so f′(2)=48−5=43f'(2) = 48 - 5 = 43.

Worked example Medium

If y=6x−2x2y = 6\sqrt{x} - \frac{2}{x^2}, what is dydx\frac{dy}{dx} at x=1x = 1?

  1. −1-1
  2. 44
  3. 77Answer
  4. 1010

How to solve it

  1. Rewrite as powers: y=6x12−2x−2y = 6x^{\frac{1}{2}} - 2x^{-2}.
  2. Power rule on each term: 6⋅12x−12=3x−126 \cdot \frac{1}{2}x^{-\frac{1}{2}} = 3x^{-\frac{1}{2}} and −2⋅(−2)x−3=4x−3-2 \cdot (-2)x^{-3} = 4x^{-3}.
  3. At x=1x = 1 every power of xx is 1, so the slope is 3+4=73 + 4 = 7.

Why each choice is right or wrong

  • A. Incorrect. −1-1 comes from 3−43 - 4: the derivative of −2x−2-2x^{-2} is +4x−3+4x^{-3}, since two minus signs make a plus.
  • B. Incorrect. 4 is the value of yy at x=1x = 1, 6−26 - 2, not the slope there.
  • C. Correct. Write y=6x12−2x−2y = 6x^{\frac{1}{2}} - 2x^{-2}. Then y′=3x−12+4x−3y' = 3x^{-\frac{1}{2}} + 4x^{-3}, which is 3+4=73 + 4 = 7 at x=1x = 1.
  • D. Incorrect. 10 differentiates 6x6\sqrt{x} as 6x−126x^{-\frac{1}{2}}, forgetting to multiply by the exponent 12\frac{1}{2}.

Explain Explain it

Task verbs: calculate, explain

Product rule: (fg)′=f′g+fg′(fg)' = f'g + fg'. Quotient rule: (fg)′=f′g−fg′g2\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}. Neither is "differentiate each piece": the derivative of a product is not the product of the derivatives.

With a table of values, plug the numbers straight into the rule. Find ff, f′f', gg and g′g' at the same input, and keep the order on top of the quotient rule.

Trap A product is not differentiated piece by piece

For x2⋅x3=x5x^2 \cdot x^3 = x^5, the derivative is 5x45x^4. Multiplying the separate derivatives gives 2x⋅3x2=6x32x \cdot 3x^2 = 6x^3, which is wrong. The product rule gives 2x⋅x3+x2⋅3x2=5x42x \cdot x^3 + x^2 \cdot 3x^2 = 5x^4, as it should.

Worked example Hard

The functions ff and gg are differentiable. Selected values are shown in the table.

Values at x=2x = 2

xxf(x)f(x)f′(x)f'(x)g(x)g(x)g′(x)g'(x)
23-145

Let h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}. What is h′(2)h'(2)?

  1. −1916-\frac{19}{16}Answer
  2. −15-\frac{1}{5}
  3. 1116\frac{11}{16}
  4. 1916\frac{19}{16}

How to solve it

  1. Read the values at 2: f=3f = 3, f′=−1f' = -1, g=4g = 4, g′=5g' = 5.
  2. Top: f′g−fg′=(−1)(4)−(3)(5)=−19f'g - fg' = (-1)(4) - (3)(5) = -19. Bottom: g2=16g^2 = 16.
  3. So h′(2)=−1916h'(2) = -\frac{19}{16}.

Why each choice is right or wrong

  • A. Correct. h′(2)=f′(2)g(2)−f(2)g′(2)g(2)2=(−1)(4)−(3)(5)16=−4−1516=−1916h'(2) = \frac{f'(2)g(2) - f(2)g'(2)}{g(2)^2} = \frac{(-1)(4) - (3)(5)}{16} = \frac{-4 - 15}{16} = -\frac{19}{16}.
  • B. Incorrect. −15-\frac{1}{5} is f′(2)g′(2)\frac{f'(2)}{g'(2)}, a quotient of the derivatives, which is not the quotient rule.
  • C. Incorrect. 1116\frac{11}{16} adds the two products on top, −4+15-4 + 15, but the quotient rule subtracts them.
  • D. Incorrect. 1916\frac{19}{16} reverses the order on top, fg′−f′gfg' - f'g, which flips the sign of the answer.

Calculate Your turn

A student says the derivative of x3exx^3 e^x is 3x2ex3x^2 e^x. Find the correct derivative, and name the term that the student's answer is missing.

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Compare with a model answer

A full answer does this

  • Gives the correct derivative, 3x2ex+x3ex3x^2e^x + x^3e^x
  • Names the missing term, x3exx^3e^x

Model answer

By the product rule, ddx(x3ex)=3x2⋅ex+x3⋅ex=x2ex(3+x)\frac{d}{dx}(x^3e^x) = 3x^2 \cdot e^x + x^3 \cdot e^x = x^2e^x(3 + x). The student's answer has only the first term. It is missing x3exx^3e^x, the term that comes from keeping x3x^3 and differentiating exe^x.

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