AP Calculus AB · Lesson 1

The derivative as a limit

Know Know it

The average rate of change of ff on [a,b][a, b] is f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, the slope of the secant line through the two endpoints. Shrink the interval and the secant slopes approach the slope of the tangent line at aa. That limit is the derivative: f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}, or equivalently f′(a)=lim⁡x→af(x)−f(a)x−af'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}.

f′(a)f'(a) is the instantaneous rate of change of ff at aa, in units of output per unit of input. The tangent line at aa is y=f(a)+f′(a)(x−a)y = f(a) + f'(a)(x - a).

From a table you can only estimate a derivative. Use the two inputs closest to aa, one on each side, and find the slope between them.

Worked example Easy

A drone rises straight up. Its height, in meters, is recorded at a few times, in seconds.

Height of the drone

tt (seconds)0258
H(t)H(t) (meters)0124562

What is the average rate of change of the drone's height from t=2t = 2 to t=5t = 5 seconds?

  1. 6 meters per second
  2. 9 meters per second
  3. 11 meters per secondAnswer
  4. 33 meters per second

How to solve it

  1. Change in height: H(5)−H(2)=45−12=33H(5) - H(2) = 45 - 12 = 33 meters.
  2. Change in time: 5−2=35 - 2 = 3 seconds.
  3. Average rate: 333=11\frac{33}{3} = 11 meters per second.

Why each choice is right or wrong

  • A. Incorrect. 6 meters per second is 122\frac{12}{2}, the average rate from t=0t = 0 to t=2t = 2, not from 2 to 5.
  • B. Incorrect. 9 meters per second is 455\frac{45}{5}, the average rate from t=0t = 0 to t=5t = 5.
  • C. Correct. The height changes by 45−12=3345 - 12 = 33 meters over 5−2=35 - 2 = 3 seconds, and 333=11\frac{33}{3} = 11 meters per second.
  • D. Incorrect. 33 is the change in height in meters; it still has to be divided by the 3 seconds it took.

Worked example Medium

What is the value of lim⁡h→0(2+h)3−8h\lim_{h \to 0}\frac{(2 + h)^3 - 8}{h}?

  1. 00
  2. 66
  3. 88
  4. 1212Answer

How to solve it

  1. Match the pattern f(a+h)−f(a)h\frac{f(a + h) - f(a)}{h}: here f(x)=x3f(x) = x^3, a=2a = 2 and f(2)=8f(2) = 8.
  2. So the limit is f′(2)f'(2). By the power rule f′(x)=3x2f'(x) = 3x^2.
  3. f′(2)=3⋅4=12f'(2) = 3 \cdot 4 = 12. (Expanding (2+h)3=8+12h+6h2+h3(2 + h)^3 = 8 + 12h + 6h^2 + h^3 and dividing by hh gives the same 12.)

Why each choice is right or wrong

  • A. Incorrect. 0 comes from the top alone going to 0; the bottom hh also goes to 0, so the quotient needs work.
  • B. Incorrect. 6 is 3⋅23 \cdot 2: it lowers the power but forgets to square the 2, since the derivative of x3x^3 is 3x23x^2.
  • C. Incorrect. 8 is 232^3, the value of x3x^3 at 2, not the slope there.
  • D. Correct. This is the definition of f′(2)f'(2) for f(x)=x3f(x) = x^3, since f(2)=8f(2) = 8. With f′(x)=3x2f'(x) = 3x^2, the limit is 3⋅22=123 \cdot 2^2 = 12.

Explain Explain it

Task verbs: explain, justify

If ff is differentiable at aa, then ff is continuous at aa. The reverse is false. A function can be continuous at aa and still have no derivative there.

There are four ways a derivative fails to exist. At a corner the slopes from the left and from the right differ. At a cusp the slopes from the two sides blow up in opposite directions. At a vertical tangent the slope grows without bound. At a discontinuity there is no derivative at all.

To decide whether a piecewise function is differentiable at a breakpoint, first check continuity there. Then compare the slope of the left piece with the slope of the right piece at that point. Both must match.

Trap Continuous is not enough

The graph of ∣x−5∣|x - 5| has no break at x=5x = 5, so it is continuous there. But its slope is −1-1 just left of 5 and 11 just right of 5, a corner, so it has no derivative at 5. Check slopes as well as continuity.

Worked example Hard

Let f(x)=x2f(x) = x^2 for x≤1x \le 1 and f(x)=2x−1f(x) = 2x - 1 for x>1x > 1. Which statement about ff at x=1x = 1 is true?

  1. ff is differentiable at x=1x = 1Answer
  2. ff is continuous but not differentiable at x=1x = 1
  3. ff is not continuous at x=1x = 1
  4. ff has a vertical tangent at x=1x = 1

How to solve it

  1. Continuity: x2x^2 gives 1 at x=1x = 1, and 2x−12x - 1 approaches 1, so the pieces meet.
  2. Slopes: the left piece has derivative 2x2x, which is 2 at 1; the right piece has slope 2.
  3. The pieces meet and their slopes match, so ff is differentiable at 1 with f′(1)=2f'(1) = 2.

Why each choice is right or wrong

  • A. Correct. Both pieces equal 1 at x=1x = 1, so ff is continuous there. The left piece has slope 2x=22x = 2 at 1 and the right piece has slope 2, so the slopes match and f′(1)=2f'(1) = 2.
  • B. Incorrect. Continuous but not differentiable needs a corner, but the slopes from both sides are 2, so there is no corner.
  • C. Incorrect. The left piece gives 12=11^2 = 1 and the right piece approaches 2(1)−1=12(1) - 1 = 1, so ff is continuous at 1.
  • D. Incorrect. A vertical tangent needs the slope to grow without bound, but both pieces have slope 2 near x=1x = 1.

Justify Your turn

Let h(x)=3xh(x) = 3x for x<1x < 1 and h(x)=x+2h(x) = x + 2 for x≥1x \ge 1. Show that hh is continuous at x=1x = 1, then explain why hh is not differentiable at x=1x = 1.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Shows continuity at 1: the limits from both sides are 3, and h(1)=3h(1) = 3
  • Says the slopes from the two sides, 3 and 1, are different

Model answer

From the left, 3x3x approaches 3(1)=33(1) = 3. From the right, x+2x + 2 approaches 1+2=31 + 2 = 3, and h(1)=1+2=3h(1) = 1 + 2 = 3. So lim⁡x→1h(x)=3=h(1)\lim_{x \to 1} h(x) = 3 = h(1), and hh is continuous at 1. But the left piece has slope 3 and the right piece has slope 1. The slopes from the two sides differ, so the graph has a corner at x=1x = 1 and h′(1)h'(1) does not exist.

Apply Apply it

Practice these topics. AP Practice questions on exactly what this lesson covers, with an explanation for every answer.

Practice these topics