Unit 3 · Topic 3.6 Beta

Second and Higher Derivatives

The second derivative f″ is the derivative of f′ : it tells how fast the rate of change is changing.

Practice 1: Implementing Mathematical ProcessesPractice 4: Communication and Notation

Question set for this topic

Part 1 · Hook

Why this matters

A car's speedometer shows how fast the position changes. But you also feel how fast the speed itself changes when you are pressed back into your seat. That is a derivative of a derivative. Second derivatives describe how a rate changes, and they show up all over Units 4 and 5.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the derivative of 3x2−5x3x^2 - 5x?

  1. 6x−56x - 5
  2. 6x6x
  3. 3x−53x - 5
  4. x3−52x2x^3 - \frac{5}{2}x^2
Show the answer

Power rule on each term.

  • Correct: 6x−56x - 5:
  • 6x6x:
  • 3x−53x - 5:
  • x3−52x2x^3 - \frac{5}{2}x^2:

2. For x2+y2=25x^2 + y^2 = 25, dydx\frac{dy}{dx} is

  1. −xy-\frac{x}{y}
  2. xy\frac{x}{y}
  3. −yx-\frac{y}{x}
  4. −x-x
Show the answer

2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.

  • Correct: −xy-\frac{x}{y}:
  • xy\frac{x}{y}:
  • −yx-\frac{y}{x}:
  • −x-x:

Part 4 · See it

See it first

Three stacked graphs on the same x-axis: f(x) = x³ − 3x (crimson) with level points at x = −1 and 1; f′(x) = 3x² − 3 (blue), zero at x = ±1; and f″(x) = 6x (green), the slope of f′, negative left of 0 and positive right of 0.
f(x)=x3−3xf(x) = x^3 - 3x, its derivative f′(x)=3x2−3f'(x) = 3x^2 - 3 and its second derivative f″(x)=6xf''(x) = 6x. Each graph shows the slopes of the one above it. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. f′f' is itself a functionit can be differentiated: f″=(f′)′f'' = (f')'
  2. f″f'' is the slope of the graph of f′f'f″>0f'' > 0 where f′f' rises, f″<0f'' < 0 where f′f' falls
  3. Keep differentiatingf‴f''', f(4)f^{(4)}, ..., f(n)f^{(n)}; for a polynomial of degree nn, f(n+1)=0f^{(n+1)} = 0
  4. For an implicit curve, differentiate dydx\frac{dy}{dx} againthen substitute the expression for dydx\frac{dy}{dx} wherever it appears

Part 6 · Key ideas

Key ideas

  • Notation: f″(x)f''(x), y″y'', d2ydx2\frac{d^2y}{dx^2}; higher ones are f‴(x)f'''(x) and f(n)(x)f^{(n)}(x).
  • Read f″f'' from a graph of f′f' as its slope, not its height.
  • If f′f' is a product or composition, differentiating again needs the product or chain rule again.
  • Implicit second derivative: differentiate dydx\frac{dy}{dx} (quotient rule, yy still depends on xx), then substitute dydx\frac{dy}{dx}.

Part 7 · Misconception

A common mistake

The wrong idea: From a graph of f′f', f″(a)f''(a) is the height of the graph at x=ax = a.

What actually happens: The height of the graph of f′f' is f′(a)f'(a). The second derivative is the slope of that graph at x=ax = a.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of f′f'

12345−3−2−112345xyf'

1. The graph shows f′f', the derivative of ff. Find f″(1)f''(1).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f″f'' is the slope of the graph of f′f'. On 0<x<20 < x < 2 the graph of f′f' is a segment of slope −2−22−0=−2\frac{-2 - 2}{2 - 0} = -2, so f″(1)=−2f''(1) = -2.

  • Answer: -2

2. On which interval is f″(x)<0f''(x) < 0?

  1. 0<x<20 < x < 2
  2. 1<x<31 < x < 3
  3. 0<x<10 < x < 1
  4. 2<x<52 < x < 5
Show the answer

f″<0f'' < 0 where the graph of f′f' is falling: on 0<x<20 < x < 2, where its slope is −2-2.

  • Correct: 0<x<20 < x < 2: The graph of f′f' falls there, so its slope f″f'' is negative.
  • 1<x<31 < x < 3: There f′f' itself is negative; the sign of f″f'' comes from the slope of f′f'.
  • 0<x<10 < x < 1: There f′f' is positive, but f″f'' is negative on all of 0<x<20 < x < 2.
  • 2<x<52 < x < 5: The graph of f′f' rises there, so f″=2>0f'' = 2 > 0.

3. What can be said about f″(2)f''(2)?

  1. It equals −2-2, the value of f′(2)f'(2)
  2. It does not exist, because the graph of f′f' has a corner at x=2x = 2
  3. It equals 0, because the graph of f′f' turns around
  4. It equals 2, the slope just to the right of x=2x = 2
Show the answer

f″(2)f''(2) is the derivative of f′f' at 2. The one-sided slopes, −2-2 and 22, differ, so f′f' is not differentiable at 2.

  • It equals −2-2, the value of f′(2)f'(2): That is the height of the graph of f′f', not its slope.
  • Correct: It does not exist, because the graph of f′f' has a corner at x=2x = 2: The slope of f′f' is −2-2 just left of 2 and 22 just right of it.
  • It equals 0, because the graph of f′f' turns around: A sharp corner has no single slope, so there is no derivative of f′f' there.
  • It equals 2, the slope just to the right of x=2x = 2: A derivative needs the slopes from both sides to agree; they are −2-2 and 22.

4. Let h(x)=xf′(x)h(x) = x f'(x). Find h′(4)h'(4).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Product rule: h′(x)=f′(x)+xf″(x)h'(x) = f'(x) + x f''(x). From the graph, f′(4)=2f'(4) = 2 and f″(4)=2f''(4) = 2 (the slope of the second segment). So h′(4)=2+4(2)=10h'(4) = 2 + 4(2) = 10.

  • Answer: 10

5. Let f(x)=sin⁡(3x)f(x) = \sin(3x). Find f″(x)f''(x).

Type an expression for f″(x)=f''(x) =.

Show the answer

f′(x)=3cos⁡(3x)f'(x) = 3\cos(3x) and f″(x)=−9sin⁡(3x)f''(x) = -9\sin(3x): each derivative brings a factor 3 from the chain rule.

  • Answer: f″(x)=f''(x) = −9sin⁡(3x)-9\,\sin(3x)

6. Let f(x)=e−2xf(x) = e^{-2x}. Find f(5)(0)f^{(5)}(0), the fifth derivative at 0.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Each derivative multiplies by −2-2: f(n)(x)f^{(n)}(x)=(−2)ne−2x= (-2)^n e^{-2x}. So f(5)(0)=(−2)5=−32f^{(5)}(0) = (-2)^5 = -32.

  • Answer: -32

Part 9 · Summary

Summary

The second derivative f″f'' is the derivative of f′f': it tells how fast the rate of change is changing. Find it by differentiating again, with the product and chain rules as needed; from a graph of f′f', read it as a slope. For implicit curves, substitute dydx\frac{dy}{dx} after the second differentiation.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections