Unit 2 · Topic 2.5 Beta

The Power Rule

The power rule says d/dx (x^r) = rx^(r − 1) for every real exponent r .

Practice 1: Implementing Mathematical Processes

Question set for this topic

Part 1 · Hook

Why this matters

Working out every derivative from the limit definition is like walking everywhere: it works, but it is slow. After doing x2x^2, x3x^3 and 1x\frac{1}{x} by hand, a pattern jumps out. Turned into a rule, that pattern gives the derivative of any power of xx in one line.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. 1x3\frac{1}{x^3} written as a power of xx is

  1. x−3x^{-3}
  2. x3x^{3}
  3. −x3-x^{3}
  4. x1/3x^{1/3}
Show the answer

A power in the denominator moves up with its sign changed: 1xn=x−n\frac{1}{x^n} = x^{-n}.

  • Correct: x−3x^{-3}:
  • x3x^{3}:
  • −x3-x^{3}:
  • x1/3x^{1/3}:

2. x23\sqrt[3]{x^2} written as a power of xx is

  1. x2/3x^{2/3}
  2. x3/2x^{3/2}
  3. x6x^{6}
  4. x−2/3x^{-2/3}
Show the answer

The root is the denominator of the exponent: xmn=xm/n\sqrt[n]{x^m} = x^{m/n}.

  • Correct: x2/3x^{2/3}:
  • x3/2x^{3/2}:
  • x6x^{6}:
  • x−2/3x^{-2/3}:

3. By the definition of the derivative, ddx(x2)\frac{d}{dx}\left( x^2 \right) is

  1. 2x2x
  2. xx
  3. 2x22x^2
  4. x2x^2
Show the answer

(x+h)2−x2h\frac{(x + h)^2 - x^2}{h}=2x+h→2x= 2x + h \to 2x.

  • Correct: 2x2x:
  • xx:
  • 2x22x^2:
  • x2x^2:

Part 4 · See it

See it first

Top: the graph of f(x) = x cubed with short tangent segments at x = minus 1.4, minus 0.7, 0, 0.7 and 1.4. Bottom, on the same x scale: the graph of f prime of x = 3x squared, with dots at heights 5.88, 1.47, 0, 1.47 and 5.88, the slopes of those tangent lines. The slope is 0 at x = 0 and the same at x and minus x.
The slopes of f(x)=x3f(x) = x^3 at five points, and the graph of f′(x)=3x2f'(x) = 3x^2 below, through the same five slopes. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. The definition gives x2→2xx^2 \to 2x, x3→3x2x^3 \to 3x^2 and x−1→−x−2x^{-1} \to -x^{-2}in each case the exponent comes down in front and the new exponent is one less
  2. That pattern holds for every real exponent rrthe power rule: ddx(xr)=rxr−1\frac{d}{dx}\left( x^r \right) = r x^{r - 1}
  3. A root or a power in a denominator is still a power of xxrewrite it with a fractional or negative exponent before using the rule
  4. A variable sits in the exponent, as in 2x2^xit is not a power of xx, and the power rule does not apply

Part 6 · Key ideas

Key ideas

  • The power rule: ddx(xr)=rxr−1\frac{d}{dx}\left( x^r \right) = r x^{r - 1} for any real number rr. Bring the exponent down, then subtract 1 from it.
  • Rewrite first: x=x1/2\sqrt{x} = x^{1/2}, 1x4=x−4\frac{1}{x^4} = x^{-4}, 1x3=x−1/3\frac{1}{\sqrt[3]{x}} = x^{-1/3}.
  • With negative exponents, "one less" means more negative: −4−1=−5-4 - 1 = -5.
  • Special cases: ddx(x)=1\frac{d}{dx}(x) = 1, and the slope of x2x^2 at xx is 2x2x.
  • The rule needs the variable in the base and a fixed number in the exponent.

Part 7 · Misconception

A common mistake

The wrong idea: ddx(1x4)=14x3\frac{d}{dx}\left( \frac{1}{x^4} \right) = \frac{1}{4x^3}: differentiate the denominator.

What actually happens: Rewrite first: 1x4=x−4\frac{1}{x^4} = x^{-4}. The power rule gives −4x−5=−4x5-4x^{-5} = -\frac{4}{x^5}. The new exponent is −4−1=−5-4 - 1 = -5, not −3-3, and the sign is negative because the graph falls for x>0x > 0.

Part 8 · Check yourself

Check yourself

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

Graph

The graph of f(x)=x3f(x) = x^3 and its tangent line at x=1x = 1

−2−112−5−4−3−2−112345xyf(1, 1)

1. Use the power rule to find f′(−2)f'(-2) for f(x)=x3f(x) = x^3.

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

Show the answer

f′(x)=3x2f'(x) = 3x^2, so f′(−2)=3(4)=12f'(-2) = 3(4) = 12.

  • Answer: 12

2. The cubic f(x)=x3f(x) = x^3 passes through (1,1)(1, 1). Using the power rule for the slope there, which line touches the cubic at that point?

  1. y=1+3(x−1)y = 1 + 3(x - 1)
  2. y=3+1(x−1)y = 3 + 1(x - 1)
  3. y=1+3x2(x−1)y = 1 + 3x^2(x - 1)
  4. y=1+3(x+1)y = 1 + 3(x + 1)
Show the answer

Point (1,1)(1, 1) and slope f′(1)=3(1)2=3f'(1) = 3(1)^2 = 3. It is y=3x−2y = 3x - 2, crossing the yy-axis at −2-2 as in the graph.

  • Correct: y=1+3(x−1)y = 1 + 3(x - 1): Height 1, slope 3.
  • y=3+1(x−1)y = 3 + 1(x - 1): The height and the slope are swapped.
  • y=1+3x2(x−1)y = 1 + 3x^2(x - 1): The slope of a line is a number: evaluate f′(x)=3x2f'(x) = 3x^2 at x=1x = 1.
  • y=1+3(x+1)y = 1 + 3(x + 1): Point-slope form uses x−1x - 1.

3. At which values of xx is the slope of the graph of f(x)=x3f(x) = x^3 equal to 12?

  1. x=−2x = -2 and x=2x = 2
  2. x=2x = 2
  3. x=−4x = -4 and x=4x = 4
  4. x=123x = \sqrt[3]{12}
Show the answer

Solve f′(x)=3x2=12f'(x) = 3x^2 = 12: x2=4x^2 = 4, so x=±2x = \pm 2. The graph is equally steep at −2-2 and 2.

  • Correct: x=−2x = -2 and x=2x = 2: Both square roots of 4.
  • x=2x = 2: x=−2x = -2 works too: 3(−2)2=123(-2)^2 = 12.
  • x=−4x = -4 and x=4x = 4: From 3x2=123x^2 = 12, x2=4x^2 = 4; take the square root.
  • x=123x = \sqrt[3]{12}: That solves x3=12x^3 = 12, a height of 12, not a slope of 12.

4. How do the slopes of the graph of f(x)=x3f(x) = x^3 at x=−1x = -1 and x=1x = 1 compare?

  1. They are equal: both are 3.
  2. They are opposites: −3-3 and 3.
  3. The slope at x=−1x = -1 is less, because the graph is lower there.
  4. They are equal: both are 1.
Show the answer

f′(x)=3x2f'(x) = 3x^2 gives 3(−1)2=33(-1)^2 = 3 and 3(1)2=33(1)^2 = 3. The heights −1-1 and 1 are opposites; the slopes are not.

  • Correct: They are equal: both are 3.: Squaring removes the sign of xx.
  • They are opposites: −3-3 and 3.: The heights are opposites; f′f' squares xx, so the slopes match.
  • The slope at x=−1x = -1 is less, because the graph is lower there.: Height does not set slope; the graph rises at both points with slope 3.
  • They are equal: both are 1.: That is x2x^2 at ±1\pm 1; the power rule also brings down the 3.

5. A student writes ddx(1x4)=14x3\frac{d}{dx}\left( \frac{1}{x^4} \right) = \frac{1}{4x^3}. What is the correct derivative?

  1. −4x5-\frac{4}{x^5}
  2. −4x3-\frac{4}{x^3}
  3. 4x5\frac{4}{x^5}
  4. −14x5-\frac{1}{4x^5}
Show the answer

Rewrite 1x4=x−4\frac{1}{x^4} = x^{-4}; the power rule gives −4x−5=−4x5-4x^{-5} = -\frac{4}{x^5}. The student differentiated the denominator instead.

  • Correct: −4x5-\frac{4}{x^5}: Coefficient −4-4, exponent −5-5.
  • −4x3-\frac{4}{x^3}: −4−1=−5-4 - 1 = -5, not −3-3.
  • 4x5\frac{4}{x^5}: The coefficient is the exponent, −4-4, so the derivative is negative.
  • −14x5-\frac{1}{4x^5}: The exponent comes down as a factor: −4-4, not −14-\frac{1}{4}.

6. Let f(x)=x3⋅x4x2f(x) = \frac{x^3 \cdot x^4}{x^2} for x≠0x \ne 0. Find f′(x)f'(x).

Type an expression for f′(x)=f'(x) =.

Show the answer

Exponent rules first: x3+4x2=x5\frac{x^{3 + 4}}{x^2} = x^5. Then f′(x)=5x4f'(x) = 5x^4.

  • Answer: f′(x)=f'(x) = 5x45x^{4}

Part 9 · Summary

Summary

The power rule says ddx(xr)=rxr−1\frac{d}{dx}\left( x^r \right) = r x^{r - 1} for every real exponent rr. Rewrite roots and fractions as powers of xx first, subtract 1 carefully with negative and fractional exponents, and remember the rule needs the variable in the base, not in the exponent.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections