Wait, That’s It?

The Power Rule: Why the Power Comes Down

Multiply out the rise, (x + h)ⁿ − xⁿ, and every term but one carries an extra h; the limit keeps only n·xⁿ⁻¹.

Chapters

  1. Why is the slope of x³ three x²?
  2. A slope at one point: the derivative
  3. The question, and a cube that grows: seven pieces
  4. Rise over run: one h comes off every piece
  5. Check it at x = 2
  6. x⁴: count, don't draw
  7. Any power n: n pieces with one h
  8. Divide by h, then the limit
  9. The power rule, and d/dx
  10. x¹⁰⁰: 101 terms, and the slope is the second
  11. Past whole numbers: the square root
  12. The one sentence, and what's next

Calculus 1.2: The power rule, the companion

Wait, That's It? · the notebook for "The Power Rule: Why the Power Comes Down | Calculus 1.2"

The episode in one sentence: multiply out the rise, (x+h)n−xn(x + h)^n - x^n, and every term but one carries an extra hh; the limit keeps only n xn−1n\,x^{n-1}.

Where this sits: Calculus 1.2, the power rule (Stewart, Calculus: Early Transcendentals, §3.1). It uses one thing from Calculus 1.1: the slope at a point is where rise over run heads as the gap hh shrinks to zero.

How to use it: run the first code cell (Shift + Enter, or Runtime → Run all), then drag the sliders. In order:

  1. Play: grow the cube and watch the three shares of the slope; check it at 2; count the picks for any nn; then the square root.
  2. FAQ: the questions students ask about this section.
  3. Practice: five textbook-style problems, easiest first, answers hidden until you click.
  4. One open question to think about.
  5. Optional side trips: Pascal's triangle, a real one-liter cube, and where the other powers get proved.

1. Play

Grow the cube

A cube of side xx has volume x3x^3. Step the gap: every side grows to x+hx + h. The new cube is the old one plus seven pieces: three slabs (x⋅x⋅hx \cdot x \cdot h, yellow), three rods (x⋅h⋅hx \cdot h \cdot h, orange) and one corner (h⋅h⋅hh \cdot h \cdot h, chalk).

The rise is those seven pieces: 3x2h+3xh2+h33x^2h + 3xh^2 + h^3. Divide by the run, hh, and one hh comes off every piece:

slope=3x2+3xh+h2.\text{slope} = 3x^2 + 3xh + h^2 .

Drag the gap toward 0. The slabs' share never moves; the other two shrink with hh.

A cube of side 2 and, drawn a little apart from it, the seven pieces it gains when every side grows by 0.50: three slabs, three rods and one corner.

3 slabs (yellow), 3 rods (orange), 1 corner, around the old cube

Each kind's share of the slope: slabs 12.00, rods 3.00, corner 0.2500.slabs3x²rods3xhcornerh²510151230.25
slope
15.2500
0.50
2

Check it at 2

At x=2x = 2 the rule says 3⋅22=123 \cdot 2^2 = 12. Here is rise over run on y=x3y = x^3 for three gaps, with each kind's share. The slabs give the 12 every time: it is the term that had exactly one hh before the division.

gap hslabs 3x²rods 3xhcorner h²rise over run
0.1120.60.0112.61
0.01120.060.000112.0601
0.001120.0060.00000112.006001

Predict, then reveal

Now y=x4y = x^4 at the same point, x=2x = 2. Nobody can draw the four-dimensional cube, but you can still count. Before you run the next cell, write down the number you think rise over run heads for. Then run it.

Show the answer
gap hrise over run on y = x⁴ at 2
0.134.481
0.0132.240801
0.00132.024008001

It heads for 32=4⋅2332 = 4 \cdot 2^3. (x+h)4(x + h)^4 is four factors. A piece with exactly one hh takes its hh from factor 1, 2, 3 or 4 and xx from the other three: four pieces, each x3hx^3h, together 4x3h4x^3h. Divide by hh: 4x34x^3. Everything else still carries an hh and shrinks away:

(x+h)4−x4h=4x3+6x2h+4xh2+h3.\frac{(x+h)^4 - x^4}{h} = 4x^3 + 6x^2h + 4xh^2 + h^3 .

At x=2x = 2 that is 32+24h+8h2+h332 + 24h + 8h^2 + h^3: 34.481, 32.240801, 32.024008001.

Count the picks, for any n

(x+h)n(x + h)^n is nn factors. Each piece of the product takes one letter from every factor: xx or hh. The cell below lists every way of picking and sorts the pieces by how many hh's they took.

  • No hh: one piece, xnx^n (the old one; the rise takes it away).
  • Exactly one hh: nn pieces, one for each factor the hh can come from. Together: n xn−1hn\,x^{n-1}h.
  • Two or more hh's: every other piece. After dividing by hh they still carry an hh.

Drag nn. The yellow bar is always nn.

The 8 pieces of (x + h) to the power 3, counted by how many h's each picked: 1, 3, 3, 1.0123123pieceshow many h's the piece picked1331
pieces of (x + h)3
8
with exactly one h
3

the pieces with one h: x·x·h, x·h·x, h·x·x

3

Rise over run against the rule

The same triangle as Calculus 1.1, on y=xny = x^n at x=1.2x = 1.2. Pick a power, then drag the gap down. Rise over run (the yellow line's slope) heads for n xn−1n\,x^{n-1}.

The curve y = x to the power 3, with the line through its points at x = 1.2 and x = 1.700. The run is the gap, the rise is the second height minus the first.0.51.01.524
n·xn−1 at x = 1.2
4.3200
rise over run
6.3700
3
0.500

Past whole numbers: the square root

x=x1/2\sqrt{x} = x^{1/2}. There is no "half a factor", so the counting argument says nothing here. The rule still works (its proof comes later in the course): bring the 12\tfrac12 down and subtract one from the power,

ddx x1/2=12 x−1/2=12x,at x=4:12⋅2=14.\frac{d}{dx}\,x^{1/2} = \tfrac12\,x^{-1/2} = \frac{1}{2\sqrt{x}} , \qquad \text{at } x = 4:\quad \frac{1}{2 \cdot 2} = \frac14 .

Check it the honest way: rise over run from (4,2)(4, 2), with the gap shrinking.

The curve y = square root of x, with the line through its points at x = 4 and x = 7.00. The run is the gap, the rise is the second height minus the first.2468123
the rule
0.25
rise over run
0.2153
3.00
gaprise over run
0.10.2485
0.010.2498
0.00010.2500

2. FAQ: where students get stuck

  1. Why does multiplying brackets out mean "pick one letter from each"?
  2. Where did xnx^n go before we divided by hh?
  3. Why does only the term with one hh survive? And why can't hh just be zero?
  4. Is a "piece" the same as a "term"?
  5. "The slope is the second." The second what?
  6. Why is 4−1/24^{-1/2} one half? And why does the rule work for x\sqrt{x} with no "half a factor"?
  7. What is ddx\frac{d}{dx}, and how is it different from dydx\frac{dy}{dx}?
  8. Do I have to multiply out (x+h)n(x + h)^n every time? What about 5x35x^3 or x2+xx^2 + x?

1. Why does multiplying brackets out mean "pick one letter from each"?

That is all multiplying out ever is. Take two brackets: (x+h)(x+h)=x⋅x+x⋅h+h⋅x+h⋅h(x + h)(x + h) = x\cdot x + x\cdot h + h\cdot x + h\cdot h. Each of the four pieces took one letter from the first bracket and one from the second, and every way of choosing shows up once. The hh was picked once in two of them (x⋅hx\cdot h and h⋅xh\cdot x): that is Calculus 1.1's 2xh2xh. With three brackets there are 2⋅2⋅2=82 \cdot 2 \cdot 2 = 8 pieces, and on the cube each bracket is a direction: pick xx and the piece is thick that way, pick hh and it is thin. With nn brackets nothing changes but the count.

2. Where did xnx^n go before we divided by hh?

The rise is the new height minus the old one: (x+h)n−xn(x + h)^n - x^n. The piece with no hh at all is xnx^n, so the subtraction removes it. (On the cube: the new cube minus the old cube leaves the seven pieces.) Only then do we divide by hh. If xnx^n were still there, xn/hx^n / h would blow up as hh shrinks.

3. Why does only the term with one hh survive? And why can't hh just be zero?

Dividing by hh takes one hh off every term. The term that had exactly one is left with none: n xn−1n\,x^{n-1}. A term that had two or more still carries an hh, so it shrinks to zero when hh does. That is what the one sentence means by an "extra" hh: extra beyond the one the division removes. And hh can't be zero while you divide: rise over run would be 00\frac{0}{0}. You divide while hh is still a gap, then let it shrink (the limit of Calculus 1.1).

4. Is a "piece" the same as a "term"?

A piece is one product of picks, like x⋅h⋅xx\cdot h\cdot x: one chunk of the cube. A term is the like pieces added up: three slabs make the term 3x2h3x^2h. The number in front of a term counts its pieces.

5. "The slope is the second." The second what?

The second term. (x+h)100(x + h)^{100} multiplied out has 101 terms, from x100x^{100} to h100h^{100}. The first is taken away by the subtraction; the second is 100 x99h100\,x^{99}h, the one with exactly one hh; divide by hh and it is 100 x99100\,x^{99}, the slope. The other 99 terms still carry an hh and vanish in the limit.

6. Why is 4−1/24^{-1/2} one half? And why does the rule work for x\sqrt{x} with no "half a factor"?

Two school rules for exponents: a power of 12\tfrac12 is a square root, and a minus sign in the exponent means "one over". So 4−1/2=141/2=14=124^{-1/2} = \dfrac{1}{4^{1/2}} = \dfrac{1}{\sqrt 4} = \dfrac12. For x=x1/2\sqrt{x} = x^{1/2} the rule says: bring 12\tfrac12 down, and 12−1=−12\tfrac12 - 1 = -\tfrac12, so the slope is 12 x−1/2=12x\tfrac12\,x^{-1/2} = \dfrac{1}{2\sqrt x}; at 4 that is 12⋅12=14\tfrac12 \cdot \tfrac12 = \tfrac14. The counting proof in the film only covers n=1,2,3,…n = 1, 2, 3, \dots For other powers the same formula is true, but it needs a different proof, later in the course. Until then the numbers back it up: rise over run at 4 reads 0.2485, 0.2498, … heading for 0.25.

7. What is ddx\frac{d}{dx}, and how is it different from dydx\frac{dy}{dx}?

ddx\frac{d}{dx} is an instruction: "take the derivative of what follows". ddx x3=3x2\frac{d}{dx}\,x^3 = 3x^2. If you name the curve y=x3y = x^3, then ddx\frac{d}{dx} of yy is written dydx\frac{dy}{dx}: Calculus 1.1's notation, the same thing with the yy put in.

8. Do I have to multiply out (x+h)n(x + h)^n every time? What about 5x35x^3 or x2+xx^2 + x?

No: that is the point of a rule. You proved it once; now x7→7x6x^7 \to 7x^6 in one line. And you never needed the whole expansion, only its first two terms. Constants in front and sums (5x35x^3, x2+xx^2 + x) come with the next rules, in Calculus 3.

3. Practice

Five problems in the textbook's notation, easiest first. Try each on paper, then click to check.

1.

Find (a) ddx x7\dfrac{d}{dx}\,x^7 and (b) ddx x\dfrac{d}{dx}\,x.

Show the answer
  • (a) The power rule with n=7n = 7: the 7 comes down in front, and the power drops by one: 7x67x^6.
  • (b) x=x1x = x^1, so n=1n = 1: 1⋅x0=11 \cdot x^0 = 1 (any nonzero number to the power 0 is 1). It makes sense: y=xy = x is a straight line that climbs one for every one across, so its slope is 1 everywhere.

2.

A square of side xx has area x2x^2. Its side grows to x+hx + h. Name the pieces the square gains, write the rise, and find the slope of x2x^2 from them.

Show the answer
  • The grown square is the old square plus two strips, each xx by hh, and one small corner, hh by hh.
  • Rise: (x+h)2−x2=2xh+h2(x + h)^2 - x^2 = 2xh + h^2.
  • Divide by the run hh (not zero): 2x+h2x + h. The strips had one hh, so none is left; the corner still has one.
  • Let hh shrink: 2x2x. The 2 in 2x2x counts the strips, as the 3 in 3x23x^2 counts the slabs.

3.

Find the slope of y=xy = \sqrt{x} at x=9x = 9.

Show the answer
  • Write the root as a power: x=x1/2\sqrt{x} = x^{1/2}.
  • The power rule: ddx x1/2=12 x−1/2=12x\dfrac{d}{dx}\,x^{1/2} = \tfrac12\,x^{-1/2} = \dfrac{1}{2\sqrt{x}} (bring the 12\tfrac12 down; 12−1=−12\tfrac12 - 1 = -\tfrac12; a negative power means "one over").
  • At x=9x = 9: 129=12⋅3=16\dfrac{1}{2\sqrt 9} = \dfrac{1}{2 \cdot 3} = \dfrac16.

4.

Find the slope of y=1xy = \dfrac{1}{x} at x=2x = 2.

Show the answer
  • Write it as a power: 1x=x−1\dfrac1x = x^{-1}.
  • The power rule with n=−1n = -1: −1⋅x−2=−1x2-1 \cdot x^{-2} = -\dfrac{1}{x^2}.
  • At x=2x = 2: −14-\dfrac14. Negative, because 1x\dfrac1x goes downhill as xx grows.

5.

Use the definition (rise over run, then the limit) to find the slope of y=x4y = x^4. One line a step.

Show the answer
  • Multiply out: (x+h)4=x4+4x3h+6x2h2+4xh3+h4(x + h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 (one piece with no hh, four with one, six with two, four with three, one with four: 16 pieces).
  • Rise: (x+h)4−x4=4x3h+6x2h2+4xh3+h4(x + h)^4 - x^4 = 4x^3h + 6x^2h^2 + 4xh^3 + h^4 (the subtraction removes x4x^4).
  • Divide by hh, which is not zero: 4x3+6x2h+4xh2+h34x^3 + 6x^2h + 4xh^2 + h^3.
  • Let h→0h \to 0: every term but the first still carries an hh, so the limit is 4x34x^3.

4. One open question

At the point (3,9)(3, 9) on y=x2y = x^2 the slope is 2⋅3=62 \cdot 3 = 6. Zoom in on the curve there and it looks like a straight line. Which line? What is its equation, and how far from the curve is it at x=3.1x = 3.1? (That is Calculus 1.3.)

Optional: all the coefficients (Pascal's triangle)

A side trip: the film needs only the first two terms of (x+h)n(x + h)^n. Here are all of them.

The number of pieces with exactly kk hh's is the number of ways to choose which kk of the nn factors give their hh. Row nn below lists those counts for k=0,1,…,nk = 0, 1, \dots, n. The second entry of every row is nn: the power that comes down.

n = 0:1
n = 1:11
n = 2:121
n = 3:1331
n = 4:14641
n = 5:15101051
n = 6:1615201561
n = 7:172135352171

Optional: a real cube, one liter

A side trip into a use: how good is the rule's estimate for a real, small change?

A cube 10 cm on a side holds exactly one liter (1,000 cm³). Make every side a millimeter longer. The slope of x3x^3 at 10 is 3⋅102=3003 \cdot 10^2 = 300 cm³ per cm, so a gap of 0.1 cm should add about 300×0.1=30300 \times 0.1 = 30 cm³. The true gain is a little more, by exactly the rods and the corner.

the rule’s estimate, 3·x²·h:30.0 cm³
the true gain, (x + h)³ − x³:30.301 cm³
slabs30.0
rods0.3
corner0.001

Optional: where the other powers get proved

A side trip: a map, not a proof.

The counting argument needs nn factors, so it proves the rule for n=1,2,3,…n = 1, 2, 3, \dots only. The film uses the rule for x1/2x^{1/2} and names x−1x^{-1}, with "proofs: Calculus 3, 4" on the board:

  • Negative whole powers (x−1x^{-1}, x−2x^{-2}, …) follow from the quotient rule, in Calculus 3.
  • Fractional powers (x1/2x^{1/2}, x2/3x^{2/3}, …) follow from the chain rule and implicit differentiation, in Calculus 4.

In every case the formula is the same: ddx xn=n xn−1\dfrac{d}{dx}\,x^n = n\,x^{n-1}.

From the film

Where this sits, in order: Calculus 1, the Big Idea (what a derivative is) https://youtu.be/ZBqtb7uhrkI then Calculus 1.1, the limit https://youtu.be/hPWb3ZwGbhk then Calculus 1.2, the power rule (this part), then Calculus 1.3, the tangent line. Playlist DEEP DIVE: Calculus.

Textbook: James Stewart, Calculus: Early Transcendentals, 9th ed., §3.1 (Derivatives of Polynomials and Exponential Functions).

Next: at the point (3, 9) on y = x² the slope is 6. Zoom in, and the curve looks like a line. Which line? That is Calculus 1.3.

Sources

  • The power rule for a positive whole number n, proved by multiplying out (x + h)ⁿ ("All other terms contain powers of h that are two or greater"), and the note that it is extended later to negative and then to rational powers: OpenStax, Calculus Volume 1, §3.3 Differentiation Rules (Theorem 3.3, The Power Rule, and its proof).https://openstax.org/books/calculus-volume-1/pages/3-3-differentiation-rules
  • The method (a gap h, rise over run, the limit) and x² turning into 2x: Calculus 1.1, linked above.
  • What is proved here: the rule for n = 1, 2, 3, and every whole number after. For the square root and for negative powers the rule is stated and checked in numbers; the board says where the proofs come (Calculus 3 and 4).
  • Every equation and number on screen is checked with SymPy in the project's check scripts.

The student is drawn with generative AI; the cube, the graphs and the equations are made by code. The narration is a synthetic voice.