Single-Variable-Calculus · Unit 1 · Video 5 · Interactive Practice

Newton and Leibniz Notation, and the Power Rule

IKey Formulas

Formula	Name	What it tells you
$\dfrac{\Delta y}{\Delta x} = \dfrac{\Delta f}{\Delta x} = \dfrac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$	Difference quotient	Average slope across $\Delta x$
$f'(x_0), \quad y'$	Newton's notation	The base point can be written in: $f'(x_0)$
$\dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{d}{dx}f = \dfrac{d}{dx}y$	Leibniz's notation	Same derivative, base point omitted
$(x + \Delta x)^n = x^n + n(\Delta x)x^{n-1} + O\big((\Delta x)^2\big)$	Binomial expansion	Two terms matter, the rest is junk
$\dfrac{d}{dx}x^n = nx^{n-1}$	Power rule	Every positive whole-number power at once

Key Insight: One calculation, run a single time, differentiates every positive whole-number power — and therefore every polynomial, term by term.

IIVisualization 1 — One Derivative, Many Names

Newton's prime and Leibniz's $dy/dx$ report the same number at every point of $y = x^2$ .

Base point x₀ 1.5

💡 Mathematics does this in many places: the notation drops something important, and the reader is expected to fill it in from context.

IIIVisualization 2 — Where the Coefficient $n$ Comes From

Each of the $n$ factors of $(x + \Delta x)^n$ donates either its $x$ or its $\Delta x$ to a term.

Exponent n 4

IVVisualization 3 — Why the Junk Vanishes

The difference quotient is $nx^{n-1}$ plus junk, and every junk term still carries a factor $\Delta x$ .

Power Δx 1.000

💡 The binomial theorem itself needs no limit — it is exact for every $\Delta x$ . The junk dropped out because we chose to send $\Delta x \to 0$ .

VQuiz Questions

Problem 1 · Power Rule, Straight Off

Given: $y = x^7$ — find $\dfrac{dy}{dx}$ .

✅ Correct! The exponent comes down as the coefficient and the power drops by one:

7x^6

❌ Not quite. The exponent came down as a coefficient, but it must also drop by one in the power.

❌ Not quite. The power dropped correctly, but the coefficient is the original exponent

n = 7

, not the new one.

❌ That is the antiderivative. The power went up to

8

and the coefficient became

1/(n+1)

. Differentiating sends the power the other way: down to

6

, with the original

7

out front.

❌ Not quite. The power rule sends

x^n

nx^{n-1}

— the exponent falls in front, the power decreases.

Show solution

Apply $\dfrac{d}{dx}x^n = nx^{n-1}$ with $n = 7$ :

\frac{d}{dx}x^7 = 7x^{7-1} = 7x^6

Both moves happen at once: the $7$ arcs down in front, and the power becomes $6$ .

Problem 2 · Which Name Is Not the Derivative?

Given: $y = f(x)$ . Which expression is not another name for the derivative?

✅ Correct!

\dfrac{\Delta y}{\Delta x}

is the difference quotient — the derivative is its limit as

\Delta x \to 0

❌ Not quite. That is Leibniz's spelling with the function slipped down out of the fraction:

\dfrac{d}{dx}y

means exactly

\dfrac{dy}{dx}

❌ Not quite. Newton wrote

f'(x_0)

and

y'

; Leibniz wrote

\dfrac{dy}{dx}

\dfrac{df}{dx}

\dfrac{d}{dx}f

and

\dfrac{d}{dx}y

. All six name the same object.

Show solution

Since $y = f(x)$ , the change in the output has two names, $\Delta y = \Delta f$ , so the quotient has two names too:

\frac{\Delta y}{\Delta x} = \frac{\Delta f}{\Delta x} = \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}

This is an average slope over a finite step $\Delta x$ — a number that still depends on how big the step is. Only after taking the limit do the derivative's names appear:

\lim_{\Delta x \to 0}\frac{\Delta f}{\Delta x} = f'(x_0) \qquad \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}f = \frac{d}{dx}y

Newton's prime is written $f'(x_0)$ when the base point is stated and $y'$ when it is left to context — so $y'$ omits the point exactly as the Leibniz forms do.

So $\dfrac{\Delta y}{\Delta x}$ is the only one on the list that is not the derivative. Note also that $\dfrac{dy}{dx}$ is not a fraction of two small numbers — at this stage it is a single symbol for a limit, not a ratio you may cancel.

Problem 3 · A Polynomial, Term by Term

Given: $y = 4x^5 + x^{12}$ — find $\dfrac{dy}{dx}$ .

✅ Correct!

5 \cdot 4 = 20

on the first term, and the second term needs no constant at all.

❌ Not quite. The constant

4

rides along and multiplies the exponent: the coefficient is

4 \cdot 5

, not

4

❌ Close. The first term is right, but the second power must drop from

12

11

❌ Not quite. On

4x^5

the exponent

5

comes down alongside the

4

that was already there — the

4

does not disappear.

❌ Not quite. Differentiate term by term with

\dfrac{d}{dx}x^n = nx^{n-1}

, keeping every constant factor.

Show solution

First term. The power rule on $x^5$ gives $5x^4$ , and the constant $4$ rides along:

\frac{d}{dx}\,4x^5 = 4 \cdot 5x^4 = 20x^4

Second term. The power rule on $x^{12}$ :

\frac{d}{dx}\,x^{12} = 12x^{11}

Together:

\frac{dy}{dx} = 20x^4 + 12x^{11}

No limit, no difference quotient, no binomial theorem — the power rule did all the work on sight.

Problem 4 · Supply the Missing Base Point

Given: $y = x^4$ — find the value of $\dfrac{dy}{dx}$ at $x = -2$ .

✅ Correct!

\dfrac{dy}{dx} = 4x^3

, and

4(-2)^3 = 4(-8) = -32

❌ Close, but check the sign. An odd power keeps the minus sign:

(-2)^3 = -8

, not

+8

❌ Not quite. That is

x^3

x = -2

; the power rule also puts the exponent

4

in front.

❌ Not quite. The exponent came down as a coefficient but the power never dropped — this evaluates

4x^4

, not

4x^3

❌ Not quite. Differentiate first, then substitute

x = -2

into the derivative.

Show solution

Step 1 — differentiate. The power rule with $n = 4$ :

\frac{dy}{dx} = \frac{d}{dx}x^4 = 4x^3

Step 2 — supply the base point. Leibniz's $\dfrac{dy}{dx}$ does not say where the derivative is taken, so the question has to say it: evaluate at $x = -2$ .

4(-2)^3 = 4(-8) = -32

In Newton's notation the point is built into the symbol, and no reminder is needed:

f'(-2) = -32

The slope is negative, which fits: to the left of the origin, $y = x^4$ is falling.

Solved: 0 / 4

Newton and Leibniz Notation, and the Power Rule

IKey Formulas

IIVisualization 1 — One Derivative, Many Names

IIIVisualization 2 — Where the Coefficient nnn Comes From

IVVisualization 3 — Why the Junk Vanishes

VQuiz Questions

IIIVisualization 2 — Where the Coefficient $n$ Comes From