Single-Variable-Calculus · Unit 6 · Video 2 · Interactive Practice

Defining e by Slope One, and the Derivative of ln x

IKey Formulas

Formula	Name	What it says
$M(e) = \displaystyle\lim_{h \to 0} \frac{e^h - 1}{h} = 1$	Definition of $e$	$e$ is the base whose graph has slope $1$ at $(0, 1)$
$\dfrac{d}{dx}\,e^x = e^x$	Derivative of $e^x$	$\dfrac{d}{dx}\,a^x = M(a)\,a^x$ with $M(e) = 1$
$\dfrac{d}{dx}\,2^{kx}\Big\|_{x=0} = k\,M(2)$ , so $e = 2^{1/M(2)}$	Why $e$ exists	Rescaling $x$ by $k$ multiplies the slope at $0$ by $k$
$\dfrac{d}{dx}\ln x = \dfrac{1}{x}\quad (x > 0)$	Derivative of $\ln x$	From $e^w = x$ by implicit differentiation

Key Insight: $\ln x$ is $e^x$ reflected across $y = x$ . The reflection swaps rise and run, so a slope $s$ at $(t, e^t)$ becomes $1/s$ at $(e^t, t)$ — and slope $e^t$ becomes $1/e^t = 1/x$ .

IIVisualization 1 — Squeezing $2^x$ Until the Slope Is One

Rescaling $x$ by $k$ turns $2^x$ into $b^x$ with $b = 2^k$ , and multiplies the slope at $0$ by $k$ .

Rescale factor k 1.00

💡 Challenge: find the $k$ that gives slope $1$ at $(0, 1)$ — the base $b = 2^k$ it produces is $e$ . (To draw to scale, the plot uses the numerical value $M(2) \approx 0.6931$ ; the video leaves it unknown, and the next video computes it.)

IIIVisualization 2 — Reflection Swaps Rise and Run

Mirroring across $y = x$ sends slope $e^t$ at $(t, e^t)$ to slope $1/e^t$ at $(e^t, t)$ on $\ln x$ .

t 0.00

IVVisualization 3 — Implicit Differentiation of $w = \ln x$

Exponentiating $w = \ln x$ and differentiating both sides gives $\tfrac{d}{dx}\ln x = \tfrac{1}{x}$ .

Point Step 1

Step 1 — Exponentiate both sides

w = \ln x \;\Longrightarrow\; e^w = x

Step 2 — Differentiate with respect to $x$

\frac{d}{dx}\big(e^w\big) = \frac{d}{dx}(x) = 1

Step 3 — Chain rule: $w$ depends on $x$

\frac{d}{dw}\big(e^w\big)\cdot\frac{dw}{dx} = e^w\,\frac{dw}{dx} = 1

using

\dfrac{d}{dw}\,e^w = e^w

Step 4 — Solve, then substitute $e^w = x$

\frac{dw}{dx} = \frac{1}{e^w} = \frac{1}{x}

VQuiz Questions

Problem 1 · Slope of the Natural Log

Given: $y = \ln x$ — find the slope of its graph at $x = 4$ .

✅ Correct!

\dfrac{d}{dx}\ln x = \dfrac{1}{x}

, so the slope at

x = 4

\tfrac{1}{4}

— the curve flattens as

x

grows.

❌ Not quite.

\ln 4 \approx 1.386

is the height of the curve at

x = 4

, not its slope.

❌ Not quite.

e^x

is its own derivative, but

\ln x

is not: its slope is

1/x

❌ Not quite. The slope of

\ln x

1/x

— the reciprocal of

x

, not

x

itself.

Show solution

With $w = \ln x$ , implicit differentiation of $e^w = x$ gives $e^w\,\dfrac{dw}{dx} = 1$ , so

\frac{d}{dx}\ln x = \frac{1}{e^w} = \frac{1}{x}.

At $x = 4$ the slope is $\dfrac{1}{4}$ .

Problem 2 · Rescaling Changes the Slope at Zero

Given: $f(x) = 2^x$ has slope $M(2)$ at $x = 0$ . Find the slope of $y = 8^x$ at $x = 0$ , in terms of $M(2)$ .

✅ Correct!

8^x = 2^{3x} = f(3x)

, and the chain rule multiplies the slope at

0

by the rescale factor

k = 3

❌ Not quite. The slope is multiplied by the rescale factor

k

b = 2^k

, not by the base:

8 = 2^3

, so

k = 3

❌ Not quite. Cubing the base does not cube the slope. The chain rule brings down the factor

3

from

2^{3x}

as a multiplier.

❌ Not quite. Rescaling by

k > 1

squeezes the graph toward the vertical axis, making it steeper at

(0, 1)

, not flatter.

Show solution

Write the base as a power of $2$ : $8 = 2^3$ , so by the power-of-a-power rule

8^x = \big(2^3\big)^x = 2^{3x} = f(3x).

By the chain rule,

\frac{d}{dx}\,f(3x) = 3\,f'(3x) \quad\Longrightarrow\quad \text{at } x = 0:\ 3\,f'(0) = 3\,M(2).

So $M(8) = 3\,M(2)$ .

Problem 3 · Tangent Line to ln x at x = e

Given: the curve $y = \ln x$ and the point $(e, 1)$ on it — find the slope there and the equation of the tangent line.

What is the slope at $(e, 1)$ ?

What is the tangent line?

✅ Correct! The tangent

y = x/e

has slope

1/e

and passes through the origin.

❌ Not quite.

e

is the slope of

e^x

at the mirror point

(1, e)

. Reflection swaps rise and run, so the slope of

\ln x

(e, 1)

is the reciprocal.

❌ Not quite.

1

is the height

\ln e

. The slope is

1/x

evaluated at

x = e

❌ Check your slope. Use

\dfrac{d}{dx}\ln x = \dfrac{1}{x}

x = e

❌ Not quite. Point-slope form is

y - 1 = \tfrac{1}{e}(x - e)

; expanding, the

-\tfrac{1}{e}\cdot e = -1

cancels the

+1

❌ Check the line. It must have slope

1/e

and pass through

(e, 1)

: use

y - 1 = \tfrac{1}{e}(x - e)

Show solution

Slope: $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ , so at $x = e$ the slope is $\dfrac{1}{e}$ .

Point: $\ln e = 1$ , so the point is $(e, 1)$ .

Point-slope form:

y - 1 = \frac{1}{e}(x - e) = \frac{x}{e} - 1 \quad\Longrightarrow\quad y = \frac{x}{e}.

The tangent line passes through the origin.

Problem 4 · Rescaling from Base e

Given: $M(e) = 1$ . Find the base $b$ whose graph $y = b^x$ has slope exactly $2$ at $(0, 1)$ , i.e. $M(b) = 2$ .

✅ Correct!

(e^2)^x = e^{2x}

, whose slope at

0

2\,M(e) = 2

❌ Not quite. Doubling the base does not double the slope. Doubling the rescale factor does:

b = e^k

has

M(b) = k

❌ Close, but inverted.

\sqrt{e} = e^{1/2}

rescales by

k = \tfrac{1}{2}

, giving slope

\tfrac{1}{2}

0

❌ Not quite. The slope at

0

is not the base:

y = 2^x

has slope

M(2)

there, not

2

. Write

b = e^k

and use the rescaling rule.

Show solution

Write the base as a power of $e$ : $b = e^k$ . Then $b^x = e^{kx}$ , and by the chain rule

\frac{d}{dx}\,e^{kx} = k\,e^{kx} \quad\Longrightarrow\quad \text{at } x = 0:\ k\,M(e) = k.

So $M(e^k) = k$ . Slope $2$ needs $k = 2$ : $\ b = e^2 \approx 7.389$ .

Solved: 0 / 4

Defining e by Slope One, and the Derivative of ln x

IKey Formulas

IIVisualization 1 — Squeezing 2x2^x2x Until the Slope Is One

IIIVisualization 2 — Reflection Swaps Rise and Run

IVVisualization 3 — Implicit Differentiation of w=ln⁡xw = \ln xw=lnx

VQuiz Questions

IIVisualization 1 — Squeezing $2^x$ Until the Slope Is One

IVVisualization 3 — Implicit Differentiation of $w = \ln x$