Natural exponential function eˣ — graph, derivative preview: Visual walkthrough

We need three plain-word ideas before we start. Let me anchor each to a picture as it appears.

Step 1 — What "slope" even means (the picture behind the whole idea)

WHAT. Pick any smooth curve. The slope at a point is how steep the curve is right there — how many units it rises for every one unit you step to the right.

WHY. We cannot talk about "the function is its own slope" until we can measure slope. And the honest way to measure the steepness at a single point is to look at a tiny straight segment joining that point to a neighbour a distance $h$ away, then shrink $h$ .

PICTURE. Look at figure below. Two points sit on the curve: one at horizontal position $x$ , the other a small gap $h$ to its right at $x + h$ . The red line joins them. As the gap $h$ shrinks toward $0$ , that red line pivots until it just kisses the curve — that touching line is the tangent, and its steepness is the slope we want.

Figure — Natural exponential function eˣ — graph, derivative preview

The slope at the point is what this ratio settles down to as $h$ becomes tiny. We write that "settling down" with the word $lim$ (short for limit — "the value it heads toward"):

$f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$

Here $f^{'} (x)$ (read " $f$ -prime of $x$ ") is just a name for the slope at position $x$ . Nothing else is assumed. This is our only tool.

Step 2 — Feed an exponential into the slope machine

WHAT. Take a general exponential $f (x) = a^{x}$ , where $a$ is any fixed positive number (say $2$ , or $3$ , or $10$ ), and $x$ is the variable we move along the ground. Drop it into the slope formula from Step 1.

WHY. We don't yet know which base is special. So we keep the base as a letter $a$ and let the algebra tell us what's special about it.

PICTURE. Figure below shows three exponential curves — $2^{x}$ , $3^{x}$ , and a steeper one — all crossing the $y$ -axis at height $1$ . The red curve is our working curve $a^{x}$ . Notice they all pass through $(0, 1)$ but leave that point at different steepnesses. That difference in launch-steepness is the whole story.

$f^{'} (x) = lim_{h \to 0} \frac{a ^{x + h} height a step ahead - a ^{x} height now}{h}$

$a^{x + h}$ is the height a little gap $h$ to the right.
$a^{x}$ is the height where we stand.
$h$ is that little gap.

Step 3 — The one algebra trick: split the exponent

WHAT. Rewrite $a^{x + h}$ as $a^{x} \cdot a^{h}$ , then pull the common $a^{x}$ out of the fraction and out of the limit.

WHY. The index law $a^{x + h} = a^{x} \cdot a^{h}$ lets us separate where we stand ( $a^{x}$ ) from the tiny step ( $a^{h}$ ). And since $a^{x}$ does not contain $h$ at all, the limit — which only cares about $h \to 0$ — treats $a^{x}$ as a fixed spectator we can carry outside.

PICTURE. In the figure, the tall red bar is the standing height $a^{x}$ ; the small bracket beside it is the fractional extra growth caused by the tiny step. The picture says: total slope $=$ (how tall you are) $\times$ (a step-factor that does not depend on how tall you are).

$f^{'} (x) = lim_{h \to 0} \frac{a ^{x} a ^{h} - a ^{x}}{h} = a^{x} \cdot call this k (a) h \to 0 lim \frac{a ^{h} - 1}{h}$

$a^{x} a^{h} - a^{x} = a^{x} (a^{h} - 1)$ — factor the standing height out.
$a^{x}$ has no $h$ , so it slides outside the limit untouched.
What's left, $h \to 0 lim \frac{a ^{h} - 1}{h}$ , is a pure number depending only on the base $a$ . Name it $k (a)$ .

Step 4 — What is $k (a)$ , really? (read it off the graph)

WHAT. Notice $k (a)$ is exactly the slope of $a^{x}$ at $x = 0$ . Plug $x = 0$ into Step 3: $f^{'} (0) = k (a) \cdot a^{0} = k (a) \cdot 1 = k (a)$ .

WHY. This turns an abstract limit into something you can see: $k (a)$ is just how steeply the curve $a^{x}$ leaves the point $(0, 1)$ . No calculation of the limit needed — measure the launch angle.

PICTURE. Figure below zooms in on the crossing point $(0, 1)$ . The red tangent line at that point has a slope; that slope-number is $k (a)$ . For a shallow base the red line is nearly flat; for a steep base it rockets up.

$k (a) = f^{'} (0) = slope of a^{x} where it crosses the y -axis$

Step 5 — $k (a)$ grows as $a$ grows — so it must pass through $1$

WHAT. Test values. For $a = 2$ , the launch slope is $k (2) \approx 0.693$ (shallow). For $a = 3$ , it's $k (3) \approx 1.099$ (already too steep). So somewhere between $2$ and $3$ there is a base whose launch slope is exactly $1$ .

WHY. As we crank the base up from $2$ toward $3$ , the curve gets steeper at the crossing point, so $k (a)$ rises smoothly from $0.693$ up past $1$ to $1.099$ . A quantity that climbs continuously from below $1$ to above $1$ must equal $1$ somewhere — that's the intermediate-value idea, just common sense on a smooth graph.

PICTURE. Figure below plots the launch slope $k (a)$ against the base $a$ . The curve rises; the red dot marks the exact spot where $k (a) = 1$ . Its horizontal coordinate is a specific number — we christen it $e$ .

Step 6 — Collapse everything: the self-slope law

WHAT. Put $a = e$ into the master formula from Step 3.

WHY. We deliberately chose $e$ to make the scaling constant equal $1$ . So the "scaled by $k$ " shrinks to "scaled by $1$ " — i.e. no scaling at all.

PICTURE. The figure shows the $e^{x}$ curve with tangent arrows drawn at several points. At each point the red arrow's steepness equals the curve's height there — short arrow low down, tall steep arrow high up. Height and steepness are locked together.

$\frac{d}{d x} e^{x} = k (e) e^{x} = 1 \cdot e^{x} = e^{x}$

$k (e) = 1$ by our choice of base — this is the whole payoff.
The function returns unchanged: slope $=$ height, everywhere.

Step 7 — Edge & degenerate cases (so you're never surprised)

WHAT. Check the corners the main story skipped.

WHY. A derivation is only trustworthy if it survives the extremes: negative $x$ , the crossing point, and the far-left tail.

PICTURE. Figure below marks three spots on $e^{x}$ : the far-left tail ( $x \to - \infty$ ), the crossing at $(0, 1)$ , and a point with negative $x$ . The red curve stays strictly above the horizontal axis the entire way.

At $x = 0$ : height $= e^{0} = 1$ , slope $= e^{0} = 1$ . The tangent is the line $y = x + 1$ .
For negative $x$ : $e^{x} = 1/ e^{∣ x ∣}$ is a positive over positive, so it's a small positive number — never zero, never negative. Slope there is also that same small positive height, so the curve is barely rising, hugging the axis.
As $x \to - \infty$ : height $\to 0^{+}$ and slope $\to 0^{+}$ together — the curve flattens onto the asymptote $y = 0$ but never touches it.
As $x \to + \infty$ : height and slope both blow up together, faster than any polynomial.
Degenerate base $a = 1$ : then $1^{x} = 1$ for all $x$ , a flat line, and $k (1) = lim_{h \to 0} \frac{1 - 1}{h} = 0$ — slope zero everywhere. That's why $a = 1$ is not a useful exponential and why the interesting base sits above it.

The one-picture summary

The figure compresses all seven steps: the slope machine (secant → tangent), the family of exponentials fanning out through $(0, 1)$ , the launch-slope $k (a)$ climbing through $1$ to select $e$ , and the final $e^{x}$ whose red tangent length equals its height at every point.

Recall Feynman retelling — the whole walkthrough in plain words

To measure steepness at a point, join it to a nearby dot and shrink the gap — that's the slope machine (Step 1). Feed in $a^{x}$ and use the index law to split "how tall you are" from "the tiny extra step" (Steps 2–3). What's left over is a pure number $k (a)$ — and that number is just how steeply the curve leaves the point $(0, 1)$ (Step 4). Crank the base up: at $2$ the launch is too shallow ( $0.69$ ), at $3$ too steep ( $1.10$ ), so somewhere between them there's a base that launches at exactly slope $1$ . Call it $e \approx 2.718$ (Step 5). For that special base the scaling factor is $1$ , so differentiating gives the function straight back: slope $=$ height everywhere (Step 6). And it behaves at every corner — always positive, flattening to zero on the left, exploding on the right (Step 7). The snowball whose speed is its size.

Recall Quick self-test

What is $k (a)$ geometrically? ::: The slope of $a^{x}$ where it crosses the $y$ -axis, i.e. $f^{'} (0)$ . Why can we pull $a^{x}$ outside the limit in Step 3? ::: Because $a^{x}$ contains no $h$ , so it's a constant spectator as $h \to 0$ . What forces $e$ to exist between $2$ and $3$ ? ::: $k (2) \approx 0.69 < 1$ and $k (3) \approx 1.10 > 1$ , and $k$ rises smoothly, so it must equal $1$ somewhere between. Why is $\frac{d}{d x} e^{x} = e^{x}$ once we know $k (e) = 1$ ? ::: The general law is $k (a) a^{x}$ ; with $k (e) = 1$ it collapses to $1 \cdot e^{x} = e^{x}$ .

Connections

Natural exponential function $e^x$ — graph, derivative preview — the parent result proved here.
Euler's number e — definition & limit — the same $e$ from the growth-limit side.
General exponential a^x and its derivative — the $k (a) a^{x}$ law from Step 3.
Natural logarithm ln x as inverse of e^x — $k (a)$ turns out to be $ln a$ .
Chain rule — extends this to $e^{g (x)}$ .
Taylor series of e^x — another route to the same self-slope fact.
Exponential growth and decay models — where "slope = height" is used in the wild.