Derivatives of eˣ and aˣ — proofs: Visual walkthrough

Before we start, three plain-word anchors we will need:

Step 1 — What "slope" even means: rise over run

WHAT. To measure the steepness of $a^{x}$ at a point $x$ , we take a second point a tiny step $h$ to the right, at $x + h$ . We draw the straight line through the two points and measure its tilt.

WHY. A curve has no single "tilt" like a straight line does — it bends. So we cheat: pick two nearby points, measure the tilt of the straight line joining them, then let the second point slide toward the first. That is the only honest way to talk about the slope of something curved.

PICTURE. In the figure, the height at $x$ is the lower dot; the height at $x + h$ is the upper dot. The run (magenta) is the horizontal gap $h$ . The rise (orange) is the vertical gap $a^{x + h} - a^{x}$ . Their ratio is the tilt.

Figure — Derivatives of eˣ and aˣ — proofs

slope of the joining line = tilt of the straight line \frac{a ^{x + h} - a ^{x} rise (orange)}{run (magenta) h}

As $h$ shrinks to $0$ the two dots merge and the joining line becomes the true tangent — the exact slope. That shrinking is written $lim_{h \to 0}$ , "the value this ratio homes in on as $h$ gets tiny." This is the first-principles definition:

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

Here $f^{'} (x)$ (read " $f$ -prime") is just a name for the slope function.

Step 2 — The magic split: the shape repeats itself

WHAT. We feed $f (x) = a^{x}$ into that ratio and use one exponent law: $a^{x + h} = a^{x} \cdot a^{h}$ .

WHY. Adding in the exponent = multiplying the powers. This law is the engine of the whole proof: it lets us peel the $x$ -part away from the $h$ -part.

PICTURE. Look at the figure: the tall bar is $a^{x}$ , the taller bar is $a^{x + h}$ . The second bar is just the first bar stretched by the factor $a^{h}$ — same shape, scaled. That scaling factor is what carries all the $h$ -information.

\frac{a ^{x + h} - a ^{x}}{h} = \frac{a ^{x} a ^{h} stretched bar - a ^{x} original bar}{h} = a^{x} \cdot depends on h only \frac{a ^{h} - 1}{h}

We factored $a^{x}$ out of both terms in the top. What's left inside, $\frac{a ^{h} - 1}{h}$ , has no $x$ in it at all — only the base $a$ and the shrinking step $h$ .

Step 3 — The constant that survives: $M (a)$

WHAT. Now let $h \to 0$ . The factor $a^{x}$ out front does not contain $h$ , so it just sits there. The leftover limit becomes a single number we name $M (a)$ .

WHY. Because $a^{x}$ is constant as far as $h$ is concerned (only $h$ is moving), it slides straight out of the limit. Whatever the limit of the leftover is, it depends on the base $a$ but not on where you are ( $x$ ). One number, fixed per base.

PICTURE. The figure zooms in near $x = 0$ . There the height is $a^{0} = 1$ , and the tangent's steepness there is exactly $M (a)$ — the slope-at-the-start. Three curves ( $2^{x}$ violet, $e^{x}$ magenta, $3^{x}$ orange) leave the point $(0, 1)$ at three different tilts.

f^{'} (x) = a^{x} \cdot M (a) = slope at x = 0 h \to 0 lim \frac{a ^{h} - 1}{h} ⟹ \frac{d}{d x} a^{x} = M (a) a^{x}

Read the box in words: the slope of an exponential is its own height times a personality number $M (a)$ . The whole rest of the page is: find $M (a)$ .

Step 4 — Choosing the base that makes $M = 1$ : this defines $e$

WHAT. Different bases give different $M (a)$ (Step 3's three tilts). We hunt for the one base whose starting slope is exactly $1$ . We name that base $e$ .

WHY. If $M = 1$ , the box in Step 3 collapses to $f^{'} (x) = 1 \cdot a^{x} = a^{x}$ — the curve becomes its own slope, with no leftover factor. That's the cleanest possible derivative, so this base earns the title "natural."

PICTURE. The figure overlays the line $y = x + 1$ (the tangent of slope $1$ through $(0, 1)$ ). The curve $2^{x}$ starts below that line's tilt (too gentle, $M < 1$ ); $3^{x}$ starts above it (too steep, $M > 1$ ). Somewhere between sits the perfect curve that kisses the line — leaves $(0, 1)$ tangent to it. That curve is $e^{x}$ .

Feeding $M (e) = 1$ into Step 3's box:

\frac{d}{d x} e^{x} = e^{x}

Step 5 — Pinning down $M (a)$ for every base: it is $ln a$

WHAT. Leaving $M (a)$ as a mystery limit is unsatisfying. We rewrite any base $a > 0$ using $e$ and differentiate with the Chain Rule.

WHY. We already fully understand $e^{(stuff)}$ . So if we can dress up $a^{x}$ as $e^{(something) x}$ , we can reuse the clean $e$ -result. The tool that lets us differentiate " $e$ of an inner function" is the chain rule: outer slope times inner slope.

PICTURE. The figure shows the rewrite as a re-labelling of the horizontal axis. Every base $a$ is $e$ raised to the power $ln a$ (this is what $\ln a$ means: the exponent you put on $e$ to get $a$ ). So $a^{x}$ is really $e^{x}$ walked along faster or slower by the factor $ln a$ .

Step by step:

a = e^{l n a} ⟹ a^{x} = (e^{l n a})^{x} = e^{(ln a) inner slope x}

\frac{d}{d x} e^{(l n a) x} = outer slope = itself e^{(l n a) x} \cdot inner slope (ln a) = a^{x} ln a

Step 6 — Every case: steep, gentle, flat, and falling

WHAT. We check the formula $M (a) = ln a$ against all kinds of bases $a > 0$ , including the degenerate ones. A rule you can't test at its edges is a rule you can't trust.

WHY. The reader must never meet a base we didn't show. So: base bigger than $e$ , base equal to $e$ , base between $1$ and $e$ , base equal to $1$ , and base below $1$ (decay).

PICTURE. The figure plots the personality number $M (a) = ln a$ against the base $a$ . Watch where the curve crosses key values.

Base $a$	$ln a$ = start-slope $M (a)$	What the curve does
$a > e$ (e.g. $3$ )	$ln 3 \approx 1.10 > 1$	steeper than $e^{x}$ at the start
$a = e$	$ln e = 1$	self-derivative — slope $=$ height
$1 < a < e$ (e.g. $2$ )	$ln 2 \approx 0.69 < 1$	gentler than $e^{x}$ at the start
$a = 1$	$ln 1 = 0$	flat line $1^{x} = 1$ ; slope $0$ ✓
$0 < a < 1$ (e.g. $\frac{1}{2}$ )	$ln \frac{1}{2} \approx - 0.69 < 0$	decay — slope negative, curve falls

The one-picture summary

Everything above, compressed: start at first principles, split with the exponent law, isolate the personality number $M (a)$ , choose the base that makes it $1$ (that's $e$ ), and rewrite all other bases through $e$ to discover $M (a) = ln a$ .

Step 1 h \to 0 lim \frac{a ^{x + h} - a ^{x}}{h} = Steps 2-3 a^{x} h \to 0 lim \frac{a ^{h} - 1}{h} = M (e) = 1 \Rightarrow e^{x} is its own slope a^{x} M (a) = Steps 4-5 a^{x} ln a

Recall Feynman retelling — the whole walk in plain words

Only positive bases $a > 0$ make an unbroken curve to slope, so that's all we allow. To find how steep $a^{x}$ is, I jump a tiny step $h$ to the right and measure rise over run. Jumping right just multiplies the height by $a^{h}$ (adding in the exponent is multiplying), so the height at $x$ pops straight out of the calculation. What's left is one number per base — the tilt right at the start, at $x = 0$ , where the height is $1$ . Because the curve is smooth with no corner there, those joining lines settle onto one tangent, so that number really exists; call it the base's "personality." As I turn the base up, the personality rises steadily and smoothly from below $1$ to above $1$ , so it must pass through exactly $1$ once — that base is $e \approx 2.718$ , and $e^{x}$ copies itself when you differentiate. Finally, since any base $a$ is just $e$ raised to $ln a$ (and a power of a power multiplies the exponents), every exponential is secretly $e^{x}$ sped up by the factor $ln a$ — so its personality number is $ln a$ . Big base $\to$ big positive $ln$ (steep climb); base $1 \to ln 1 = 0$ (flat line); base below $1 \to$ negative $ln$ (it falls). One tidy function, $ln a$ , gets every case right.

Active recall

Why must the base

a

be strictly positive?

a \leq 0

, values like

a^{1/2} = a

aren't real, so

a^{x}

isn't an unbroken curve; only

a > 0

gives a real value for every real

x

Why does the limit

M (a) = lim_{h \to 0} (a^{h} - 1) / h

actually exist?

The ratio is the tilt of joining lines near

x = 0

; on the smooth exponential these settle onto one tangent from both sides, so the limit is a genuine number.

Why is there exactly one base

e

with

M (e) = 1

M (a) = ln a

rises steadily and continuously from below

1

to above

1

a

grows, so by the intermediate-value idea it equals

1

at exactly one base.

What does

M (a)

represent geometrically?

The slope of

a^{x}

x = 0

, where the height is

a^{0} = 1

Which exponent rule turns

(e^{l n a})^{x}

into

e^{(l n a) x}

A power of a power multiplies the exponents:

(b^{c})^{d} = b^{c d}

What is

M (a)

equal to, finally?

ln a

What is

M (a)

when

a = 1

, and what does the curve look like?

ln 1 = 0

;

1^{x} = 1

is a flat line with slope

0

What sign does

M (a)

have for

0 < a < 1

, and what does that mean?

Negative; the curve decays (falls).

Connections

Derivatives of eˣ and aˣ — proofs — the parent result this page derives visually.
Limits — first principles definition of derivative — Step 1's rise-over-run engine.
Chain Rule — Step 5's outer-slope-times-inner-slope tool.
Natural Logarithm ln x — $ln a$ is "the power on $e$ that makes $a$ ."
The number e — definitions — the $(1 + \frac{1}{n})^{n}$ view of the same $e$ .
Exponential Growth and Decay — Step 6's sign story in the wild.
Power Rule — the rule NOT to use here (moving base vs moving exponent).