Derivatives of eˣ and aˣ — proofs: Worked examples

This page is the drill-hall for the parent proofs. The parent showed why $\frac{d}{d x} e^{x} = e^{x}$ and $\frac{d}{d x} a^{x} = a^{x} ln a$ . Here we hunt down every kind of exponential a problem can throw at you and solve one of each — including the sneaky degenerate cases (base $= 1$ , decaying bases, negative exponents) that people forget exist.

The scenario matrix

Every exponential-derivative problem falls into one of these boxes. We will solve at least one example per row.

#	Case class	What's tricky about it	Covered by
1	Base $= e$ , exponent $= x$	the pure self-derivative	Ex 1
2	Base $a > 1$ , exponent $= x$	picks up an $ln a$ factor	Ex 2
3	Base $0 < a < 1$ (decay)	$ln a < 0$ → negative slope	Ex 3
4	Degenerate base $a = 1$	$ln 1 = 0$ → derivative is zero	Ex 4
5	Exponent is a linear inside $k x$	chain rule → extra constant $k$	Ex 5
6	Exponent is negative / non-linear	sign and shape of $u^{'} (x)$	Ex 6
7	General base with a function exponent	must convert to base $e$ first	Ex 7
8	Slope at a point (geometry)	evaluate derivative, read a tangent	Ex 8
9	Real-world word problem (growth rate)	attach units, interpret the number	Ex 9
10	Exam twist: solve for the base	run the derivative rule backwards	Ex 10

Case 1 — the pure self-derivative

Figure — Derivatives of eˣ and aˣ — proofs

Case 2 — a bigger base wears an $ln a$ coat

Case 3 — a shrinking base gives a NEGATIVE slope

Worked example Ex 3 (cell 3,

decay). Differentiate $h (x) = (\frac{1}{2})^{x}$ and find its slope at $x = 0$ . Forecast: this curve falls as $x$ grows. So its slope should be… negative. Let's confirm the sign comes out right.

$h^{'} (x) = (\frac{1}{2})^{x} ln \frac{1}{2}$ . Why this step? Same rule $a^{x} ln a$ ; nothing about it required $a > 1$ — only $a > 0$ , and $\frac{1}{2} > 0$ . ✓
Note $ln \frac{1}{2} = - ln 2 \approx - 0.6931$ . Why this step? $ln$ of a number below $1$ is negative — that negative is what makes the whole derivative negative, encoding "the curve goes down."
At $x = 0$ : $h^{'} (0) = 1 \cdot (- ln 2) = - ln 2 \approx - 0.6931$ . Why this step? $(\frac{1}{2})^{0} = 1$ ; the slope at $x = 0$ equals the personality number, here negative.

Verify: the function decreases everywhere, so a negative slope is correct. Notice its size $0.6931$ equals the slope of $2^{x}$ — mirror bases $2$ and $\frac{1}{2}$ have opposite-sign personality numbers because $ln \frac{1}{2} = - ln 2$ . ✓ The decay curve and its downward tangent are shown below. See Exponential Growth and Decay.

Case 4 — the degenerate base $a = 1$

Case 5 — a linear inside function

Worked example Ex 5 (cell 5,

chain rule / linear inside). Differentiate $e^{5 x}$ . Forecast: the answer is not $e^{5 x}$ . What extra factor sneaks in? (The slope of the inside, $5$ .)

Identify inside $u = 5 x$ , so $u^{'} = 5$ . Why this step? The exponent isn't a bare $x$ ; the Chain Rule needs the inside's slope.
Outside derivative of $e^{u}$ is $e^{u}$ ; multiply by $u^{'}$ : $\frac{d}{d x} e^{5 x} = e^{5 x} \cdot 5 = 5 e^{5 x}$ . Why this step? Chain rule = (outside slope) $\times$ (inside slope). Here the outside is self-derivative, so only the $5$ is added.

Verify (limit sanity): by the first-principles definition, the derivative is $lim_{h \to 0} \frac{e ^{5 (x + h)} - e ^{5 x}}{h} = lim_{h \to 0} \frac{e ^{5 x} ( e ^{5 h} - 1 )}{h}$ . Here we use the linearization $e^{t} \approx 1 + t$ for tiny $t$ (this is exactly the parent's defining fact $lim_{t \to 0} \frac{e ^{t} - 1}{t} = 1$ , i.e. $e^{t} - 1 \approx t$ ). With $t = 5 h$ : $e^{5 h} - 1 \approx 5 h$ , so the quotient $\approx \frac{e ^{5 x} \cdot 5 h}{h} = 5 e^{5 x}$ . ✓ Matches.

Case 6 — negative / non-linear inside

Worked example Ex 6 (cell 6). Differentiate

$q (x) = e^{- x^{2}}$ (the bell-curve shape). Forecast: the inside $- x^{2}$ has slope $- 2 x$ . So the answer should carry a factor $- 2 x$ — meaning the slope is $0$ at $x = 0$ and negative for $x > 0$ . Predict the sign before computing.

Inside $u = - x^{2}$ , so $u^{'} = - 2 x$ . Why this step? Non-linear exponent → its derivative is itself a function of $x$ , not a constant. Use the Power Rule on $- x^{2}$ .
Chain rule: $q^{'} (x) = e^{- x^{2}} \cdot (- 2 x) = - 2 x e^{- x^{2}}$ . Why this step? Outside self-derivative $e^{u}$ times the inside slope $- 2 x$ .
At $x = 1$ : $q^{'} (1) = - 2 (1) e^{- 1} = - 2/ e \approx - 0.7358$ . Why this step? Plug the point to get the actual tangent slope.

Verify: at $x = 0$ , $q^{'} (0) = - 2 (0) e^{0} = 0$ — the peak of the bell, where the tangent is flat. ✓ For $x > 0$ the factor $- 2 x$ is negative, so the curve descends on the right — exactly the bell's right side, drawn below.

Case 7 — general base with a function exponent

Worked example Ex 7 (cell 7). Differentiate

$r (x) = 2^{x^{2}}$ . Forecast: two things fight here — a base that isn't $e$ and a non-linear exponent. What's the safe move? Convert to base $e$ first, then chain-rule.

Rewrite: $2^{x^{2}} = e^{(l n 2) x^{2}}$ . Why this step? The clean chain rule lives in base $e$ ; any positive base becomes $a = e^{l n a}$ , so $2^{x^{2}} = e^{x^{2} l n 2}$ .
Inside $u = (ln 2) x^{2}$ , slope $u^{'} = 2 x ln 2$ . Why this step? $ln 2$ is a constant multiplier; differentiate $x^{2}$ by the power rule → $2 x$ , keep the constant.
$r^{'} (x) = e^{x^{2} l n 2} \cdot 2 x ln 2 = 2^{x^{2}} (2 x) ln 2$ . Why this step? This is the chain rule again — outside derivative times inside derivative: the outside $e^{u}$ differentiates to itself $e^{x^{2} l n 2}$ , and we multiply by the inside slope $u^{'} = 2 x ln 2$ from step 2. We then rewrite $e^{x^{2} l n 2}$ back as $2^{x^{2}}$ for a tidy answer.
At $x = 1$ : $r^{'} (1) = 2^{1} \cdot 2 \cdot ln 2 = 4 ln 2 \approx 2.7726$ . Why this step? Evaluate to a number for a concrete check.

Verify: at $x = 0$ the factor $2 x$ is $0$ , so $r^{'} (0) = 0$ — and $2^{x^{2}}$ has a flat minimum at $x = 0$ (its lowest point, value $1$ ). ✓

Case 8 — slope-at-a-point, read geometrically

Figure (Ex 8): the blue curve $y = e^{x}$ and its orange tangent $y = x + 1$ meet at the red point $(0, 1)$ . The tangent's slope ( $1$ ) equals the curve's height there ( $e^{0} = 1$ ) — the self-derivative property made visible. Away from $x = 0$ the straight line peels off, since the true curve keeps steepening.

Case 9 — real-world growth rate (with units!)

Worked example Ex 9 (cell 9,

word problem). A bacteria population is $P (t) = 100 e^{0.4 t}$ , where $t$ is in hours. How fast is it growing at $t = 3$ hours? Forecast: "how fast" means a derivative. The units of the answer will be bacteria per hour, not bacteria. Guess: will the rate be bigger than the current count or smaller? (Rate $= 0.4 \times$ count, so smaller.)

$P^{'} (t) = 100 e^{0.4 t} \cdot 0.4 = 40 e^{0.4 t}$ . Why this step? Inside $u = 0.4 t$ has slope $0.4$ ; chain rule multiplies the self-derivative by $0.4$ . The constant $100$ rides along untouched.
At $t = 3$ : $P^{'} (3) = 40 e^{1.2} \approx 40 \times 3.3201 \approx 132.8$ . Why this step? Evaluate at the requested time to get the instantaneous rate.

Verify (units + structure): $P^{'} (3) \approx 132.8$ bacteria per hour. Cross-check: the current count is $P (3) = 100 e^{1.2} \approx 332.0$ , and $P^{'} (3) = 0.4 \times P (3) = 0.4 \times 332.0 = 132.8$ . ✓ For any self-scaling exponential, growth rate $=$ (growth constant) $\times$ (current size) — the signature of Exponential Growth and Decay.

Case 10 — exam twist: run the rule backwards

Recall Case-checklist for exams

When you see an exponential to differentiate, ask in order: Is the base $e$ or some other positive number? ::: If not $e$ , pull out the personality number $ln a$ (or convert to $e^{(l n a) u}$ first). Is the exponent a bare $x$ or an inside function $u (x)$ ? ::: If an inside function, multiply by $u^{'} (x)$ — the chain rule. Is $0 < a < 1$ ? ::: Then $ln a < 0$ , so the slope is negative — this is decay. Is $a = 1$ ? ::: Then $ln 1 = 0$ , so the derivative is $0$ (it's just a constant). Do they want a number (slope at a point)? ::: Differentiate first, THEN substitute the point.

Connections

Derivatives of eˣ and aˣ — proofs — the parent this page drills.
Chain Rule — powers cells 5, 6, 7, 9.
Power Rule — used to differentiate the inside exponents like $- x^{2}$ , $x^{2}$ .
Natural Logarithm ln x — inverting $ln a = 2$ in Ex 10.
Exponential Growth and Decay — cells 3 and 9.
The number e — definitions — the $e^{x} \approx 1 + x$ tangent of Ex 8.
Limits — first principles definition of derivative — the sanity check in Ex 5.