Natural exponential function eˣ — graph, derivative preview: Worked examples

This page is a case-hunt. The parent note built the graph and the slope-equals-height fact. Here we ask: what kinds of question can this topic actually throw at you? We list them all, then kill each one with a fully worked example.

Before we start, one reminder of the only two facts we use:

$e^{x}$ is always positive — a positive base to any power never dips to zero or below.
The slope at any point equals the height there: $\frac{d}{d x} e^{x} = e^{x}$ .

Everything below is just those two facts wearing different costumes.

The scenario matrix

Every problem in this topic lands in one of these cells. The right column names the example that clears it.

Cell	The scenario	Cleared by
A. Positive exponent	evaluate/interpret $e^{x}$ for $x > 0$	Ex 1
B. Negative exponent	$e^{- x}$ — reciprocal, stays positive	Ex 2
C. Zero / degenerate input	$e^{0}$ , slope at $x = 0$ , tangent there	Ex 3
D. Limiting behaviour	$x \to + \infty$ and $x \to - \infty$	Ex 4
E. Scaled exponent (chain)	$e^{k x}$ , including negative $k$	Ex 5
F. Comparing slopes geometrically	which point is steeper, tangent lines	Ex 6
G. Real-world word problem	continuous growth/decay, units	Ex 7
H. Exam-style twist	solve $e^{x} =$ something, or a trap	Ex 8
I. Horizontal translation	$e^{x + b}$ — a sideways shift = vertical stretch	Ex 9

Worked Examples

Cell A — positive exponent

Cell B — negative exponent

Cell C — zero / degenerate input

Cell D — limiting behaviour

Cell E — scaled exponent (chain rule), both signs of $k$

Worked example 5 — Differentiate

$y = e^{3 x}$ and $y = e^{- 0.5 x}$ Forecast: For $e^{3 x}$ , does the derivative pick up a factor of $3$ , or $\frac{1}{3}$ , or nothing?

Steps ( $e^{3 x}$ ):

Inner function $u = 3 x$ , outer $y = e^{u}$ . Why this step? Chain rule needs us to see the composition: an exponential of a linear function.
$\frac{d y}{d u} = e^{u}$ , $\frac{d u}{d x} = 3$ . Why this step? Outer derivative is itself; inner derivative of $3 x$ is the constant $3$ .
Multiply: $\frac{d y}{d x} = e^{3 x} \cdot 3 = 3 e^{3 x}$ . Why this step? The chain rule multiplies the two rates.

Steps ( $e^{- 0.5 x}$ ):

Set inner function $u = - 0.5 x$ , so $\frac{d u}{d x} = - 0.5$ . Why this step? Same decomposition as before — we split the composite into inner ( $- 0.5 x$ ) and outer ( $e^{u}$ ) so the chain rule can act; the point is to test what a negative inner slope does.
$\frac{d y}{d x} = e^{- 0.5 x} \cdot (- 0.5) = - 0.5 e^{- 0.5 x}$ . Why this step? A negative inner slope makes the whole derivative negative — this function is decaying, its graph slopes downward.

Verify: at $x = 0$ , slope of $e^{3 x}$ is $3 e^{0} = 3$ ; slope of $e^{- 0.5 x}$ is $- 0.5 e^{0} = - 0.5$ . Positive vs negative — a rising curve vs a falling one. ✓

Cell F — comparing slopes geometrically

Cell G — real-world word problem (units!)

Worked example 7 — Bacteria growing continuously

A colony starts at $500$ cells and grows continuously so its count is $N (t) = 500 e^{0.2 t}$ where $t$ is in hours. Find the count and the growth rate at $t = 3$ h.

Forecast: Will the growth rate (cells per hour) be bigger or smaller than the initial $500$ ? Guess before computing.

Steps:

Count: $N (3) = 500 e^{0.2 \times 3} = 500 e^{0.6} \approx 500 \times 1.822 = 911.1$ cells. Why this step? Substitute $t = 3$ ; $e^{0.6}$ is the growth factor over 3 hours.
Rate: $N^{'} (t) = 500 \cdot 0.2 e^{0.2 t} = 100 e^{0.2 t}$ (chain rule, Exponential growth and decay models). Why this step? The derivative of $e^{0.2 t}$ pulls out the $0.2$ ; multiply by the $500$ out front.
$N^{'} (3) = 100 e^{0.6} \approx 100 \times 1.822 = 182.2$ cells per hour. Why this step? Evaluate the rate function at $t = 3$ ; units are cells/hour because we differentiated with respect to $t$ in hours.

Verify: $N^{'} (3) = 0.2 \times N (3) = 0.2 \times 911.1 = 182.2$ ✓ — the rate is exactly $0.2$ times the current population, the signature of continuous growth. And $182.2 < 500$ : at $t = 3$ the count has grown but its hourly rate is smaller than the initial population — the "slope = a fixed fraction of height" logic.

Cell H — exam-style twist (solve an equation / spot the trap)

Worked example 8 — Solve

$e^{2 x} = 7.389$ , and spot the differentiation trap Forecast: From Example 1 we found $e^{2} \approx 7.389$ . So what must $2 x$ equal? Guess $x$ before solving.

Steps (solve):

Recognise $7.389 \approx e^{2}$ , so the equation is $e^{2 x} = e^{2}$ . Why this step? Matching the right side to a power of $e$ lets us equate exponents.
Equal bases ⇒ equal exponents: $2 x = 2$ , so $x = 1$ . Why this step? $e^{x}$ is one-to-one (it never repeats a value), so $e^{a} = e^{b} \Rightarrow a = b$ — the job of Natural logarithm ln x as inverse of e^x.

The trap: a student differentiates $y = e^{2 x}$ as " $2 x e^{2 x - 1}$ " using the power rule. 3. Why the trap is wrong: the power rule needs a variable base, constant exponent ( $x^{n}$ ). Here the variable is in the exponent, so we use the chain rule: $\frac{d}{d x} e^{2 x} = 2 e^{2 x}$ .

Verify: solution check $e^{2 \times 1} = e^{2} = 7.389$ ✓. Derivative check at $x = 0$ : correct answer gives $2 e^{0} = 2$ ; the wrong "power-rule" answer gives $2 (0) e^{- 1} = 0$ — clearly absurd for a curve rising at $x = 0$ . ✓ Trap avoided.

Cell I — horizontal translation

Worked example 9 — What does

$e^{x + 2}$ look like versus $e^{x}$ ? Forecast: Shifting the input $x \to x + 2$ moves the graph left by $2$ . But there is a slicker way to see it. Guess: is $e^{x + 2}$ just $e^{x}$ moved, or also stretched?

Steps:

Split the exponent: $e^{x + 2} = e^{x} \cdot e^{2}$ . Why this step? The index law $e^{a + b} = e^{a} e^{b}$ turns a sideways shift into a constant multiplier $e^{2} \approx 7.389$ out front.
So $e^{x + 2} = 7.389 e^{x}$ — the same curve stretched vertically by $\approx 7.389$ . Why this step? A shift-left of an exponential is indistinguishable from a vertical stretch — a special quirk of $e^{x}$ that no ordinary curve shares.
Its derivative: $\frac{d}{d x} e^{x + 2} = e^{x + 2}$ (still its own derivative), or equally $7.389 e^{x}$ . Why this step? Chain rule with inner slope $1$ leaves the derivative unchanged; the constant $e^{2}$ just rides along.

Verify: at $x = 0$ , $e^{0 + 2} = e^{2} \approx 7.389$ , and $7.389 e^{0} = 7.389$ ✓ — the "shift left by 2" and the "stretch by $e^{2}$ " give the identical value everywhere.

Active Recall

Recall Which cell does "estimate

$e^{- 2}$ " belong to, and what's the answer? Cell B (negative exponent). $e^{- 2} = 1/ e^{2} \approx 0.135$ , positive.

Slope of

e^{x}

x = 2

versus

x = 1

— which is steeper and by what factor?

x = 2

, by a factor of

e \approx 2.718

Solve

e^{2 x} = e^{2}

x = 1

(equal bases ⇒ equal exponents)

Growth rate of

N = 500 e^{0.2 t}

as a fraction of

N

0.2 N

(rate is

0.2

times current size)

Why is

e^{- x}

never negative?

It equals

1/ e^{x}

, positive over positive

Derivative of

e^{- 0.5 x}

- 0.5 e^{- 0.5 x}

(chain rule; negative slope = decay)

What does a shift

e^{x + 2}

equal as a stretch?

e^{2} e^{x} \approx 7.389 e^{x}

(shift left = vertical stretch)

Connections

Natural exponential function eˣ — graph, derivative preview (parent)
Euler's number e — definition & limit
General exponential a^x and its derivative
Natural logarithm ln x as inverse of e^x
Chain rule
Exponential growth and decay models
Taylor series of e^x

Worked examples — Natural exponential function eˣ — graph, derivative preview