Derivatives of eˣ and aˣ — proofs: Question bank

Before you start, three tools are assumed. If any feels shaky, revisit them:

The derivative from first principles — slope as a limit — from Limits — first principles definition of derivative.
The chain rule — differentiating a function inside a function — from Chain Rule.
The natural log $ln$ — the inverse of $e^{(\cdot)}$ — from Natural Logarithm ln x.

The one picture behind every trap

Almost every trap on this page turns on one number. Let me build it here so you never have to leave the page.

Here is why that one number controls the whole derivative. Using the exponent law $a^{x + h} = a^{x} a^{h}$ ,

\frac{d}{d x} a^{x} = h \to 0 lim \frac{a ^{x} a ^{h} - a ^{x}}{h} = a^{x} M (a) h \to 0 lim \frac{a ^{h} - 1}{h} = M (a) a^{x} .

The $a^{x}$ slides out because it has no $h$ in it. So the derivative of $a^{x}$ is just $a^{x}$ scaled by its slope-at-zero. And that scaling number turns out to be exactly the natural log:

\frac{d}{d x} a^{x} = a^{x} ln a, M (a) = ln a

because writing $a = e^{l n a}$ gives $a^{x} = e^{(l n a) x}$ , whose derivative (chain rule) is $e^{(l n a) x} \cdot ln a = a^{x} ln a$ .

The figure below shows what $M (a) = ln a$ means geometrically: the tangent line drawn at $(0, 1)$ tilts up more for bigger bases.

Figure — Derivatives of eˣ and aˣ — proofs

The second figure shows curvature — why decay curves ( $0 < a < 1$ ) still bend upward.

True or false — justify

Each line: decide true or false, then give the reason. Bare "true/false" earns nothing.

e^{x}

is the only function equal to its own derivative.

False. Any constant multiple

C e^{x}

also satisfies

\frac{d}{d x} (C e^{x}) = C e^{x}

; the family

C e^{x}

(including

C = 0

) is the full set, but

e^{x}

is not alone.

\frac{d}{d x} a^{x} = a^{x} ln a

, then a bigger base always gives a steeper curve everywhere.

True.

ln a

grows with

a

and

a^{x} > 0

always, so a larger base makes the slope factor larger at every

x

— the whole curve is steeper.

For

0 < a < 1

the function

a^{x}

is increasing.

False. Then

ln a < 0

, so

\frac{d}{d x} a^{x} = a^{x} ln a < 0

everywhere (since

a^{x} > 0

): the curve falls — this is decay.

The slope of

a^{x}

x = 0

equals its height there.

False in general. Height at

x = 0

a^{0} = 1

, but slope is

M (a) = ln a

(the slope constant built above); they are equal only when

ln a = 1

, i.e.

a = e

1^{x}

has derivative

1^{x} ln 1 = 0

True.

1^{x} = 1

is a flat horizontal line, and

ln 1 = 0

, so its slope is

0

everywhere — perfectly consistent.

The constant

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

depends on both

a

and

x

False. The variable

x

never enters this limit;

M (a)

is a pure number fixed by the base

a

alone (it turns out to be

ln a

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

False. Inner function is

- x

with derivative

- 1

, so by the chain rule

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

— a decreasing curve.

e^{x}

can be negative for large negative

x

False.

e^{x} > 0

for every real

x

; as

x \to - \infty

it approaches

0

from above but never touches or crosses it.

Spot the error

Each line states a "proof" or claim with a hidden flaw. Name the flaw in one sentence.

\frac{d}{d x} a^{x} = x a^{x - 1}

, by the power rule."

The power rule

\frac{d}{d x} x^{n} = n x^{n - 1}

needs a fixed exponent and varying base; here the base is fixed and the exponent varies, the opposite situation, so it does not apply — the answer is

a^{x} ln a

\frac{d}{d x} e^{5 x} = e^{5 x}

because

e^{anything}

is its own derivative."

The self-derivative rule holds only when the exponent is exactly

x

; here the inner function is

5 x

, so the chain rule adds a factor

5

, giving

5 e^{5 x}

"Since

a = e^{l n a}

, we get

M (a) = lo g_{10} a

The rewrite uses the natural log (

a = e^{l n a}

is only true with

ln

), so the constant must be

ln a

, not

lo g_{10} a

\frac{d}{d x} 2^{x^{2}} = 2^{x^{2}} ln 2

, just applying

a^{x} ln a

The exponent here is

x^{2}

, not

x

, so this is not the plain rule; converting to

e^{x^{2} l n 2}

and using the chain rule gives an extra inner derivative

2 x ln 2

, so the answer is

2^{x^{2}} (2 x) ln 2

e^{x}

is its own derivative, so its second derivative must be something new."

Differentiating

e^{x}

again just returns

e^{x}

; every derivative of

e^{x}

e^{x}

, so nothing new ever appears.

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

0

because the numerator

\to 0

The denominator

h \to 0

too, so this is a

\frac{0}{0}

indeterminate form; the ratio settles to

ln a

, not

0

— you cannot read it off from the numerator alone.

"Because

\frac{d}{d x} a^{x} = a^{x} ln a

, differentiating

a^{x}

twice gives

a^{x} (ln a)

again."

Each differentiation multiplies by another factor

ln a

, so the second derivative is

a^{x} (ln a)^{2}

, not

a^{x} ln a

"Rewriting

a^{x} = (e^{l n a})^{x} = e^{l n a} \cdot x

Powers do not turn into products;

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

by multiplying exponents, not

e^{l n a} \cdot x

Why questions

Answer each in one or two sentences of genuine reasoning.

Why does

a^{x}

slide out of the difference-quotient limit while

a^{h} - 1

stays inside?

Because

a^{x + h} = a^{x} a^{h}

, and

a^{x}

carries no

h

— it is constant as

h \to 0

— so it factors out, leaving only the

h

-dependent piece

\frac{a ^{h} - 1}{h}

inside the limit.

Why is

e

called the natural base rather than, say,

10

Because

e

is the unique base whose slope-at-height-1 constant

M (a)

equals exactly

1

, making

e^{x}

its own derivative with no extra factor — the cleanest possible growth law.

Why must we convert

a^{x}

to base

e

before differentiating?

The chain rule only gives us clean derivatives through the known rule

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

; writing

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

puts the expression in that form so the derivative machinery applies.

Why does a base slightly larger than

e

give a slope constant slightly larger than

1

Because

M (a) = ln a

and

ln

is increasing with

ln e = 1

; a base just above

e

has

ln a

just above

1

, so the curve is a touch steeper at

x = 0

Why can't the power rule and the exponential rule ever agree for

2^{x}

The power rule would give

x 2^{x - 1}

, which grows without bound relative to

2^{x}

, whereas the true derivative

2^{x} ln 2

stays a fixed multiple of the function — they describe fundamentally different behaviours.

Why does the limit definition of

e

(

M (e) = 1

) match the compound-interest definition

e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}

M (e) = 1

means

e^{h} \approx 1 + h

for small

h

, so

e \approx (1 + h)^{1/ h}

; setting

h = 1/ n

and letting

n \to \infty

recovers exactly the compound-interest limit. See The number e — definitions.

Why is the constant

M (a)

the same as "the slope of

a^{x}

x = 0

Plugging

x = 0

into the general derivative

\frac{d}{d x} a^{x} = M (a) a^{x}

gives

M (a) \cdot a^{0} = M (a) \cdot 1 = M (a)

, so

M (a)

literally is the slope at the origin point

(0, 1)

Edge cases

The degenerate and boundary scenarios the rules must survive.

What is

\frac{d}{d x} a^{x}

when

a = 1

1^{x} = 1

is constant, and

ln 1 = 0

, so the derivative is

1^{x} \cdot 0 = 0

— matching the flat line, no contradiction.

\frac{d}{d x} a^{x} = a^{x} ln a

valid for

a = 0

No;

0^{x}

is not a well-defined positive exponential (undefined at

x \leq 0

and

ln 0

diverges to

- \infty

), so the rule requires

a > 0

What happens to

\frac{d}{d x} a^{x}

a \to e

from either side?

The slope factor

ln a \to ln e = 1

continuously, so the derivative smoothly approaches

e^{x}

—

e

sits as the single balance point where

M = 1

For

a

between

0

and

1

, is the second derivative of

a^{x}

positive or negative?

Positive:

\frac{d ^{2}}{d x ^{2}} a^{x} = a^{x} (ln a)^{2}

, and squaring

ln a

makes it positive even though

ln a < 0

, so decay curves are convex (they flatten out, curving upward).

x = 0

, which is steeper,

e^{- x}

(1/2)^{x}

e^{- x}

has slope

ln (e^{- 1}) = - 1

;

(1/2)^{x}

has slope

ln (1/2) \approx - 0.693

. So

e^{- x}

falls faster (

- 1

is more negative than

- 0.693

), i.e. it is steeper downward at the origin.

x \to - \infty

, what is the slope of

e^{x}

and does the curve ever flatten completely?

The slope equals the height

e^{x}

, which

\to 0^{+}

; the curve flattens toward horizontal but never reaches zero slope at any finite

x

Does

\frac{d}{d x} a^{x} = a^{x} ln a

ever give a slope of exactly

1

at a point other than through

a = e

Yes; for any base

a > 1

, set

a^{x} ln a = 1

and solve — some

x

satisfies it, since

a^{x} ln a

ranges over all positive values, but only

a = e

makes it happen at $x = 0$ .

Connections

Derivatives of eˣ and aˣ — proofs — the parent proof these traps stress-test.
Chain Rule — the tool behind every "forgot the inner derivative" trap.
Natural Logarithm ln x — why $M (a)$ must be $ln$ , not $lo g_{10}$ .
The number e — definitions — the two definitions cross-checked in the "why" section.
Exponential Growth and Decay — the sign-of- $ln a$ edge cases in context.
Power Rule — the rule these traps repeatedly warn not to misapply.