Derivatives of ln x and logₐ(x): Question bank

Intuition How to use this page

Every item below is a trap — a statement that sounds right but hides a subtlety about which base, the domain, the chain rule, or the limiting behaviour of the log slope. Cover the answer, commit to a verdict, then reveal. If your reason differs from ours even when your verdict matches, you have found a gap. These test understanding, not arithmetic — for heavy computation see the other deep dives.

Throughout, when a log has something more complicated than a bare $x$ inside it, we call that inner expression the inside function and write it $u = u (x)$ ; its derivative is $u^{'} = \frac{d u}{d x}$ . Keeping $u$ named up front is what makes the chain rule traps below visible.

The one fact everything hangs on: $\frac{d}{d x} ln x = \frac{1}{x}$ , and for any other base you pay a toll of $\frac{1}{ln a}$ . See Derivatives of ln x and logₐ(x) for the full derivation.

The three pictures behind every trap

Before the questions, fix three images in your mind — most traps are just one of these three pictures misread.

Picture 1 — the slope $\frac{1}{x}$ is the height of $ln x$ 's steepness. As you slide right, $ln x$ climbs but the tangent line tilts down toward flat.

Figure — Derivatives of ln x and logₐ(x)

Picture 2 — changing the base only stretches the curve vertically. A base above $e$ squashes it (gentler slope); a base between $1$ and $e$ stretches it (steeper slope); a base below $1$ flips it upside down (negative slope).

Picture 3 — the standard limit $lim_{t \to 0} \frac{l n ( 1 + t )}{t} = 1$ is just the slope of $ln x$ at $x = 1$ . Near the point $(1, 0)$ the log curve hugs the line $y = x - 1$ , whose slope is exactly $1$ .

True or false — justify

The slope of

ln x

is always positive

True —

ln x

is only defined for

x > 0

, and there

\frac{1}{x} > 0

, so the curve is forever climbing (never flat, never falling).

ln x

gets steeper and steeper as

x \to \infty

False — the slope

\frac{1}{x}

shrinks toward

0

, so the curve flattens; it keeps rising but ever more gently.

\frac{d}{d x} lo g_{a} x = \frac{1}{x}

for every base

a

False — only for

a = e

. In general it is

\frac{1}{x l n a}

; the extra factor

\frac{1}{l n a}

1

only when

ln a = 1

, i.e.

a = e

lo g_{a} x

and

ln x

are the same curve up to a vertical stretch

True —

lo g_{a} x = \frac{1}{l n a} ln x

, a constant multiple, so it is

ln x

scaled vertically by

\frac{1}{l n a}

; that is exactly why their slopes differ by that same constant.

The slope of

ln x

equals exactly

1

at some point

True — at

x = 1

, since

\frac{1}{1} = 1

. This is the only place where

ln x

rises at

4 5^{\circ}

For

a > 1

the slope of

lo g_{a} x

is smaller than that of

ln x

at every

x

False — this only holds when

a > e

. For

1 < a < e

we have

0 < ln a < 1

, so

\frac{1}{x l n a} > \frac{1}{x}

: the slope is actually larger. The break-even base is exactly

a = e

, where

ln a = 1

\frac{d}{d x} ln (3 x)

is bigger than

\frac{d}{d x} ln x

because of the factor

3

False —

ln (3 x) = ln 3 + ln x

, and the constant

ln 3

vanishes on differentiating, so both derivatives equal

\frac{1}{x}

\frac{d}{d x} ln x = \frac{1}{x}

holds for negative

x

too

False —

ln x

is undefined for

x \leq 0

, so its derivative simply does not exist there. (It is

ln ∣ x ∣

that extends to

x < 0

\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}

for all

x \neq = 0

True — on

x < 0

ln ∣ x ∣ = ln (- x)

, and by the chain rule

\frac{- 1}{- x} = \frac{1}{x}

; on

x > 0

it is plainly

\frac{1}{x}

. So one clean formula covers both sides.

ln x

has an inflection point somewhere on its domain

False — its second derivative is

- \frac{1}{x ^{2}}

, which is negative for every

x > 0

and never zero. The concavity never switches, so there is no inflection point; the curve is concave-down everywhere.

Spot the error

\frac{d}{d x} lo g_{2} x = \frac{1}{x}

because logs differentiate to one-over-

x

The base is wrong. Only base

e

gives

\frac{1}{x}

; here

\frac{d}{d x} lo g_{2} x = \frac{1}{x l n 2}

. The claim forgot the toll

\frac{1}{l n a}

\frac{d}{d x} ln (5 x) = \frac{5}{5 x} = \frac{5}{x}

The final cancellation was botched:

\frac{5}{5 x} = \frac{1}{x}

, not

\frac{5}{x}

. Here the inside function is

u = 5 x

, so

u^{'} = 5

cancels the

5

inside.

\frac{d}{d x} ln (x^{2}) = \frac{1}{x ^{2}}

The chain rule was dropped. With inside function

u = x^{2}

\frac{u ^{'}}{u} = \frac{2 x}{x ^{2}} = \frac{2}{x}

. (Equivalently

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

"Since

lo g_{a} x = \frac{l n x}{l n a}

, its derivative is

\frac{1/ x}{1/ a} = \frac{a}{x}

ln a

is a fixed number, not a function of

x

, so it is not differentiated to

\frac{1}{a}

. Treat

\frac{1}{l n a}

as a constant multiplier: the answer is

\frac{1}{x l n a}

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

This confuses

(ln x)^{2}

with

ln (x^{2})

. For the square of

ln x

use the chain rule:

2 ln x \cdot \frac{1}{x} = \frac{2 l n x}{x}

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

You cannot use the power rule when the exponent is a variable. Use logarithmic differentiation:

ln y = x ln x \Rightarrow y^{'} = x^{x} (ln x + 1)

. See Logarithmic Differentiation.

\frac{d}{d x} ln (cos x) = \frac{1}{cos x}

The chain rule was cut short. With inside function

u = cos x

, its derivative

u^{'} = - sin x

is missing. Correct:

\frac{- s i n x}{c o s x} = - tan x

Why questions

Why does base

e

give the simplest derivative?

Because

ln e = 1

, so the toll factor

\frac{1}{l n a}

becomes

1

and disappears, leaving the bare

\frac{1}{x}

. See Derivative of e^x and a^x.

Why does the constant inside

ln (3 x)

not affect the derivative?

By Logarithm Laws,

ln (3 x) = ln 3 + ln x

; a constant adds a flat vertical shift, which has zero slope, so it contributes nothing to the derivative.

Why do we use implicit differentiation to find

\frac{d}{d x} ln x

Because we only know the slope of

e^{x}

(it is

e^{x}

), so we start from

e^{y} = x

and let Implicit Differentiation transfer that known slope into the slope of the inverse.

Why is

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

and not just

\frac{1}{u}

Here

u = u (x)

is the inside function and

u^{'}

its rate of change. Because

u

itself depends on

x

, the Chain Rule multiplies the outer slope

\frac{1}{u}

by the inner slope

u^{'}

to account for how fast the inside is moving.

Why does the standard limit

lim_{t \to 0} \frac{l n ( 1 + t )}{t} = 1

appear in the first-principles proof?

It is the derivative of

ln

x = 1

, and it equals

1

because

(1 + t)^{1/ t} \to e

and

ln e = 1

. See Standard Limit (1+t)^{1/t} → e.

Why can two different-looking methods (algebra vs. chain rule) give the same derivative for

ln (3 x)

They describe the same function two ways; correct maths must agree. The chain rule gives

\frac{3}{3 x} = \frac{1}{x}

; the log law gives

\frac{d}{d x} (ln 3 + ln x) = \frac{1}{x}

. Any disagreement signals an algebra slip.

Why is the slope of

lo g_{10} x

so tiny for large

x

Because

ln 10 \approx 2.30

, so the slope

\frac{1}{x l n 10}

is already divided by

2.30

and then further crushed by large

x

— the base-

10

curve is a gentle, heavily-flattened version of

ln x

Why is

ln x

concave down everywhere?

Its second derivative is

- \frac{1}{x ^{2}}

, which is negative for all

x > 0

; a negative second derivative means the slope is always decreasing, so the tangent lines keep tilting down and the curve bends downward throughout.

Edge cases

What is

\frac{d}{d x} ln x

x \to 0^{+}

The slope

\frac{1}{x} \to + \infty

: the curve rises almost vertically just to the right of

0

, matching how

ln x \to - \infty

there — an infinitely steep drop-off at the domain edge.

Is there any

x

where

ln x

has zero slope?

No —

\frac{1}{x}

is never

0

for finite

x > 0

. The slope only approaches

0

x \to \infty

but never reaches it, so

ln x

has no horizontal tangent and no maximum.

Does

\frac{d}{d x} lo g_{a} x

change sign for a base

0 < a < 1

Yes — then

ln a < 0

, so

\frac{1}{x l n a} < 0

: the log is decreasing. A base below

1

flips the curve, and the negative slope reflects that.

What happens to

\frac{1}{x l n a}

as the base

a \to 1

ln a \to 0

, so the derivative blows up to

\pm \infty

. This mirrors the rule that

lo g_{a}

is undefined at

a = 1

— the closer the base is to

1

, the wilder the log's steepness.

x = 1

, do

ln x

and

lo g_{a} x

have the same slope?

No (unless

a = e

). Both pass through the point

(1, 0)

, but their slopes there are

1

and

\frac{1}{l n a}

respectively — same height, different steepness.

How does the slope

\frac{1}{x}

ln ∣ x ∣

behave on the two sides of

0

From the right (

x \to 0^{+}

) it runs to

+ \infty

, but from the left (

x \to 0^{-}

) it runs to

- \infty

— the two one-sided slopes head to opposite infinities, so they cannot join up at

x = 0

Does

ln ∣ x ∣

have a derivative at

x = 0

No —

ln ∣ x ∣

is undefined at

x = 0

(and blows to

- \infty

nearby), so no slope exists there. The formula

\frac{1}{x}

holds only for

x \neq = 0

Does the concavity of

lo g_{a} x

ever flip for different bases?

For

a > 1

it is concave down (

- \frac{1}{x ^{2} l n a} < 0

); for

0 < a < 1

ln a < 0

makes the second derivative positive, so the flipped curve is concave up — but in neither case is there an inflection point, since the sign never changes on

x > 0

Recall One-line self-check

Toll for base $a$ ? ::: The factor $\frac{1}{l n a}$ multiplying $\frac{1}{x}$ .
Only base with no toll? ::: $a = e$ , since $ln e = 1$ .
Allowed values of the base $a$ ? ::: $a > 0$ and $a \neq = 1$ .
Slope of $ln x$ as $x \to \infty$ ? ::: Shrinks to $0$ (flattens, never negative).
Second derivative of $ln x$ ? ::: $- \frac{1}{x ^{2}}$ , negative everywhere — concave down, no inflection.
Formula valid for $x < 0$ ? ::: Only $ln ∣ x ∣$ , giving $\frac{1}{x}$ for all $x \neq = 0$ .
Base making $lo g_{a} x$ steeper than $ln x$ ? ::: Any $1 < a < e$ , since then $0 < ln a < 1$ .

Connections

Derivatives of ln x and logₐ(x) — the parent, with full derivations.
Derivative of e^x and a^x — the inverses whose slopes seed everything here.
Chain Rule — the tool most of these traps abuse or forget.
Implicit Differentiation — why base $e$ gives $\frac{1}{x}$ .
Logarithm Laws — why constants inside a log vanish.
Logarithmic Differentiation — the right way to attack $x^{x}$ .
Standard Limit (1+t)^{1/t} → e — behind the first-principles proof.