Derivative from first principles — difference quotient definition: Question bank

This is the conceptual companion to the parent note. Nothing here needs heavy algebra; it is all about what the symbols mean.

Spot the error

Each line contains one flawed step. Name the mistake.

f (x) = x^{2}

, so

f (x + h) = x^{2} + h

." What went wrong?

Only the first

x

was replaced. Every

x

becomes the block

(x + h)

f (x + h) = (x + h)^{2} = x^{2} + 2 x h + h^{2}

\frac{2 x h + h ^{2}}{h} = 2 x h + h

." What went wrong?

Only one term was divided by

h

. Divide both terms:

\frac{2 x h}{h} + \frac{h ^{2}}{h} = 2 x + h

\frac{x + h - x}{h}

has no

h

on top, so the limit doesn't exist." What went wrong?

The

h

is hidden. Multiply by the conjugate to surface it:

\frac{( x + h ) - x}{h ( x + h + x )} = \frac{1}{x + h + x}

, which has a fine limit.

"For

f (x) = 1/ x

we get

\frac{- h}{x ( x + h )}

, and dividing by

h

leaves

\frac{- h}{x ( x + h )}

still." What went wrong?

Dividing by

h

cancels the

h

\frac{1}{h} \cdot \frac{- h}{x ( x + h )} = \frac{- 1}{x ( x + h )}

. The

h

in the numerator must be crossed out.

"Since the two points merge, the run is

0

, so we can never define a slope." What went wrong?

The points never actually merge —

h

approaches

0

but stays nonzero, so the run stays nonzero throughout the limit. The limit captures the trend, not the forbidden endpoint.

lim_{h \to 0} (2 x + h) = 2 x + 0 = 2 x

, but only because we plugged

h = 0

." What is subtly wrong with the wording?

The substitution shortcut is legal only because

2 x + h

is now a continuous, well-defined expression (no

h

in a denominator). It's the safe re-plug of Step 4, not the illegal Step-1 plug.

Why questions

Say why, in the sense the topic demands.

Why do we introduce a second point at

x + h

instead of using calculus "at" one point directly?

Slope is defined as rise-over-run, which needs a change; one point gives no change. The buddy point manufactures a run we can later shrink. See Average vs Instantaneous Rate of Change.

Why must we factor and cancel the

h

before taking the limit?

Before cancelling, plugging

h = 0

gives

\frac{0}{0}

, which carries no information. Cancelling removes the shared zero and reveals the finite limit hiding underneath — the whole point of Indeterminate forms 0 over 0.

Why does the secant line "become" the tangent as

h \to 0

As the buddy point slides toward

x

, the cutting line pivots about

(x, f (x))

until it only touches the curve. Its slope, the difference quotient, settles on the tangent's slope — the derivative.

Why is the derivative called an instantaneous rate of change even though it uses two points?

The two points are a scaffold; the limit erases the gap between them, so the surviving number describes the rate at a single instant, not over an interval. See Average vs Instantaneous Rate of Change.

Why does the conjugate trick work only for square-root (and similar) expressions?

It exploits

(a - b) (a + b) = a - b

, which turns a stubborn difference of roots into a plain difference that contains the

h

we need to cancel. Without roots there is nothing for the conjugate to rationalise.

Why does a straight line's first-principles slope not depend on

h

The rise of a line is exactly proportional to its run, so

\frac{rise}{h}

simplifies to the constant slope

m

before any limit is taken — every secant is the tangent.

Edge cases

Boundary and degenerate inputs the algorithm must survive.

What happens to the difference quotient of

f (x) = 1/ x

x \to 0

f^{'} (x) = - 1/ x^{2}

blows up to

- \infty

; the function isn't even defined at

x = 0

, so no derivative exists there. The formula silently excludes

x = 0

What is

f^{'} (0)

for

f (x) = x

, and why is it special?

f^{'} (x) = \frac{1}{2 x} \to + \infty

x \to 0^{+}

; the tangent is vertical, so no finite slope exists at

0

even though

f

is defined there. See Continuity and Differentiability.

Can

h

approach

0

from only one side for a one-sided derivative?

Yes — at an endpoint of a domain (e.g.

x

x = 0

) only

h \to 0^{+}

is available, giving a one-sided derivative via a one-sided limit.

For

f (x) = ∣ x ∣

, what do the left and right difference quotients give at

x = 0

The right side (

h > 0

) gives

+ 1

, the left side (

h < 0

) gives

- 1

; since they disagree, the two-sided limit fails and

f^{'} (0)

does not exist — a corner.

What is the derivative of

f (x) = x^{2}

exactly at the vertex

x = 0

f^{'} (0) = 2 (0) = 0

— the slope is zero because the bottom of the parabola is momentarily flat, matching the geometry.

f (x + h) - f (x)

does not go to

0

h \to 0

, what does that tell you?

f

is discontinuous at

x

; then

\frac{nonzero}{0}

blows up and no derivative exists — a concrete illustration of "differentiable

\Rightarrow

continuous." See Continuity and Differentiability.

Is the difference quotient ever exactly equal to the derivative for some nonzero

h

Yes for linear

f

(it equals the slope for all

h

); for curved

f

a specific nonzero

h

can match by the Mean Value Theorem, but in general the quotient only approaches the derivative. See Average vs Instantaneous Rate of Change.

Connections

Derivative from first principles — difference quotient definition — the parent this bank drills.
Limits — formal definition and one-sided limits — one-sided limits power the edge cases.
Indeterminate forms 0 over 0 — the reason cancellation must precede substitution.
Secant and Tangent lines — the geometry behind the true/false items.
Continuity and Differentiability — corners and vertical tangents.
Average vs Instantaneous Rate of Change — secant vs. tangent framing.

Question bank — Derivative from first principles — difference quotient definition

True or false — justify