Derivatives from difference quotients

A short trip gives an average rate, not yet an instantaneous one

A moving object's position is s(t)=t² metres, where t≥0 is measured in seconds. At t=1 its position is 1 metre; at t=3/2 it is 9/4 metres. The position change is 5/4 metres over 1/2 second, so the average velocity is (5/4)/(1/2)=5/2 metres per second.

Take a shorter interval ending at t=5/4. The change from t=1 is 25/16−1=9/16 metres over 1/4 second, giving 9/4 metres per second. These two averages differ. To ask for velocity at the single time 1, we need the limiting average as the other time approaches 1.

A positive increment h moves the second time after 1; a negative h moves it before 1, provided 1+h≥0. For s, the exact quotient ((1+h)²−1)/h=2+h holds for h≠0. As h becomes arbitrarily small from either permitted side, the quotient tends to 2 metres per second.

Review functions and their actual values, signed square expansion and real roots using the preparation links. First follow this example, the square and the corner; return to the quantified definition and continuity argument afterward. Every limit concept needed here is recalled locally, so an unpublished calculus route is not a prerequisite.

Two permitted points define one signed secant rate

Let f:D→ℝ, with a∈D. Choose h≠0 such that a+h∈D. The output change is f(a+h)−f(a), the input change is h, and their quotient is Qₐ(h)=(f(a+h)−f(a))/h. It is the slope of the secant line through the two graph points (a,f(a)) and (a+h,f(a+h)).

An increment is signed. For the square at a=1 with h=−1/2, the output change is 1/4−1=−3/4 and the quotient is (−3/4)/(−1/2)=3/2. Reversing the endpoints changes both differences' signs and leaves their ratio unchanged; it does not make the slope negative automatically.

The quotient has output units divided by input units. Metres divided by seconds give metres per second; cost in euros divided by number of items gives euros per item. If a position derivative is negative, motion follows the negative coordinate direction. Speed is the absolute value of that signed velocity.

The equivalent input form is (f(x)−f(a))/(x−a), with x∈D and x≠a. Put h=x−a to pass between the forms. Neither original quotient is defined at the zero denominator. Cancelling a nonzero factor can simplify permitted values without defining that missing quotient.

A derivative requires one finite limit for every sufficiently small increment

We call a a two-sided domain accumulation point when inputs of D approach a from both below and above: for every δ>0 there is a permitted x in each of (a−δ,a) and (a,a+δ). This condition prevents a derivative statement from being vacuous when no nearby input exists. The value f(a) must also exist.

In this lesson, the finite two-sided derivative f′(a)=L means that both domain sides are available and Qₐ(h) tends to the same real number L as the permitted nonzero increments tend to zero. On an open interval around a, this is the usual derivative. The domain-relative formulation also permits gaps, but tests every nearby input actually in D.

Explicitly, for every output-per-input error ε>0 there must be an input tolerance δ>0 such that every h with a+h∈D and 0<|h|<δ satisfies |Qₐ(h)−L|<ε. One δ must work for all those increments. It may depend on ε, a and the rule, but cannot be chosen separately after each h is seen. The finite limit is unique: if L₁≠L₂, use error |L₁−L₂|/3 for both and a common permitted small h; the triangle inequality would force |L₁−L₂|<2|L₁−L₂|/3, a contradiction.

The symbols h→0 describe an approach using nonzero increments; they do not instruct us to substitute h=0 into Qₐ. A finite derivative is a real number. If quotients grow beyond every finite bound or keep differing between the sides, no such finite derivative exists.

We may also write dy/dx at x=a for f′(a), when y=f(x). Here that notation names a limit of difference quotients, not an ordinary quotient of two numbers both equal to zero. The derivative function assigns x↦f′(x) only at the inputs where this finite derivative exists.

Algebra proves the square and cube derivatives

For f(x)=x² on ℝ, expand (a+h)²−a²=2ah+h²=h(2a+h). Since h≠0, Qₐ(h)=2a+h. Its error from 2a is exactly |h|. Given ε>0, choose δ=ε; every 0<|h|<δ gives error<ε. Therefore f′(a)=2a for every real a.

For a constant rule f(x)=c, every permitted quotient is 0, so its derivative is 0. For a line f(x)=mx+b on ℝ, the change is mh and Qₐ(h)=m, so f′(a)=m. A rate describes how outputs change, not the size of f(a).

For f(x)=x³, expansion gives Qₐ(h)=3a²+3ah+h². If |h|<1, the error from 3a² is at most (3|a|+1)|h|. Choose δ=min(1,ε/(3|a|+1)); this proves f′(a)=3a² without using a general power rule.

At a=0 the cubic quotient is h² and the derivative is 0, although the cubic is not constant in any interval around zero. A zero slope at one point does not assert that the function is constant everywhere or that the point is a maximum or minimum. General differentiation and shape rules come later.

Preserve the original domain while simplifying

For f(x)=1/x and a≠0, require h≠0 and a+h≠0. Subtracting fractions gives Qₐ(h)=−1/(a(a+h)). Its candidate limit is −1/a². Choose |h|<|a|/2 so |a+h|>|a|/2; the denominator then stays away from zero.

The exact error is |Qₐ(h)+1/a²|=|h|/(|a|²|a+h|)<2|h|/|a|³. Thus δ=min(|a|/2,ε|a|³/2) proves derivative −1/a². The excluded input a=0 has neither a function value nor a derivative; cancellation cannot remove that condition.

For f(x)=√x at a>0, rationalise the numerator for h≠0 with a+h≥0: Qₐ(h)=1/(√(a+h)+√a). The candidate derivative is 1/(2√a). Its error is |h|/[2√a(√(a+h)+√a)²], at most |h|/(2a√a).

Taking δ=min(a/2,2εa√a) keeps a+h>0 and makes the error strictly less than ε. This proves the finite two-sided derivative at positive a. At a=0 the same simplified quotient becomes 1/√h for h>0; that endpoint question must be treated separately rather than substituting into the positive-input formula.

A finite derivative supplies the tangent's slope and point

When the finite derivative exists, define its tangent line through the actual point by y=f(a)+f′(a)(x−a). For the square at a=1, the point is (1,1) and the slope is 2, so the tangent is y=2x−1. A slope alone is insufficient: many parallel lines have slope 2.

The tangent need not intersect a curve only once or stay on one side. The cubic at 0 has tangent y=0 and passes from below that line to above it. Our definition uses the limiting secant slope rather than a visual rule about touching without crossing.

The finite-change identity f(a+h)−f(a)=hQₐ(h) is exact. Replacing Qₐ(h) by f′(a) generally gives an approximation, not an exact equality for nonzero h. At a=1 the square change is 2h+h², not merely 2h. Quantitative linear approximation belongs to a later lesson.

For position s(t)=t² metres, s′(1)=2 metres per second is instantaneous velocity, whereas the 1/2-second average is 5/2 metres per second. The generic coordinate diagrams illustrate slopes; their numerical axes are not automatically physical distances and times for every displayed rule.

Available sides and equal finite slopes are separate requirements

The finite right derivative f′₊(a)=L uses the same quotient and error condition but only 0<h<δ; its domain must accumulate from the right. The finite left derivative f′₋(a)=L uses −δ<h<0 and left accumulation. Both still use the actual f(a).

When both sides are available, the finite two-sided derivative exists exactly when both finite one-sided derivatives exist and are equal. A two-sided guarantee immediately applies to each side. Conversely, for a given ε use the smaller of the two side tolerances, which covers every permitted sign with the same common L.

For x² restricted to [0,+∞), the right derivative at 0 is 0 because Q₀(h)=h for h>0. There is no left approach in that domain, so our convention does not call this a two-sided derivative. Some texts use endpoint derivatives as shorthand for one-sided derivatives; we name the side explicitly.

For √x on [0,+∞), Q₀(h)=1/√h. Given any M>0, every 0<h<1/M² gives Q₀(h)>M. Hence it has no finite right derivative, and no left approach is available. The vertical tangent x=0 describes this unbounded slope; +∞ is not a real derivative value.

For a singleton domain D={0} with f(0)=2, no permitted nonzero increment approaches zero. Neither side derivative is defined in our convention. The function is nevertheless continuous on its domain: the only sufficiently near permitted input is 0 itself, whose output error is zero.

A corner or a changed point value can destroy a derivative

For f(x)=|x| at 0, Q₀(h)=|h|/h. It is −1 for h<0 and 1 for h>0. Both side derivatives are finite but unequal, so the two-sided derivative does not exist. The function is continuous at 0 because ||x|−0|=|x|; continuity does not require the side slopes to agree.

At a=1, the same absolute-value rule has quotient 1 for |h|<1, because nearby 1+h is positive. Its derivative there is 1. A corner at one input does not prevent differentiability at another input.

Now define g(x)=x² for x≠0 but g(0)=1. At 0 its quotient is (h²−1)/h=h−1/h. The actual point value cannot be replaced by the nearby polynomial's missing value 0. For 0<h<1/2, the quotient is less than 1/2−1/h and becomes arbitrarily negative; for negative h it has the opposite sign behaviour.

The rule g is not continuous at 0: nearby outputs h² tend to 0, while g(0)=1. Removing the point 0 from its domain would instead make the derivative at 0 inapplicable, because f(a) must exist. Altering, excluding and correctly retaining a point are three different situations.

A finite derivative forces continuity at its actual point

Suppose the finite derivative at a is L. Apply the quotient guarantee with error 1: for some η>0, every permitted 0<|h|<η has |Qₐ(h)|≤|L|+|Qₐ(h)−L|<|L|+1. Thus the nearby quotient has one finite bound.

Given an output error ε>0, choose δ=min(η,ε/(|L|+1)). For a permitted nonzero |h|<δ, the exact identity gives |f(a+h)−f(a)|=|h||Qₐ(h)|<ε. At h=0 the difference is already zero. This is domain-relative continuity at a, proved without assuming it in advance.

The same argument proves continuity from the corresponding side when a finite side derivative exists. A right derivative alone does not control a separately available left side. On a one-sided endpoint domain it controls every sufficiently near permitted input and proves endpoint continuity.

The converse is false: the absolute-value corner is continuous but lacks a two-sided derivative. A discontinuous point cannot have a finite two-sided derivative. Do not turn that necessary continuity condition into a sufficient test for differentiability.

Identical finite tables can conceal different derivatives

Consider z(x)=0 and p(x)=x(x²−1)(x²−1/4)(x²−1/16), both on ℝ with value 0 at the base input 0. At the six increments ±1, ±1/2 and ±1/4, both difference quotients are 0. Any derivative conclusion based only on those six matching rows would give the same answer for both functions.

For z the quotient is always 0, so z′(0)=0. For p and h≠0, cancel only h to obtain Q₀(h)=(h²−1)(h²−1/4)(h²−1/16). Its limit at 0 is (−1)(−1/4)(−1/16)=−1/64, so p′(0)=−1/64. The product uses the actual formula, not the sampled zeros.

To justify that product limit locally, each factor tends to its constant term because h²→0. Products of two factors tending to finite A and B tend to AB: write uv−AB=(u−A)v+A(v−B), bound |v|≤|B|+1 near zero and make both errors small. Repeating this finite argument proves the displayed three-factor limit.

At the extra increment 1/8 the product quotient is −2835/262144, rather than 0. One extra row reveals the finite-table ambiguity here, but that extra row still is not the limit proof. A table can suggest a candidate or refute a claim at a tested input; only the all-input argument establishes the derivative.

For the square at a=1, the exact identity |Q₁(h)−2|=|h| supplies the whole guarantee. With ε=1/4, the increments ±1/4 give equality, outside the strict error condition; ±1/8 satisfy it. More generally δ=ε works for every positive tolerance, not just the one drawn.

Use the definition in an order that preserves every condition

Name the real domain, base input and actual point value. Check which domain sides accumulate there. Write the quotient for permitted h≠0, simplify only under those restrictions and justify a finite common limit or the precise obstruction.

Retain units when interpreting a rate. A finite derivative gives the tangent through (a,f(a)); it does not make a nonzero finite change exact or prove global shape. When working from a table, separate an estimate from the definition's complete guarantee.

Return to the square's explicit error bound, the reciprocal's denominator bound, the corner's two slopes and the continuity proof. These are useful patterns for later differentiation rules; they stand independently here, with all limit and domain conditions stated.

Worked examples

Try it yourself

1. Find the derivative of f(x)=5x−7 at any real a using a quotient.

   Show the answer and reasoning

   f′(a)=5.

   The numerator is 5h, so for h≠0 the quotient is exactly 5; every quotient error is zero.

2. For f(x)=−8 on ℝ, is the derivative −8 or 0? Explain.

   Show the answer and reasoning

   It is 0.

   The two output values are equal, so their difference and every permitted quotient are 0. Output value and rate are different quantities.

3. Find f′(3) and the tangent for f(x)=x².

   Show the answer and reasoning

   f′(3)=6; y=6x−9.

   The quotient is 6+h, whose error from 6 is |h|. Use the actual point (3,9) in y−9=6(x−3).

4. For x² at a=1 and h=−1/4, find the signed changes and quotient.

   Show the answer and reasoning

   Input change −1/4; output change −7/16; quotient 7/4.

   The second input is 3/4 and its output is 9/16. Subtract the actual base output 1, then divide by the signed increment.

5. Prove the derivative of x³ at a=2 without a power rule.

   Show the answer and reasoning

   It is 12.

   The quotient is 12+6h+h². For |h|<1 its error is at most 7|h|, so δ=min(1,ε/7) proves the limit 12.

6. May you evaluate the original square difference quotient at h=0 after cancelling h?

   Show the answer and reasoning

   No; the simplified rule can have a limit at 0 while the original quotient stays undefined there.

   Cancellation was justified only for h≠0. The derivative names a limiting value and does not create a denominator-zero quotient.

7. Does 1/x have a derivative at 0 because its derivative formula is known away from 0?

   Show the answer and reasoning

   No.

   The original function has no value at 0. The derivative definition requires f(a), and the formula −1/a² applies only for a≠0.

8. Use rationalisation to find the derivative of √x at a=9.

   Show the answer and reasoning

   It is 1/6.

   The quotient is 1/(√(9+h)+3) for permitted h≠0. Its error from 1/6 is at most |h|/54; δ=min(9/2,54ε) proves the finite two-sided limit.

9. Does continuity of |x| at 0 establish its derivative there?

   Show the answer and reasoning

   No; the side derivatives −1 and 1 disagree.

   Continuity controls outputs approaching the actual point. Differentiability additionally asks the divided changes to approach one common finite slope.

10. For x² on [0,2], identify derivative sides at both endpoints.

    Show the answer and reasoning

    Right derivative at 0 is 0; left derivative at 2 is 4. Neither endpoint has both available sides.

    The quotient is 2a+h. At 0 only positive h occur; at 2 only negative h occur. Their corresponding finite limits follow the same square error bound.

11. Is f(4)=10 on D={4} differentiable at 4 under this lesson's convention?

    Show the answer and reasoning

    No derivative side is available, although it is continuous on that domain.

    No nonzero increment remains in the domain near 4. The accumulation condition prevents a vacuous slope statement.

12. For f(x)=x when x≠0 and f(0)=2, find the actual quotient at 0 and decide finite differentiability.

    Show the answer and reasoning

    Q₀(h)=1−2/h; no finite derivative exists.

    The numerator is h−2, using the actual base value. The quotient is unbounded with opposite signs on the two sides; the point is also discontinuous.

13. If C(n) is cost in euros and n is a continuously modelled number of items, what are the derivative units? What extra issue arises for a domain of integers only?

    Show the answer and reasoning

    Euros per item; an integer-only domain has no accumulating nearby inputs at an integer.

    Output-per-input units come from the quotient. Treating n as continuous is a modelling extension; an integer-only function does not meet this lesson's accumulation requirement for a derivative.

14. Position is s(t)=10−3t metres for t≥0. Find velocity at t=2 and speed there.

    Show the answer and reasoning

    Velocity −3 metres per second; speed 3 metres per second.

    Every permitted quotient is −3. The sign gives coordinate direction; speed is the magnitude. Both nearby time sides are available at 2.

15. For the square at 1 and h=1/2, is f(1+h)−f(1) exactly f′(1)h?

    Show the answer and reasoning

    No: the change is 5/4, while the derivative times h is 1.

    The exact change is 2h+h². The derivative describes the limiting rate, and the omitted h² remains nonzero at this finite increment.

16. Suppose finite left and right quotient limits both equal L. Prove one tolerance covers the full permitted two-sided approach.

    Show the answer and reasoning

    For each ε>0 choose δ=min(δ₋,δ₊).

    Each nonzero permitted h with |h|<δ has one sign and falls under that side's error guarantee. Both sides use the same L, and both must accumulate.

17. For the matching product p, calculate the quotient at h=1/8 and state its derivative at 0.

    Show the answer and reasoning

    Quotient −2835/262144; derivative −1/64.

    The three factors are −63/64, −15/64 and −3/64. Their product gives the finite value; their limits at zero give (−1)(−1/4)(−1/16).

18. For the square at 1 with ε=1/4, do h=±1/4 meet the strict quotient-error guarantee? Give permitted examples instead.

    Show the answer and reasoning

    No. For instance h=±1/8 meet it.

    The exact error is |h|. Equality with ε is excluded; every 0<|h|<1/4 is covered by δ=1/4.

What to carry forward

• A difference quotient uses the actual f(a), a permitted nonzero increment and signed output/input changes.
• A finite derivative is one all-input quotient limit, not a value obtained by dividing by zero or a conclusion proved by finitely many rows.
• Two-sided derivatives require available domain approaches on both sides and equal finite side limits; endpoints and isolated inputs need explicit treatment.
• Direct algebra and error bounds prove the square, cube, line, constant, reciprocal and positive-root examples with their original exclusions.
• A finite derivative supplies the tangent slope through the actual point and retains output-per-input units.
• Differentiability implies continuity; a continuous corner shows why the converse fails.
• Finite samples and finite drawing windows illustrate a proof. Matching tables can hide different derivatives, and an unbounded side slope is not a real derivative value.