Linear approximation and differentials

A slightly wider square needs more than the two strips

A square panel has side 3 cm and area 9 cm². Increase each side by 0.1 cm. The new area is 3.1²=9.61 cm². An easy estimate adds two strips of area 3×0.1=0.3 cm² each, giving 9.6 cm². A corner square of area 0.1²=0.01 cm² completes the exact area.

The estimate keeps the part proportional to the side change and leaves out its square. More generally, a signed change h gives (3+h)²=9+6h+h². For positive side lengths, 9+6h is the local line estimate, and h² is its signed remainder. At h=−0.1 the estimate is 8.4 cm² while the actual area is 8.41 cm²; the positive remainder also occurs when the input decreases.

The schematic beside this explanation makes the small corner visible without drawing its width to scale. Its dimensions and table specify the exact model. A drawing can explain which part was omitted; the algebra establishes its value.

Read the base point, increment and local rate

The published preparation links support clear meanings and conditions, substitution, and checking an estimate's purpose. A function f assigns one output to each permitted input. The base point a is the input whose value and local rate we know; h is the signed input change, and x=a+h is the new input.

For the two-sided result here, assume f is defined on an open interval around a: some positive distance on both sides is allowed. For h≠0 in that interval, the average change per input unit is Qₐ(h)=(f(a+h)−f(a))/h. The derivative f′(a) is one finite number approached by this quotient for every sufficiently small positive and negative h.

Saying Qₐ(h)→m means that for every accuracy ε>0, there is δ>0 such that 0<|h|<δ implies |Qₐ(h)−m|<ε. The notation |u| is the nonnegative size of u. A few numerical quotients suggest this behaviour; they do not check all sufficiently small increments.

For f(x)=x² at a=3, Q₃(h)=6+h, so f′(3)=6. The square-area rate has units cm² per cm: multiplying by the input change in cm produces an area change in cm². First follow this example and the actual-change diagram; return to the general error proof and the corner after the notation is familiar.

Use both the value and the derivative

Write L(a+h)=f(a)+f′(a)h, or equivalently L(x)=f(a)+f′(a)(x−a). This is the tangent line or linearization at a. It has the correct output at a and the derivative slope there. A line with the correct slope but the wrong starting value is not this local approximation.

The word linear here includes an affine expression mx+b. L is an auxiliary rule, not a claim that the original f is a straight line everywhere. For f(x)=mx+b, L=f exactly; for the square, the missing h² term vanishes only at h=0.

Choose a near the desired input where both f(a) and f′(a) are known. For √4.04, a=4 gives f(a)=2 and f′(a)=1/4, so L(4.04)=2.01. The exact root is about 2.009975124. The actual error and the needed tolerance decide whether that estimate is useful.

Differentiability controls error relative to the input change

Define the signed remainder R(h)=f(a+h)−L(a+h), with R(0)=0. For a permitted h≠0, exact subtraction gives R(h)/h=Qₐ(h)−f′(a). If the derivative exists, the right side tends to zero. Therefore |R(h)|/|h|→0.

We write this as R(h)=o(h), spoken as little o of h. It means that for every ε>0, some δ>0 makes |R(h)|<ε|h| whenever 0<|h|<δ. It does not mean R(h)=0, and the ratio at h=0 is undefined. The quantified condition concerns nearby nonzero changes.

This proves f(a+h)=f(a)+f′(a)h+o(h). Conversely, if a line through (a,f(a)) with finite slope m has remainder satisfying |R(h)|/|h|→0, then Qₐ(h)=m+R(h)/h tends to m. Thus the derivative exists and that slope is unique. The local approximation and derivative condition describe the same first-order behaviour.

Differentiability at one point supplies no universal usable step size and no automatic bound of the form |R(h)|≤Ch². For example, f(x)=|x|^(3/2)=|x|√|x| at 0 has quotient with size √|h|, so its derivative is 0 and R=o(h), but |R(h)|/h²=1/√|h| grows without bound. A quadratic error bound needs additional information; none is assumed here.

Exact identities give the sign and scale of three errors

For the square at 3, R(h)=h². Hence |R(h)|/|h|=|h| for h≠0. Reducing the magnitude of h by a factor of ten reduces absolute error by a factor of one hundred and the local ratio by ten. These conclusions come from the identity, not from the number of plotted samples.

For √x at a=4, rationalising the difference quotient gives Q₄(h)=1/(√(4+h)+2)→1/4. In a neighbourhood with 4+h>0, rationalising again gives R(h)=−h²/[4(√(4+h)+2)²]. Its denominator is positive, so the local line lies above the root graph for every nonzero allowed h. The ratio tends to zero because the denominator tends to 64.

For 1/x at a=2, Q₂(h)=−1/[2(2+h)]→−1/4. The line is L(2+h)=1/2−h/4, and R(h)=h²/[4(2+h)]. In the positive-input neighbourhood h>−2 the remainder is nonnegative, so the line lies below the function. At h=0.1 it estimates 0.475; the exact value is 10/21 and R=1/840.

The model keeps the same chart window for each rule when the increment changes. Tiny gaps may be invisible at that scale; read the table and error ratio rather than inferring zero error from overlapping pixels. The finite curves and bars illustrate these analytic proofs.

A differential is the change on the line

At fixed a, choose the input differential dx=h and define the output differential dy=f′(a)dx. The actual finite change is Δy=f(a+h)−f(a). Their exact relationship is Δy=dy+R(h). The symbol Δ names a finite change; d in dy names the linear response at the chosen base point.

The differential dy is exactly the change L(a+h)−L(a). It generally only approximates the function change Δy. A chosen dx may be positive, negative or zero; the derivative definition still used nonzero quotients. The formula dy=f′(a)dx does not require treating dx as a mysterious nonzero infinitesimal.

For the square panel with dx=0.1 cm, dy=0.6 cm² and Δy=0.61 cm². With dx=−0.1 cm, dy=−0.6 cm² and Δy=−0.59 cm². The same positive h² remainder adjusts both signed changes. Do not replace a negative change by its magnitude until the question asks for an absolute error.

For a general differentiable rule y=f(x), the notation dy=f′(x)dx describes a response depending on both the base input x and its chosen increment dx. Our calculations fix x=a first. The derivative's output-per-input units cancel the input unit in dx; a dimensionless relative error is a different quantity.

State what the error is relative to

For approximating the new value f(a+h) by L(a+h), the signed remainder is actual minus estimate. The absolute approximation error is |R(h)|, in output units. When f(a+h)≠0, the relative approximation error is |R(h)|/|f(a+h)|; its percentage is that ratio multiplied by 100. The denominator is the actual new value's magnitude.

For the 3.1 cm square, absolute error is 0.01 cm² and relative error is 0.01/9.61=1/961, about 0.1041%. Dividing by the nominal area 9 instead would answer a different comparison. Neither denominator is the input change |h| used in the differentiability condition.

At a=0 for f(x)=x², the derivative is 0 and the line is 0. For nonzero h, absolute error h² tends to zero and |R(h)|/|h|=|h| tends to zero, but relative approximation error is h²/h²=1, or 100%. A good first-order approximation need not have a small relative error near a zero output.

At h=0 in this flat example, the absolute error is zero while relative error is undefined because the actual output is zero. Do not label 0/0 as zero error. If the true value is unknown in an application, using a nominal value in the denominator creates an estimated comparison and must be identified as such.

Separate a first-order measurement estimate from a guaranteed interval

Suppose a circle is modelled exactly by A(r)=πr² and a nominal radius is 10 cm. Assume the true radius is within 0.1 cm of that nominal value: r=10+dr and |dr|≤0.1 cm. This is a stated tolerance interval, not a probability distribution or a claim that a measured value is the exact radius.

The differential is dA=20πdr, giving the first-order estimate |dA|≤2π cm². Normalising that estimate by nominal area 100π gives 2%, also an estimate. The exact identity is A(10+dr)−100π=20πdr+πdr², so dropping the nonnegative second term loses a small but real contribution.

Because squaring increases with positive radius, 9.9≤r≤10.1 gives 98.01π≤A(r)≤102.01π cm². Thus the exact deviation from nominal area ranges from −1.99π to 2.01π cm², and its absolute size is at most 2.01π cm². The symmetric ±2π differential estimate is not a guaranteed error bound.

For nominal radius r₀>0 and 0≤u≤r₀, a tolerance |dr|≤u guarantees nonnegative radii and |A(r₀+dr)−A(r₀)|≤2πr₀u+πu². This bound follows by expanding and taking absolute values. Real modelling errors in circular shape or radius measurement are additional assumptions; differentials do not remove them.

Zero slope and missing slope are different

The square at 0 has a finite derivative 0, so a horizontal tangent is valid even though nonzero changes produce positive output. A derivative can describe first-order behaviour while higher-order behaviour supplies the entire finite change.

For f(x)=|x| at 0, Q₀(h)=1 for h>0 and −1 for h<0. There is no single two-sided derivative. The proposed horizontal line L=0 has R(h)=|h| and |R(h)|/|h|=1 for every nonzero h. A small absolute gap alone cannot establish first-order agreement.

No different finite slope repairs both sides of this corner: a slope m would need 1−m→0 on the right and −1−m→0 on the left, requiring m=1 and m=−1 simultaneously. We may name the right-hand and left-hand slopes separately, but cannot silently use either as the two-sided derivative.

Domain exclusions also matter. The reciprocal has no value at zero; √x has no real value at negative x, and its endpoint at zero has no finite derivative from the right. A formula derived near a=4 or a=2 does not supply those missing values or justify extrapolation across an excluded input.

Choose the step using a tolerance, not the drawing

For the square at 3, requiring absolute error at most 0.01 cm² gives h²≤0.01, or |h|≤0.1 cm. The original positive-side condition 3+h>0 still applies. This is a genuine finite bound from an exact remainder identity, stronger than merely saying the error tends to zero.

For the root example, keep |h|≤0.4 so 4+h≥3.6. Its remainder denominator is at least 4(√3.6+2)²; therefore |R(h)|≤h²/[4(√3.6+2)²]. That explicit interval and bound can be checked against a desired output tolerance.

A larger step can fail badly. The reciprocal's line at 2 gives L(10)=−1.5, while f(10)=0.1. Both inputs are permitted, but the local approximation was not a global agreement. A tangent can also cross a curve: x³ at 0 has derivative 0 and lies on opposite sides of its line y=0 for negative and positive inputs. Neither touching once nor staying on one side is the definition of a tangent.

Use the explorer to compare signed steps, the zero step, the flat tangent and the corner. State the base point, domain, derivative and denominator for each answer. Return to the proof for the all-increment guarantee; return to the measurement interval when a practical bound is needed.

Worked examples

Try it yourself

1. At a=3 for f(x)=x², take h=0.2. Find L, dy, Δy and R.

   Show the answer and reasoning

   L=10.2, dy=1.2, Δy=1.24 and R=0.04.

   Use L=9+6h and Δy=6h+h². The exact output is 3.2²=10.24, so Δy−dy=0.04.

2. At h=0, can the value |R(h)|/|h| be set to zero because function and line agree?

   Show the answer and reasoning

   No. The error is zero, but the ratio is undefined.

   Its denominator is zero. The limit condition considers 0<|h|<δ and does not assign a quotient at h=0.

3. Does the line y=6x approximate x² to first order at a=3 merely because its slope is 6?

   Show the answer and reasoning

   No. It misses the base point: it gives 18 rather than 9.

   The correct line is y=9+6(x−3)=6x−9. A fixed output mismatch cannot have remainder o(h).

4. Use the line at 4 to estimate √3.96. Is the estimate below the actual root?

   Show the answer and reasoning

   The estimate is 1.99; it is above, not below, the actual root.

   Here h=−0.04. The exact remainder formula is negative for either sign of nonzero h, so actual value is less than the line value.

5. For the area estimate 9.6 cm² versus actual area 9.61 cm², which denominator defines relative approximation error?

   Show the answer and reasoning

   Use 9.61 cm², giving 0.01/9.61=1/961.

   Relative approximation error compares the absolute remainder with the actual new output, not nominal area 9 or input change 0.1.

6. For x² at a=0 and h=−0.1, does dy=0 prove that the actual output change is zero?

   Show the answer and reasoning

   No. Δy=0.01 while dy=0.

   The derivative supplies the first-order term. The exact remainder h² contains the entire nonzero change in this case.

7. For |x| at 0, the horizontal line's absolute error tends to zero. Does that make it a derivative tangent?

   Show the answer and reasoning

   No. Its local error ratio is always 1 for nonzero h.

   First-order agreement requires |R|/|h|→0. The right and left quotient limits are 1 and −1, so no two-sided derivative exists.

8. What signed steps guarantee error at most 0.0025 for the square line at a=3?

   Show the answer and reasoning

   All |h|≤0.05, with the original positive-side domain condition retained.

   R=h², so h²≤0.0025 is equivalent to |h|≤0.05. In this interval 3+h is automatically positive.

9. Does a radius tolerance of ±0.1 cm at nominal radius 10 cm guarantee circle-area error no more than 2π cm²?

   Show the answer and reasoning

   No. The upper-radius endpoint gives error 2.01π cm².

   The differential drops πdr². The exact expanded formula and interval give the slightly larger guaranteed bound 2.01π cm².

10. If f′(a) is 6 cm² per cm and dx is −0.1 cm, what is dy and its unit?

    Show the answer and reasoning

    dy=−0.6 cm².

    Multiply the signed increment by the derivative. The input cm cancels, leaving area units; taking an absolute value would change the question.

11. Does differentiability alone guarantee a quadratic bound |R(h)|≤Ch² near the base point?

    Show the answer and reasoning

    No. It guarantees R=o(h), a weaker condition.

    At zero, f(x)=|x|^(3/2) has derivative 0, but |R(h)|/h²=1/√|h| is unbounded. No fixed finite C works throughout a small punctured neighbourhood.

What to carry forward

• A local line needs the actual base value and a finite derivative: L(a+h)=f(a)+f′(a)h.
• The derivative definition proves |f(a+h)−L(a+h)|/|h|→0, not exact equality or an automatic quadratic bound.
• A differential dy=f′(a)dx is the exact change on the line and an approximation to Δy on the function.
• Absolute, relative and input-normalised errors use different denominators; a zero denominator stays undefined.
• Preserve signed increments, domains and derivative units, especially at corners and zero outputs.
• A practical measurement guarantee needs an explicit interval or remainder bound; a first-order estimate alone is insufficient.