Calculus I › Applications of derivatives › full formula sheet

f(x) ≈ f(a) + f′(a)(x − a)

Say it: “f of x is approximately f of a plus f prime of a times x minus a”

Linear approximation

Borrow the tangent line as a stand-in for the function — and estimate anything with pencil and paper.

Notation: a is the easy nearby point you choose; x is the hard point you want. The ≈ means “approximately equal”.

Before this lesson: Tangent line at x = a · Definition of the derivative

Where it comes from

The problem: estimate √4.02 with no calculator. The naive move is to round to √4 = 2 — but is the answer 2.001 or 2.05? Rounding gives no error control. A sneakier naive move: use the secant between (4, 2) and (9, 3), slope 0.2, giving 2 + 0.2(0.02) = 2.004. Closer — but still a secant, still an overshoot. The right idea: near x = 4, the curve hugs its tangent, so just use the tangent’s height instead of the curve’s:

tangent at a = 4
=
y = 2 + (1/4)(x − 4)
From the tangent-line page: f(4) = 2, f′(4) = 1/4.
at x = 4.02
≈
2 + (1/4)(0.02) = 2.005
True value: √4.02 ≈ 2.0049875. Our estimate is off by about 0.00001 — five correct digits from one line of arithmetic.

But approximation is not magic — distance matters. Try the same trick on f(x) = x² with a = 1 to estimate f(2):

1 + 2(2 − 1) = 3  vs.  true 2² = 4a full 1 away from a: the tangent has drifted far from the curve — the counterexample that keeps you honest

Close to a, the tangent is nearly the curve; far from a, it isn’t. That single fact governs everything on this page.

Before reading on: you only know f(4) = 2 and f′(4) = 1/4. What is your best guess for √4.02 — and how far from 4 could you go before the guess gets embarrassing?

Derivation

There is no new theorem here — the formula is the tangent line, relabeled. The derivation is three short moves:

tangent line at a
=
y = f(a) + f′(a)(x − a)
Step 1 — the tangent. From the previous page: the line through (a, f(a)) with slope f′(a). (Requires f differentiable at a.)
for x near a
≈
f(x) ≈ y
Step 2 — the hug. Zoom in near a and the curve is visually indistinguishable from its tangent. So the curve’s height f(x) is nearly the line’s height y.
⇒
f(x) ≈ f(a) + f′(a)(x − a)
Step 3 — substitute. Replace y by its tangent expression. The ≈ carries the warning: this is exact only at x = a, and degrades as x moves away. ∎

How wrong can it be? Taylor’s theorem says the error is about ½f″(c)(x−a)² for some c between a and x. Two lessons in one line: the error is quadratic in the distance (halve the distance, quarter the error), and sharp bends (large f″) hurt. That is exactly why the x²-at-2 estimate above failed.

Before reading on: would you linearize √x at a = 0 to estimate √0.1? What’s the catch?

How to use it

The procedure, every time:

  1. Pick a. Two requirements: close to x, and f(a) and f′(a) easy to compute by hand. For roots, pick a perfect square/cube; for trig and exponentials, pick 0; for ln, pick 1.
  2. Write L(x) = f(a) + f′(a)(x − a). Build the tangent line exactly as on the previous page.
  3. Evaluate at your x. One line of arithmetic — that is the estimate.
  4. Sanity-check. Is the estimate on the correct side and in the right ballpark? If x − a isn’t tiny, distrust the answer.

Choosing a: the whole game

For √4.02, a = 4 (perfect square, distance 0.02). For sin(0.1), a = 0 (sin 0 = 0, cos 0 = 1). For ln(1.1), a = 1 (ln 1 = 0). The pattern: a is where the function is trivial.

When not to use it

Skip it when x is far from every easy a, when the curve bends sharply between a and x, or where f isn’t differentiable — there is no tangent to borrow. And never confuse the estimate with the truth: ≈ is not =.

Common mistake: picking a faraway “easy” point — e.g. a = 9 to estimate √4.02. The tangent at 9 has slope 1/6 and the point is 5 away: 3 + (1/6)(−4.98) ≈ 2.17, off by 0.17. Closeness beats easiness.

Worked examples

Four estimates, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the classic: √4.02

  1. Pick a. a = 4: a perfect square, only 0.02 from 4.02. f(x) = √x.
  2. Build L(x). f(4) = 2; f′(x) = 1/(2√x), so f′(4) = 1/4. L(x) = 2 + (1/4)(x − 4).
  3. Evaluate. L(4.02) = 2 + (1/4)(0.02) = 2 + 0.005 = 2.005.
  4. Check. True √4.02 ≈ 2.0049875 — error about 0.00001 ✓.
Common mistake: writing L(x) = 2 + (1/4)x (forgetting the − 4 shift). At x = 4 that gives 3, not 2 — always verify L(a) = f(a).
Your turn: Estimate √4.1 with a = 4.

Answer: ≈ 2.025.

L(x) = 2 + (1/4)(x − 4); L(4.1) = 2 + 0.1/4 = 2.025. True √4.1 ≈ 2.0248 ✓.

Example 2 — the famous one: sin(0.1)

  1. Pick a. a = 0: sin 0 = 0 and cos 0 = 1 are trivial, and 0.1 is small. f(x) = sin x.
  2. Build L(x). f(0) = 0; f′(x) = cos x, so f′(0) = 1. L(x) = 0 + 1·(x − 0) = x.
  3. Evaluate. sin(0.1) ≈ 0.1.
  4. Check. True sin(0.1) ≈ 0.0998334 — error about 0.0002 ✓. (This is why “sin x ≈ x for small x” appears everywhere in physics.)
Common mistake: using degrees. The approximation sin x ≈ x needs x in radians — sin(0.1°) ≈ 0.001745, not 0.1.
Your turn: Estimate sin(0.2) with a = 0.

Answer: ≈ 0.2.

L(x) = x, so sin(0.2) ≈ 0.2. True value ≈ 0.19867 ✓.

Example 3 — exponentials: e0.03

  1. Pick a. a = 0: e0 = 1 is trivial. f(x) = ex.
  2. Build L(x). f(0) = 1; f′(x) = ex, so f′(0) = 1. L(x) = 1 + 1·(x − 0) = 1 + x.
  3. Evaluate. e0.03 ≈ 1 + 0.03 = 1.03.
  4. Check. True e0.03 ≈ 1.0304545 — error about 0.0005 ✓. (This is the “ex ≈ 1 + x” behind continuous-interest estimates.)
Common mistake: writing L(x) = x (dropping the f(a) = 1). Then L(0) = 0 ≠ e0 = 1 — the L(a) = f(a) check catches it instantly.
Your turn: Estimate e0.05 with a = 0.

Answer: ≈ 1.05.

L(x) = 1 + x, so e0.05 ≈ 1.05. True value ≈ 1.0513 ✓.

Example 4 — cube roots: ∛√27.3

  1. Pick a. a = 27: a perfect cube, distance 0.3. f(x) = x1/3.
  2. Build L(x). f(27) = 3; f′(x) = (1/3)x−2/3, so f′(27) = (1/3)(1/9) = 1/27. L(x) = 3 + (1/27)(x − 27).
  3. Evaluate. L(27.3) = 3 + 0.3/27 = 3 + 0.0111… ≈ 3.0111.
  4. Check. 3.0111³ ≈ 27.3007, so ∛√27.3 ≈ 3.0111 ✓ — four good digits.
Common mistake: differentiating x1/3 as (1/3)x2/3 (forgetting to subtract 1 in the exponent). Power rule: bring down, then reduce the exponent: (1/3)x−2/3.
Your turn: Estimate ∛√26.9 with a = 27.

Answer: ≈ 2.9963.

L(x) = 3 + (1/27)(x − 27); L(26.9) = 3 − 0.1/27 ≈ 2.99630. Check: 2.9963³ ≈ 26.90 ✓.

Memorization tips

  • It is the tangent line wearing an ≈: f(x) ≈ f(a) + f′(a)(x − a). If you know the tangent formula, you know this one.
  • Close and easy: a must be near x AND make f(a), f′(a) trivial. Chant “close and easy” when choosing.
  • The big four at a = 0: sin x ≈ x, ex ≈ 1 + x, ln(1+x) ≈ x, (1+x)r ≈ 1 + rx. Memorize these; they recur for years.
  • Verify L(a) = f(a): your linearization must be exact at the base point. If it isn’t, the algebra slipped.
  • Error is quadratic in distance: halve (x − a), quarter the error. If the gap isn’t small, don’t trust the digits.
  • Radians only for trig approximations — sin x ≈ x dies in degrees.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and linear approximation is yours.

← Back to the Calculus I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the linear approximation formula?

f(x) ≈ f(a) + f′(a)(x − a): replace the function near x = a by its tangent line, choosing a close to x where f(a) and f′(a) are easy to compute.

Why must a be close to x?

The error is roughly ½f″(c)(x−a)² — quadratic in the distance. Far from a the tangent drifts from the curve: estimating 2² from a = 1 gives 3 instead of 4.

Is linear approximation the same as the tangent line?

Yes — it is the tangent line y = f(a) + f′(a)(x − a) used as a stand-in for f(x). The tangent page finds the line; this page uses it to estimate values.

When does linear approximation fail badly?

When x is far from a, when the curve bends sharply between a and x (large f″), or where f isn’t differentiable — there’s no tangent to borrow.

What are the famous small-x approximations?

Near x = 0: sin x ≈ x, ex ≈ 1 + x, ln(1+x) ≈ x, and (1+x)r ≈ 1 + rx — all are linear approximations at a = 0.

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