Linear Approximation (Linearization): Formula and Examples

Linear approximation, also known as linearization, is a calculus technique that estimates a function’s value near a specific point by using its tangent line. The formula for linear approximation of a function f(x) at a point x = a is L(x) = f(a) + f′(a)(x − a), which provides a simple, accurate estimate for values close to a.

Linear approximation (or linearization) estimates a function near a point a using its tangent line:

$$f(x) \approx L(x) = f(a) + f'(a)(x - a)$$

It works because, zoomed in closely enough, a smooth curve looks almost straight.

What Is Local Linearity and Why Does It Matter?

Linear approximation works because of a fundamental principle called local linearity. Pick any smooth curve and zoom in on one point. The more you magnify, the straighter the curve looks, until it is indistinguishable from its tangent line. Mathematicians call this local linearity, and it is really what “differentiable” means: the derivative is the slope of the line that the curve resembles up close.

So if you know the height f(a) and the slope f'(a) at one convenient point, you can walk along the tangent line a short distance and land very close to the curve. Starting height plus slope times distance traveled: that is the whole formula in words.

What Is the Formula for Linear Approximation (Linearization)?

The formula for linear approximation is derived directly from the point-slope form of a line. This tangent line passes through the point (a, f(a)) and has slope f'(a). From algebra, the point-slope form of a line says:

$$y - f(a) = f'(a)(x - a)$$

Adding f(a) to both sides and renaming y as L(x) gives the linearization formula:

$$L(x) = f(a) + f'(a)(x - a)$$

Nothing new is needed: linear approximation is simply the tangent line, used as an estimating tool rather than as a picture. Geometric secant lines approach this tangent line as the distance between points shrinks to zero, making L(x) the best linear estimate of f(x) near a.

How Do You Linearize a Multivariable Function?

For functions of multiple variables, the concept of linearization extends from a tangent line to a tangent plane (or hyperplane for more than two variables). For a function f(x,y) at a point (a,b), the linearization formula is:

$$L(x,y) = f(a,b) + (x-a)f_x(a,b) + (y-b)f_y(a,b)$$

Here, fₓ(a,b) and fᵧ(a,b) are the partial derivatives of f with respect to x and y, respectively, evaluated at the point (a,b). This formula essentially approximates the function’s surface with a flat plane that touches the surface at (a,b) and has the same “slope” in all directions at that point.

How to Do a Linear Approximation (Step-by-Step Guide)

To perform a linear approximation, you need to identify a convenient base point, find the function’s value and derivative at that point, and then apply the linearization formula.

1. Choose a base point a near x where both f(a) and f'(a) are easy to compute exactly by hand. This point should be close to the value you want to approximate.
2. Find the derivative f'(x) of your function and evaluate it at a to get the slope f'(a). This derivative represents the instantaneous rate of change of the function at point a.
3. Build the linearization equation: Using the values found in steps 1 and 2, construct the tangent line equation: L(x) = f(a) + f'(a)(x – a).
4. Evaluate L(x) at the target input by plugging in x and simplifying. The resulting value, L(x), will be your linear approximation of f(x).

Good choices of base points a depend on the function:

• Square roots: Pick the nearest perfect square (e.g., a = 4 for 4.1, or a = 9 for 9.2).
• Cube roots: Pick the nearest perfect cube (e.g., a = 27 for 27.3, or a = 8 for 8.1).
• Exponentials and logarithms: Pick a = 0 or a = 1 where values like e0 = 1 or ln(1) = 0 are known.
• Trigonometric functions: Pick standard reference angles like 0, π/6, π/4, π/3, or π/2. Ensure you work in radians for derivatives.

Linear Approximation Examples

Let’s walk through several examples to illustrate how to apply the linear approximation formula to different types of functions.

Example 1: Estimate √4.1

Estimate the value of √4.1 without a calculator.

Use f(x) = √x at the base point a = 4. First compute the exact functional value and derivative:

• f(4) = √4 = 2
• f'(x) = 1 / (2√x), so f'(4) = 1 / (2√4) = 1/4 = 0.25

Build the linearization formula:

$$L(x) = 2 + \frac{1}{4}(x - 4)$$

Now evaluate L(4.1) where the displacement x – a = 4.1 – 4 = 0.1:

$$\sqrt{4.1} \approx 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025$$

The true value of √4.1 is approximately 2.0248457.

Example 2: Estimate √9.2

Estimate √9.2 using a linear approximation.

Set f(x) = √x and select a = 9 as the base point. Calculate the base values:

• f(9) = √9 = 3
• f'(x) = 1 / (2√x), so f'(9) = 1 / (2√9) = 1/6

Form the tangent line equation:

$$L(x) = 3 + \frac{1}{6}(x - 9)$$

Plug in x = 9.2, giving x – a = 0.2:

$$\sqrt{9.2} \approx 3 + \frac{1}{6}(0.2) = 3 + \frac{0.2}{6} = 3 + \frac{1}{30} \approx 3.0333$$

The true value is approximately 3.03315. The linear approximation is remarkably close, off by roughly 0.00018.

Example 3: Estimate ln(1.05)

Estimate the natural logarithm of 1.05.

Choose f(x) = ln(x) and base point a = 1. The base evaluations are:

• f(1) = ln(1) = 0
• f'(x) = 1/x, so f'(1) = 1/1 = 1

The linearization function is:

$$L(x) = 0 + 1(x - 1) = x - 1$$

Evaluating at x = 1.05 yields:

$$\ln(1.05) \approx 1.05 - 1 = 0.05$$

The true value of ln(1.05) is approximately 0.04879. The approximation overestimate of 0.05 comes from the curvature of the logarithmic graph.

Example 4: Estimate (1.01)³

Estimate (1.01)³ using linearization.

Set f(x) = x3 and select a = 1:

• f(1) = 13 = 1
• f'(x) = 3×2, so f'(1) = 3(1)2 = 3

The linearization formula is:

$$L(x) = 1 + 3(x - 1)$$

Plug in x = 1.01 where x – a = 0.01:

$$(1.01)^3 \approx 1 + 3(0.01) = 1.03$$

The true value is 1.030301. The estimate 1.03 captures the leading change perfectly and differs from the exact answer by only 0.000301.

Example 5: A Cube Root

Estimate the cube root of 27.3.

Use f(x) = x1/3 at a = 27, because 27 is a perfect cube. The derivative is f'(x) = (1/3)x-2/3 = 1 / (3×2/3). Since 272/3 = (271/3)2 = 32 = 9, we find f'(27) = 1 / (3 × 9) = 1/27.

Setting up the linearization:

$$L(x) = 3 + \frac{1}{27}(x - 27)$$

Evaluating at x = 27.3 gives x – a = 0.3 = 3/10:

$$\sqrt[3]{27.3} \approx 3 + \frac{1}{27}(0.3) = 3 + \frac{3}{270} = 3 + \frac{1}{90} \approx 3.01111$$

The true value is about 3.01107. This estimate is slightly high because the cube root graph is concave down for positive numbers.

Example 6: Estimate tan 44° (Watch the Units)

Estimate tan(44°).

The nearest standard angle where tangent is simple is 45°, which is π/4 in radians. Calculus derivative rules assume angle measurements in radians. As (d/dx) tan(x) = sec2(x) and sec2(π/4) = (√2)2 = 2, the linearization equation at a = π/4 is:

$$L(x) = 1 + 2\left(x - \frac{\pi}{4}\right)$$

The step from 45° to 44° is -1°, which must be converted to radians: -1° = -π/180 radians. Substituting x – a = -π/180 into the linearization gives:

$$\tan 44^\circ \approx 1 + 2\left(-\frac{\pi}{180}\right) = 1 - \frac{\pi}{90} \approx 1 - 0.03491 = 0.96509$$

The true value of tan(44°) is approximately 0.96569. If you had mistakenly used degrees directly and plugged in -1 instead of -π/180, you would have calculated 1 + 2(-1) = -1, an absurd result. Derivative formulas for trigonometric functions only hold when inputs are expressed in radians.

Useful Linearizations Near 0

When approximating values near x = 0 (that is, setting a = 0), several standard functions yield simplified linearization formulas, often called small-angle approximations or Maclaurin series approximations of the first order:

Function f(x)	Linearization L(x) near x = 0
sin x	x
tan x	x
cos x	1
ex	1 + x
ln(1 + x)	x
(1 + x)k	1 + kx

These linear approximations represent the first two terms of each function’s Taylor series expanded about 0 (also known as a Maclaurin series). Adding higher-order terms like quadratic or cubic powers creates higher-degree polynomials that yield even greater accuracy across wider intervals.

The final row, (1 + x)k ≈ 1 + kx, is known as the binomial approximation and is especially useful in physics and financial calculations. For example, estimating (1.02)10 using x = 0.02 and k = 10 gives:

$$(1.02)^{10} = (1 + 0.02)^{10} \approx 1 + 10(0.02) = 1.2$$

This is extremely close to the exact value of 1.218994. Similarly, physicists rely on the small-angle approximation sin(θ) ≈ θ to simplify the nonlinear differential equation describing a swinging pendulum into a simple harmonic oscillator equation for small amplitude swings.

Will the Linear Approximation Overestimate or Underestimate the True Value?

Whether a linear approximation overestimates or underestimates the true value depends on the concavity of the function between the base point a and the target point x. Concavity is determined by the sign of the second derivative, f”(a):

• If f”(a) < 0 (concave down): The curve bends downward below its tangent line. Therefore, the tangent line lies above the actual curve, and L(x) overestimates the true value (L(x) > f(x)). This explains why our estimate of 2.025 for √4.1 is slightly larger than the true value 2.02485.
• If f”(a) > 0 (concave up): The curve bends upward above its tangent line. The tangent line lies below the actual curve, and L(x) underestimates the true value (L(x) < f(x)). For example, ex is concave up everywhere, so L(x) = 1 + x always underestimates ex for non-zero x.

You can determine concavity quickly by evaluating the second derivative at the base point, as detailed in the comprehensive guide to inflection points and concavity.

How Big Is the Error in Linear Approximation?

To know how trustworthy a linear approximation is, you can bound the magnitude of the approximation error E(x) = f(x) – L(x) using Taylor’s Remainder Theorem. If |f”(c)| ≤ M for all points c in the interval between a and x, then:

$$\big|f(x) - L(x)\big| \le \frac{M}{2}(x - a)^2$$

This bound comes directly from Taylor’s theorem with remainder, which rests on the Mean Value Theorem. The presence of the (x – a)2 term tells us that the error scales quadratically with distance: cutting the distance to the base point in half reduces the maximum potential error by a factor of four. This means that the closer your chosen point a is to the target x, the significantly more accurate your approximation will be.

Let’s calculate the error bound for our first example, estimating √4.1 at a = 4:

1. Find the second derivative: f”(x) = -1 / (4×3/2).
2. Find the maximum absolute value M of f”(x) on the interval [4, 4.1]. Since |f”(x)| decreases as x increases, its maximum occurs at the left endpoint x = 4:

   $$M = \left|-\frac{1}{4(4)^{3/2}}\right| = \frac{1}{4(8)} = \frac{1}{32} = 0.03125$$

3. Apply the error bound formula with x – a = 0.1:

   $$\text{Error} \le \frac{1/32}{2}(0.1)^2 = \frac{1}{64}(0.01) = \frac{0.01}{64} \approx 0.00015625$$

The actual error of our estimate was |2.0248457 – 2.025| = 0.0001543, which sits safely just below our theoretical maximum bound of 0.00015625.

How Do Differentials Relate to Linear Approximation?

Differentials express the linear approximation concept in terms of changes rather than absolute values. Defining dx = x – a as an independent change in the input variable, the differential of the output variable y = f(x) is defined as:

$$dy = f'(x) dx$$

Here, dx represents the actual change in x, Δy = f(a + dx) – f(a) represents the true change along the curve, and dy represents the estimated change along the tangent line. Linear approximation states that Δy ≈ dy. This means the approximate change in the function’s output is found by multiplying the derivative at the base point by the small change in the input.

Differentials are widely used in laboratory science and engineering to estimate error propagation. For instance, suppose a sphere’s radius is measured as r = 10 cm with a potential measurement error of dr = ±0.1 cm. The volume of a sphere is V = (4/3)πr3. We can estimate the maximum error in the computed volume using the differential dV:

$$dV = V'(r) dr = 4\pi r^2 dr$$

$$dV = 4\pi (10)^2 (\pm 0.1) = 4\pi (100) (0.1) = \pm 40\pi \approx \pm 125.7 \text{ cm}^3$$

While an error of 125.7 cm3 may sound large, the relative error provides better perspective. Dividing the differential error dV by total volume V gives:

$$\frac{dV}{V} = \frac{4\pi r^2 dr}{\frac{4}{3}\pi r^3} = 3 \frac{dr}{r}$$

Since the relative error in radius is dr / r = 0.1 / 10 = 0.013(0.01) = 0.03 Exponent rules dictate that volume errors compound at three times the rate of radius errors.

Example 7: Estimating a Change Using Differentials

A circular metal plate is heated so that its radius expands from 5 cm to 5.1 cm. Estimate the increase in the plate’s area.

The area formula is A = πr2. Taking the differential with respect to radius gives:

$$dA = 2\pi r \, dr$$

Substitute r = 5 cm and dr = 5.1 – 5 = 0.1 cm:

$$dA = 2\pi(5)(0.1) = \pi \approx 3.14 \text{ cm}^2$$

The exact increase in area is ΔA = π(5.12 – 52\() = π(26.01 - 25) = 1.01π ≈ 3.17 \text{ cm}^2\). The small discrepancy of \(0.01π ≈ 0.0314 \text{ cm}^2\) is the tiny outer corner area ignored by the linear approximation. Geometrically, dA = 2πr dr represents flattening the thin outer ring into a rectangle of length 2πr (circumference) and width dr (thickness).

When Does Linear Approximation Fail?

Linear approximation relies entirely on local smoothness. It produces misleading or undefined results in several specific scenarios:

• Non-differentiable points (corners and cusps): At points like x = 0 for f(x) = |x|, left and right derivatives do not match. No single tangent line exists, so linearization cannot be built.
• Vertical tangent lines: For functions like f(x) = x1/3 at a = 0, the derivative f'(0) = ∞ is undefined. A vertical line cannot function as a linear estimator L(x).
• Discontinuities: If a function jumps or breaks at or near a, the tangent line at a cannot predict behavior across the discontinuity.
• Large evaluation distances: Estimating f(x) far from a leads to significant errors because higher-order curvature dominates over large intervals. The further you move from the point of tangency, the less accurate the straight line becomes.
• Rapidly oscillating functions: Functions like f(x) = x2 sin(1/x) near 0 oscillate too quickly for a single straight line to capture nearby behavior accurately.

Where Is Linear Approximation Used?

Linear approximation serves as a foundational building block across mathematics, physics, and computational algorithms:

• Root finding: It powers Newton’s method, an iterative algorithm that repeatedly replaces a complex function curve with its tangent line to find zero crossings quickly.
• Error analysis: Physical science laboratories use differentials to calculate how uncertainties in raw measured quantities propagate into computed experimental outcomes. This is crucial for understanding the reliability of experimental results.
• Physics modeling: Complex differential equations in aerodynamics, robotics, and pendulum dynamics are linearised near equilibrium states to make analytical solutions possible. For example, the period of a pendulum for small swings can be approximated linearly.
• Multivariable calculus: The same core principle extends to functions of two variables, f(x, y), where a function near (a, b) is approximated by a 2D tangent plane:

  $$L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$$

• Control Theory: In robotics and engineering, complex non-linear systems are often linearized around an operating point to design simpler, effective controllers.
• Optics: Gaussian optics, which deals with lenses and mirrors, often uses linear approximations for small angles and distances to simplify ray tracing calculations.

Common Mistakes to Avoid When Using Linear Approximation

To ensure accurate and useful linear approximations, be aware of these common pitfalls:

• Choosing a far-away base point a: Estimating √50 using a = 36 yields L(50) = 6 + (1/12)(14) ≈ 7.1667, whereas using the closer base point a = 49 yields L(50) = 7 + (1/14)(1) ≈ 7.0714. The true value is 7.0711, making the error from a = 36 more than 200 times larger. Always select the closest point where f(a) and f'(a) are easily calculable.
• Using f'(x) instead of the constant f'(a): The slope f'(a) must be evaluated as a single constant number at the base point. Keeping f'(x) variable produces a non-linear expression rather than a straight line equation. The derivative is evaluated at the specific point a, not as a function of x.
• Forgetting to convert angles to radians: Trigonometric derivative formulas assume radian measure. Using degree steps directly produces completely incorrect answers. Always convert degrees to radians before applying derivative formulas.
• Mixing up x and (x – a): Ensure you multiply the slope f'(a) by the displacement distance (x – a) rather than the target point x itself. For √4.1, multiply f'(4) = 0.25 by 0.1, not by 4.1. The (x-a) term represents the small change from your known point.
• Confusing dy with Δy: Remember that dy is the estimated change along the linear tangent line, whereas Δy is the actual change along the function curve. While they are approximately equal for small changes, they are not identical.
• Choosing a center point that is close but hard to evaluate: The purpose of linear approximation is simplification. If calculating f(a) or f'(a) is as difficult as calculating f(x) directly, the benefit of linearization is lost.

Linearization Calculator

You can also use the full-page linear approximation calculator, or check derivatives first with the derivative calculator.

Practice Problems

Test your understanding by working through these practice problems, then review the complete step-by-step solutions below.

1. Estimate √25.2 using linearization.
2. Estimate sin(0.05) near a = 0.
3. Next, Estimate e0.1 near a = 0.
4. Estimate the cube root of 27.3.
5. Also, Estimate tan(44°) using base point a = π/4.
6. Estimate (1.02)10 using the binomial approximation formula.

Step-by-step solutions

1. Then, Estimate √25.2:

   Set f(x) = √x and a = 25. Then f(25) = 5 and f'(x) = 1 / (2√x), so f'(25) = 1/10 = 0.1.

   $$L(x) = 5 + 0.1(x - 25)$$

   $$L(25.2) = 5 + 0.1(0.2) = 5 + 0.02 = 5.02$$

   (True value: ≈ 5.01996. Since f”(25) < 0, 5.02 is a slight overestimate.)

2. Estimate sin(0.05):

   Set f(x) = sin(x) and a = 0. Then f(0) = 0 and f'(x) = cos(x), so f'(0) = cos(0) = 1.

   $$L(x) = 0 + 1(x - 0) = x$$

   $$\sin(0.05) \approx 0.05$$

   (True value: ≈ 0.049979. Concave down on (0, 0.05), so this is a slight overestimate.)

3. Estimate e0.1:

   Set f(x) = ex and a = 0. Then f(0) = 1 and f'(x) = ex, so f'(0) = 1.

   $$L(x) = 1 + 1(x - 0) = 1 + x$$

   $$e^{0.1} \approx 1 + 0.1 = 1.1$$

   (True value: ≈ 1.10517. Since ex is concave up everywhere, 1.1 is an underestimate.)

4. Estimate √[3]{27.3}:

   Set f(x) = x1/3 and a = 27. Then f(27) = 3 and failure slope f'(x) = 1 / (3×2/3), so f'(27) = 1/27.

   $$L(x) = 3 + \frac{1}{27}(x - 27)$$

   $$L(27.3) = 3 + \frac{1}{27}(0.3) = 3 + \frac{1}{90} \approx 3.01111$$

   (True value: ≈ 3.01107. Slight overestimate due to concave down curvature.)

5. Estimate tan(44°):

   Set f(x) = tan(x) and a = π/4. Then f(π/4) = 1 and f'(x) = sec2(x), so f'(π/4) = 2. Convert -1° step to radians: dx = -π/180.

   $$L(x) = 1 + 2\left(-\frac{\pi}{180}\right) = 1 - \frac{\pi}{90} \approx 0.96509$$

   (True value: ≈ 0.96569. Underestimate.)

6. Estimate (1.02)10:

   Use binomial approximation (1 + x)k ≈ 1 + kx with x = 0.02 and k = 10:

   $$(1 + 0.02)^{10} \approx 1 + 10(0.02) = 1 + 0.2 = 1.2$$

   (True value: ≈ 1.21899. Underestimate.)

Frequently Asked Questions

How accurate is linear approximation?

The accuracy depends on the distance between the target point x and base point a, as well as the magnitude of the function’s second derivative. The absolute error is bounded by (M/2)(x – a)2, where M is the maximum of |f”(c)| between a and x. Because the error shrinks quadratically as x approaches a, linear approximations are exceptionally accurate for small displacements.

Is linear approximation the same as the tangent line?

Yes, mathematically they are identical objects. The phrase “tangent line” refers to the geometric straight line touching a curve at point a, whereas “linear approximation” (or linearization) refers to using that tangent line’s equation, L(x) = f(a) + f'(a)(x – a), as a numerical tool to estimate nearby values of f(x).

What is the difference between linearization and differentials?

Linearization and differentials express the exact same mathematical relationship in two different notations. Linearization estimates the total updated value of a function, f(x) ≈ L(x). Differentials estimate the isolated change in the output variable, dy = f'(x) dx ≈ Δy. Because L(x) = f(a) + dy, you can freely convert between both forms.

How do I choose the point a?

Choose a base point a that satisfies two requirements: it must be as close as possible to your target input x, and both f(a) and f'(a) must be solvable exactly and easily by hand. Standard choices include nearby perfect squares for square roots, perfect cubes for cube roots, 0 or 1 for logarithmic/exponential functions, and standard reference angles for trigonometry.

Why use linear approximation if calculators exist?

Beyond manual estimation, linearization provides essential theoretical insight. It reveals how sensitive a system’s output is to small fluctuations in input, which forms the foundation of experimental error analysis. Also, linearization allows scientists and engineers to replace intractable nonlinear equations with simple linear models around operating equilibrium points, simplifying complex problems in physics and engineering.

Further reading

• Linear Approximations (Paul’s Online Math Notes) — detailed step-by-step examples, graphs, and small-angle applications.
• Differentials (Paul’s Online Math Notes) — comprehensive coverage of differential notation and measurement error estimations.