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.
Why it works: 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.
Where the formula comes from
The tangent line passes through the point \( (a, f(a)) \) and has slope \( f'(a) \). The point-slope form of a line says \( y - f(a) = f'(a)(x - a) \). Add \( f(a) \) to both sides and rename \( y \) as \( L(x) \), and you have the linearization. Nothing new is needed: linear approximation is the tangent line, used as an estimating tool rather than as a picture.
How to use it
- Choose a point \( a \) near \( x \) where \( f(a) \) and \( f'(a) \) are easy to compute.
- Build \( L(x) = f(a) + f'(a)(x - a) \).
- Evaluate \( L \) at the value you want.
Good choices of \( a \) are perfect squares for square roots, perfect cubes for cube roots, 0 or 1 for exponentials and logs, and standard angles such as \( \frac\pi4 \) for trig.
Example 1: estimate √4.1
Use \( f(x) = \sqrt x \) at \( a = 4 \). Then \( f(4) = 2 \) and \( f'(x) = \frac{1}{2\sqrt x} \), so \( f'(4) = \frac14 \).
$$\sqrt{4.1} \approx 2 + \frac14(0.1) = 2.025$$
The true value is \( 2.0248457\ldots \): an error of about 0.00015.
Example 2: estimate √9.2
\( a = 9 \): \( \sqrt{9.2} \approx 3 + \frac{1}{6}(0.2) = 3.0333 \). (True: 3.03315.)
Example 3: ln(1.05)
\( f(x) = \ln x \) at \( a = 1 \): \( L(x) = x - 1 \), so \( \ln 1.05 \approx 0.05 \). True value: 0.04879.
Example 4: (1.01)³
\( f(x) = x^3 \) at \( a = 1 \): \( L(x) = 1 + 3(x - 1) \), so \( (1.01)^3 \approx 1.03 \). True value: 1.030301.
Example 5: a cube root
Estimate the cube root of 27.3. Use \( f(x) = x^{1/3} \) at \( a = 27 \), because 27 is a perfect cube. The derivative is \( f'(x) = \frac13x^{-2/3} \), and \( 27^{2/3} = 9 \), so \( f'(27) = \frac{1}{27} \).
$$\sqrt[3]{27.3} \approx 3 + \frac{1}{27}(0.3) \approx 3.01111$$
The true value is about 3.01107. The estimate is slightly high, because the cube root graph is concave down.
Example 6: tan 44° (watch the units)
The nearest easy angle is 45°, which is \( \frac\pi4 \) in radians. Since \( \frac{d}{dx}\tan x = \sec^2x \) and \( \sec^2\frac\pi4 = 2 \), the linearization is \( L(x) = 1 + 2\left(x - \frac\pi4\right) \).
The step from 45° to 44° is \( -1^\circ \), which must be converted to \( -\frac{\pi}{180} \) radians:
$$\tan 44^\circ \approx 1 - \frac{2\pi}{180} \approx 0.96509$$
The true value is 0.96569. If you had used \( -1 \) instead of \( -\frac{\pi}{180} \), you would get \( -1 \), which is absurd. Derivative formulas for trig functions only hold in radians.
Useful linearizations near 0
| Function | \( L(x) \) near \( x = 0 \) |
|---|---|
| \( \sin x \) | \( x \) |
| \( \tan x \) | \( x \) |
| \( \cos x \) | \( 1 \) |
| \( e^x \) | \( 1 + x \) |
| \( \ln(1 + x) \) | \( x \) |
| \( (1+x)^k \) | \( 1 + kx \) |
These are the first two terms of each function’s Taylor series. Adding more terms gives better approximations.
The last row is worth memorizing. For example, \( (1.02)^{10} \approx 1 + 10(0.02) = 1.2 \), close to the true 1.219. Physicists use \( \sin\theta \approx \theta \) to simplify the pendulum equation for small swings.
Over- or under-estimate?
Look at concavity at \( a \):
- If \( f''(a) < 0 \) (concave down), the tangent lies above the curve, so \( L \) overestimates. That’s why 2.025 is slightly too big for \( \sqrt{4.1} \).
- If \( f''(a) > 0 \) (concave up), \( L \) underestimates.
You can find concavity quickly with the second derivative, as explained in the guide to inflection points and concavity.
How big is the error?
If \( |f''| \le M \) between \( a \) and \( x \), then
$$\big|f(x) - L(x)\big| \le \frac{M}{2}(x - a)^2$$
This bound comes from Taylor’s theorem, which itself rests on the Mean Value Theorem. For \( \sqrt{4.1} \), the second derivative is \( -\frac{1}{4}x^{-3/2} \), whose size is largest at \( x = 4 \), where it equals \( \frac{1}{32} \). The bound is \( \frac{1}{64}(0.1)^2 \approx 0.000156 \), and the actual error is 0.000154, just under it. The \( (x - a)^2 \) factor explains why halving the distance to \( a \) cuts the error by about four.
Differentials
The same idea written with changes: \( dy = f'(x)\,dx \). If a sphere’s radius \( r = 10 \) cm is measured with error \( dr = \pm0.1 \) cm, the volume error is about
$$dV = 4\pi r^2\,dr = 4\pi(100)(0.1) \approx 125.7\ \text{cm}^3$$
Often the relative error is more useful. Dividing \( dV \) by \( V = \frac43\pi r^3 \) gives \( \frac{dV}{V} = 3\frac{dr}{r} \). A 1% error in the radius becomes about a 3% error in the volume. Here \( dx \) is the change in input, \( dy \) is the change along the tangent line, and \( \Delta y \) is the true change along the curve; linear approximation says \( \Delta y \approx dy \).
Example 7: estimating a change
A circular metal plate is heated and its radius grows from 5 cm to 5.1 cm. About how much does its area increase? With \( A = \pi r^2 \), the differential is \( dA = 2\pi r\,dr = 2\pi(5)(0.1) = \pi \approx 3.14 \) square centimeters.
The exact change is \( \pi(5.1^2 - 5^2) = 1.01\pi \approx 3.17 \). The tiny difference, \( 0.01\pi \), is the thin sliver of area that the tangent line ignores. Geometrically, \( dA \) is the circumference times the thickness of the new ring, which is why it works so well for thin rings.
Common mistakes
- Choosing a far-away a. Estimating \( \sqrt{50} \) from \( a = 36 \) instead of \( a = 49 \) gives 7.1667 instead of 7.0714, an error more than 200 times larger (the true value is 7.0711).
- Using f'(x) instead of f'(a). The slope is a fixed number, computed at the base point. \( L(x) \) must be a straight line.
- Forgetting radians. As Example 6 shows, degrees give nonsense for trig functions.
- Mixing up x and x − a. Plug in the distance from \( a \), not \( x \) itself. For \( \sqrt{4.1} \), the step is 0.1, not 4.1.
- Confusing dy with Δy. \( dy \) is the linear estimate; \( \Delta y \) is the exact change. They are close, not equal.
Where it’s used
Linear approximation is the engine inside Newton’s method, which repeatedly solves the tangent line for zero. It explains error propagation in lab measurements, it underlies small-angle approximations in physics, and it is the first step toward Taylor polynomials. The same idea in several variables uses the tangent plane.
Linearization calculator
You can also use the full-page linear approximation calculator, or check derivatives first with the derivative calculator.
Practice problems
Try these before checking the answers.
- \( \sqrt{25.2} \)
- \( \sin(0.05) \)
- \( e^{0.1} \)
- The cube root of 27.3
- \( \tan 44^\circ \) using \( a = \frac\pi4 \)
- \( (1.02)^{10} \)
Answers: (1) \( \approx 5.02 \); (2) \( \approx 0.05 \); (3) \( \approx 1.1 \); (4) \( \approx 3.0111 \); (5) \( \approx 0.9651 \); (6) \( \approx 1.2 \).
FAQ
How accurate is linear approximation?
The error is roughly \( \frac{f''(a)}{2}(x - a)^2 \), so it’s excellent close to \( a \) and gets worse quadratically farther away.
Is linear approximation the same as the tangent line?
Yes. \( L(x) \) is the tangent line, just used to estimate values.
What is the difference between linearization and differentials?
They are the same idea in two notations. Linearization gives an estimate of the value \( f(x) \); differentials give an estimate of the change \( dy \). Since \( L(x) = f(a) + dy \) with \( dx = x - a \), you can switch freely between them.
How do I choose the point a?
Pick the closest point to your target where both \( f(a) \) and \( f'(a) \) can be computed exactly by hand: a perfect square, a perfect cube, 0, 1, or a standard angle.
Why use linear approximation if calculators exist?
It shows how sensitive an output is to small input changes, which is what error analysis needs. It also turns complicated equations into simple ones near a known point, which is how many physics and engineering models are built.
Further reading
- Linear Approximations (Paul’s Online Math Notes) — more examples, including the small-angle approximation
- Differentials (Paul’s Online Math Notes) — the dy and dx notation and estimating measurement errors
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