Linear Approximation and Differentials

Q: What is a linear approximation?

Using the tangent line at a known point to estimate nearby values: f(x) ≈ f(a) + f′(a)(x − a).

Q: Why does linear approximation work?

A differentiable function is locally linear: close to a point, its graph is almost indistinguishable from the tangent line there.

Q: How do I know if the estimate is too high or too low?

If the graph is concave down near a (f″ < 0), the tangent lies above it, so the estimate is too high. If concave up, too low.

Q: What is a differential?

For y = f(x), the differential dy = f′(x) dx is the change along the tangent line when x changes by dx. It approximates the true change Δy.

Q: How accurate is a linear approximation?

Very accurate close to a and worse farther away. The error shrinks roughly like the square of the distance from a.

Quick answer

Zoomed in far enough, a differentiable function looks like its tangent line. So near x = a, f(x) ≈ L(x) = f(a) + f′(a)(x − a), the linearization of f at a. It turns hard values into mental arithmetic: √4.1 ≈ 2 + ¼(0.1) = 2.025. The estimate is too high where the graph bends down (f″ < 0) and too low where it bends up (f″ > 0). In differential form, a small change dx in the input causes a change of about dy = f′(x) dx in the output.

What you'll learn

Write the linearization of a function at a point
Use it to estimate function values
Tell whether an estimate is an overestimate or an underestimate
Use differentials to estimate small changes and errors

Local linearity

Zoom in on the graph of a differentiable function at a point, and the curve straightens out until it is almost indistinguishable from its tangent line. Near that point, the tangent line is an excellent stand-in for the function.

y = √x
L(x) = 2 + ¼(x − 4)

The tangent line hugs the curve near the point

The linearization

The tangent line at $x = a$ is the linearization of $f$ at $a$ :

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

and for $x$ near $a$ , $f(x) \approx L(x)$ . Choose $a$ where $f$ and $f'$ are known exactly, close to the value you want.

Why the tangent line is the best straight-line guess

The derivative’s definition says $\tfrac{f(a + h) - f(a)}{h} \to f'(a)$ . Put differently, $f(a + h) = f(a) + f'(a)\,h + (\text{error})$ , where the error shrinks faster than $h$ itself: divided by $h$ , it still goes to $0$ .

No other line through $\big(a, f(a)\big)$ does that. A line with any other slope $m$ misses by about $(f'(a) - m)h$ , an error proportional to $h$ . The tangent line is the unique line whose error vanishes faster than the step, which is exactly what makes it the best local approximation. For smooth functions the error is roughly $\tfrac{1}{2}f''(a)\,h^2$ : halve the step and the error drops by a factor of four.

Too high or too low?

The tangent line lies above a graph that bends downward (concave down, $f'' < 0$ ), and below a graph that bends upward (concave up, $f'' > 0$ ).

For $\sqrt{x}$ , $f''(x) = -\tfrac{1}{4}x^{-3/2} < 0$ , so the curve is concave down and $2.025$ is a slight overestimate — as the true value confirms.

Differentials

For $y = f(x)$ , write $dx$ for a change in the input. The differential

dy = f'(x)\,dx

is the change along the tangent line, and it approximates the true change $\Delta y = f(x + dx) - f(x)$ . Differentials are the standard way to estimate how an error in a measurement spreads into an error in a computed result.

Worked examples

Common mistakes

Practice problems

Find the linearization of $f(x) = x^3$ at $a = 2$ .

Answer

$L(x) = 8 + 12(x - 2)$

Full solution

$f(2) = 8$ and $f'(x) = 3x^2$ , so $f'(2) = 12$ .
Use problem 1 to estimate $2.1^3$ . Is it too high or too low?

Answer

$9.2$ , too low

Full solution

$L(2.1) = 8 + 12(0.1) = 9.2$ . $f''(x) = 6x > 0$ near $2$ , so the graph is concave up and the tangent lies below: the true value, $9.261$ , is higher.
Estimate $\sqrt{9.2}$ .

Answer

About $3.0333$

Full solution

At $a = 9$ : $f(9) = 3$ , $f'(9) = \tfrac{1}{6}$ . $L(9.2) = 3 + \tfrac{0.2}{6} \approx 3.0333$ .
Estimate $\ln(1.03)$ .

Answer

About $0.03$

Full solution

At $a = 1$ : $\ln 1 = 0$ , and $(\ln x)' = \tfrac{1}{x}$ equals $1$ there. So $L(x) = x - 1$ and $\ln(1.03) \approx 0.03$ .
Estimate $\sqrt[3]{27.5}$ .

Answer

About $3.0185$

Full solution

At $a = 27$ : $f(27) = 3$ and $f'(x) = \tfrac{1}{3}x^{-2/3}$ , so $f'(27) = \tfrac{1}{27}$ . $L(27.5) = 3 + \tfrac{0.5}{27} \approx 3.0185$ .
Find the differential $dy$ for $y = x^2 \sin x$ .

Answer

$dy = (2x\sin x + x^2\cos x)\,dx$

Full solution

$dy = y'\,dx$ , with $y'$ from the product rule.
A circle’s radius is measured as $5$ cm with a possible error of $0.2$ cm. Estimate the possible error in its area.

Answer

About $2\pi \approx 6.28$ cm²

Full solution

$dA = 2\pi r\,dr = 2\pi(5)(0.2) = 2\pi$ .
Estimate $\cos(0.1)$ with the linearization of $\cos x$ at $0$ . Why is the result not very informative?

Answer

$1$ ; the tangent at $0$ is horizontal, so the line misses the first change.

Full solution

$\cos 0 = 1$ and $-\sin 0 = 0$ , so $L(x) = 1$ . The true value, $0.995$ , differs by about $\tfrac{1}{2}(0.1)^2$ , the second-order term the line leaves out.
Is the linearization of $\ln x$ at $1$ an overestimate or underestimate of $\ln 1.2$ ?

Answer

An overestimate

Full solution

$(\ln x)'' = -\tfrac{1}{x^2} < 0$ , so the graph is concave down and the tangent lies above it. $L(1.2) = 0.2$ , while $\ln 1.2 \approx 0.182$ .
A student estimates $\sqrt{4.1}$ as $L(4.1) = 2 + \tfrac{1}{2\sqrt{x}}(0.1)$ and leaves the answer in terms of $x$ . What went wrong?

Hint

The slope of the tangent line is a number.

Answer

The slope must be evaluated at $a = 4$ : $f'(4) = \tfrac{1}{4}$ , giving $2.025$ .

Full solution

$L(x) = f(a) + f'(a)(x - a)$ uses the derivative at the point of tangency.

With $f'(4) = \tfrac{1}{4}$ : $L(4.1) = 2 + \tfrac{1}{4}(0.1) = 2.025$ .

Frequently asked questions

What is a linear approximation?

Using the tangent line at a known point to estimate nearby values: f(x) ≈ f(a) + f′(a)(x − a).

Why does linear approximation work?

A differentiable function is locally linear: close to a point, its graph is almost indistinguishable from the tangent line there.

How do I know if the estimate is too high or too low?