Linearization and Differentials

What is the linearization of a function?

The linearization of f at a is the linear function L(x)=f(a)+f'(a)(x-a), whose graph is the tangent line to y=f(x) at (a,f(a)). The reason it is useful is that a differentiable curve behaves locally like its tangent line: magnify the graph around the point and it flattens, until the curve and the line cannot be told apart. For y=x^2 at (1,1) the tangent line is y=2x-1, and in a window about 0.006 wide the two graphs look the same.

The tangent line at x=a passes through (a,f(a)) with slope f'(a), so its point-slope equation, from the tangent line equation, is y=f(a)+f'(a)(x-a). As long as that line stays close to the graph when we move off the point of tangency, L(x) is a good approximation of f(x).

Definitions. If f is differentiable at x=a, then the approximating function L(x)=f(a)+f'(a)(x-a) is the linearization of f at a. The approximation f(x)≈ L(x) is the standard linear approximation of f at a, and the point x=a is the center of the approximation.

Why it works. The definition of the derivative says that the quotient (f(x)-f(a))/(x-a) is close to f'(a) when x is close to a, that is, f(x)-f(a)≈ f'(a)(x-a). Both f(x)-f(a) and f'(a)(x-a) shrink to 0 as x→ a, but the first agrees with the second to better than a multiple of x-a, and this is what "close" means here. A function with a corner or a vertical tangent at a has no derivative there, so it has no linearization at a.

What it is for. Nobody would use a linearization to find one particular square root, because a calculator is better. Its use is to replace a complicated formula by a simple one over a whole interval of values, when the small error is tolerable, and to understand how an output reacts to a small change of the input. Zoom in below and read the table: each tenfold magnification makes the gap between the curve and the line about a hundred times smaller.

How do you find a linearization, and how accurate is it?

To find a linearization, compute f(a) and f'(a) at the center and put them into L(x)=f(a)+f'(a)(x-a); it is accurate for x near a and less and less accurate as x moves away.

Example 1. Find the linearization of f(x)=√(1+x) at a=0. Since f'(x)=(1)/(2√(1+x)), we have f(0)=1 and f'(0)=\frac12, so L(x)=1+\frac12(x-0)=1+\frac x2. How good is √(1+x)≈1+\frac x2? Near 0 it is very good, and it worsens as x moves away:

For x=2 it gives 2 as the approximation of √3, which is not even accurate to one decimal place.

Example 2: choose a better center. A linearization normally loses accuracy away from its center, so near x=3 we need the linearization at a=3. With f(3)=2 and f'(3)=(1)/(2√4)=\frac14, L(x)=2+\frac14(x-3)=\frac54+\frac x4. At x=3.2 this gives √(4.2)≈\frac54+(3.2)/(4)=2.050, which differs from the true 2.04939 by less than one thousandth. The linearization at 0 gives 1+(3.2)/(2)=2.6 there, off by more than 25\%. So pick a center a where f(a) and f'(a) are easy to find and which is close to the place you need.

Example 3. For f(x)=\cos x at a=\fracπ2 we have f\left(\fracπ2\right)=0 and f'\left(\fracπ2\right)=-\sin\fracπ2=-1, so L(x)=0-1·\left(x-\fracπ2\right)=-x+\fracπ2.

The common linearizations at x=0. Working them out as in Example 1 gives:

The last row is the most useful: it is the linearization of f(x)=(1+x)^k, where f(0)=1 and f'(0)=k. It is good for x close to 0 and gives, with x replaced by something small, √(1+x)≈1+\frac x2 (1)/(1-x)≈1+x (1)/(√(1-x^2))≈1+(x^2)/(2) and, with k a power, 1+5x^4^(1/3)≈1+(5x^4)/(3). It also estimates powers of numbers close to 1: 1.0002^50≈1+50·0.0002=1.01 (the true value is 1.010049) and 1.009^(1/3)≈1+(0.009)/(3)=1.003. Try your own function, center and x below.

What are the differentials dx and dy?

The differential dx is an independent variable, a change of x that you may choose, and the differential of y=f(x) is dy=f'(x) dx, the amount that the tangent line rises or falls when x changes by dx. The notation (dy)/(dx) for the derivative looks like a ratio but is not one by its definition; the differentials make it one, because when dx\ne0 (dy)/(dx)=(f'(x) dx)/(dx)=f'(x). Unlike dx, the variable dy is always dependent: it depends on both x and dx. If dx is given a value and x is a number in the domain, these determine the number dy. Often dx is chosen to be \Delta x, the change in x.

Example 4. (a) Find dy if y=x^5+37x. (b) Find dy when x=1 and dx=0.2. Here dy=\left(5x^4+37\right)dx, and with x=1 and dx=0.2 it is (5+37)(0.2)=8.4.

The geometric meaning. Set x=a and dx=\Delta x. The true change of y=f(x) is \Delta y=f(a+dx)-f(a). The corresponding change of the tangent line L is \Delta L=L(a+dx)-L(a)=f'(a) dx=dy. So dy is the change of the linearization, the rise of the tangent line over the run dx, whereas \Delta y is the rise of the curve. They are close when dx is small, and the move of the slider below shows the difference.

Differential forms. We sometimes write df=f'(x) dx and call df the differential of f; for f(x)=3x^2-6 it is df=d\left(3x^2-6\right)=6x dx. Every differentiation rule has a differential form: (d(u+v))/(dx)=(du)/(dx)+(dv)/(dx) becomes d(u+v)=du+dv, and d(\sin u)=\cos u du.

Example 5. The chain rule and the other rules give differentials: d(\tan2x)=\sec^2(2x) d(2x) d(\tan2x)=2\sec^2(2x) dx d\left((x)/(x+1)\right)=((x+1) dx-x dx)/((x+1)^2) d\left((x)/(x+1)\right)=(dx)/((x+1)^2). The first uses the chain rule, with u=2x inside the tangent.

How do you estimate values and changes with differentials?

To estimate f(a+dx) when f(a) is known and dx is small, use f(a+dx)≈ f(a)+dy with dy=f'(a) dx; equivalently the change \Delta f is approximately the differential df. The goal is to choose a nearby point a where both f(a) and f'(a) are easy to evaluate.

Example 6: the area of a circle. The radius r of a circle increases from a=10 m to 10.1 m. Use dA to estimate the increase in the area, and the area of the enlarged circle. Since A=π r^2, the estimated increase is dA=A'(a) dr=2π a dr dA=2π(10)(0.1)=2π m^2. So A(10.1)≈ A(10)+dA A(10.1)≈100π+2π=102π m^2. Directly, A(10.1)=π(10.1)^2=102.01π m^2, so the error is 0.01π m^2, the difference \Delta A-dA.

Example 7a: a cube root. To estimate 7.97^(1/3) take f(x)=x^1/3, where dy=(1)/(3x^2/3) dx. The closest number with an easy cube root is a=8, and to get a+dx=7.97 we choose dx=-0.03: f(7.97)≈ f(8)+(1)/(3·8^2/3)(-0.03) f(7.97)≈2+(1)/(12)(-0.03)=1.9975. The true value is 1.997497 to six places.

Example 7b: a sine. To estimate \sin\left(\fracπ6+0.01\right) take a=\fracπ6, where \sin\fracπ6=\frac12 and \cos\fracπ6=(√3)/(2), and dx=0.01: \sin\left(\fracπ6+0.01\right)≈\frac12+(√3)/(2)(0.01) \sin\left(\fracπ6+0.01\right)≈0.5087. The true value is 0.508635. Computers can use this method: an algorithm stores a table of sine and cosine values at a grid of angles and fills in the values between with a differential, like Example 7b.

Absolute, relative and percentage change. When we move from a to a+dx, the change in f can be described in three ways:

In Example 6 the area changes by (2π)/(100π)=2\% (estimated) when the radius changes by (0.1)/(10)=1\%.

How large is the error, and how does a measuring error spread?

If f is differentiable at a and x changes from a to a+\Delta x, the true change is \Delta f=f'(a) \Delta x+\varepsilon \Delta x, where \varepsilon→0 as \Delta x→0, so the error of the differential estimate is \varepsilon \Delta x, a product of two small quantities. The reason is the definition of the derivative. The error is \Delta f-df=\left((f(a+\Delta x)-f(a))/(\Delta x)-f'(a)\right)\Delta x \Delta f-df=\varepsilon \Delta x, and the quantity in parentheses is a difference quotient minus its limit, which tends to 0. For the circle of Example 6, \Delta A-dA=0.01π=π(\Delta r)^2, so \varepsilon=π \Delta r=0.1π. For many common functions the error is much smaller than \Delta x: when f has a second derivative it is about \frac12f''(a) \Delta x^2, a rule of thumb that is proved later with Taylor's theorem and that explains the widget above, where the error curve is flat at a and grows like a parabola.

The linearization is the best linear approximation. Suppose another linear function g(x)=m(x-a)+c is used to approximate f with error E(x)=f(x)-g(x), and we ask for two things: E(a)=0, and (E(x))/(x-a)→0 as x→ a (the error is negligible next to x-a). The first gives c=f(a). Then (E(x))/(x-a)=(f(x)-f(a))/(x-a)-m \longrightarrow f'(a)-m, which is 0 only for m=f'(a). So L(x)=f(a)+f'(a)(x-a) is the only linear approximation whose error is zero at a and negligible compared with x-a. Every other line through (a,f(a)) leaves an error of the order of x-a.

Sensitivity to change. The equation df=f'(x) dx tells how sensitive the output of f is to a change of the input at different values of x: the larger |f'|, the greater the effect of a given dx (compare the sensitivity of a function). Example 8. The depth of a well is s=4.9t^2 meters, found by timing a dropped stone. Then ds=9.8t dt. If the time is read with an error of dt=0.1 s, then at t=2 s the error is about ds=(9.8)(2)(0.1)=1.96 m, but at t=5 s it is (9.8)(5)(0.1)=4.9 m. For the same error in the time, the depth is more uncertain when the fall is longer.

How a measuring error spreads. For a quantity y=cx^n, such as the area π r^2 (n=2) or the volume x^3 of a cube (n=3), dy=ncx^n-1dx, and dividing by y=cx^n, (dy)/(y)=n (dx)/(x). So a relative error in x becomes n times as large in y. A 2\% error in the radius of a circle gives about 2\% in the circumference (n=1) and 4\% in the area; a 0.5\% error in the edge of a cube gives 1\% in the surface area (n=2) and 1.5\% in the volume; a 1\% error in the diameter of a sphere gives 3\% in its volume. Read the other way, it gives a tolerance: to calculate the volume of a cylinder with height equal to radius, V=π h^3, to within 1\%, the height must be measured to within about \frac13\%.

The estimate is for small changes. Blood flow through an artery is V=kr^4, so a 10\% increase of the radius is estimated to raise the flow by 4·10=40\%. The exact value is 1.1^4-1=46.4\%. The estimate is first order, and the error grows like the square of the change, so the larger the change, the worse the percentage rule. Move the sliders below to see where the line and the exact curve part.

What are linearizations used for in science?

Linearizations turn a hard function into a simple one for small changes, and three classic uses are the small-angle approximation, the energy of a moving mass and Newton's method.

Small angles. The linearization of \sin x at 0 is x, so for a small angle in radians \sinθ≈θ: \sin0.1=0.0998, close to 0.1. The equation of a pendulum becomes linear with this replacement, which is why a pendulum of small swing keeps a steady period. The angle must be in radians; \sin10^\circ≈10 is wrong, while \sin10^\circ≈0.1745 is right.

Mass and energy. Newton's second law F=(d)/(dt)(mv)=ma assumes that the mass is constant, but in Einstein's corrected formula the mass is m=(m_0)/(√(1-v^2/c^2)), where m_0 is the rest mass and c≈300 000 km/s is the speed of light. For v much smaller than c the number x=(v^2)/(c^2) is near 0, and the linearization (1-x)^-1/2≈1+\frac x2 gives m≈ m_0\left(1+(v^2)/(2c^2)\right) (m-m_0)c^2≈\frac12m_0v^2. The right-hand side is the kinetic energy of Newtonian physics. So the change in kinetic energy in going from rest to the speed v is approximately the change in mass times c^2, \Delta(m)c^2≈\Delta(KE), and since c^2≈9×10^16 m^2/s^2, a small change in mass is a large change in energy. At v=0.1c the exact factor is 1.005038 and the approximation gives 1.005.

Newton's method. The linearization of f at a is a line, and a line has an easy zero. Solving L(x)=0 gives x=a-(f(a))/(f'(a)), which is usually closer to a zero of f than a is. Repeating this is Newton's method. For f(x)=x^2-2 and the start a=1.5 the next values are 1.416667, 1.414216, 1.414214: three steps for √2 to six places.

How do you compute linearizations and differentials in code?

With SymPy a linearization is one line: differentiate, evaluate at the center and build f(a)+f'(a)(x-a). The function below reproduces the linearizations of this article exactly, and the lines after it check numbers with plain Python: the table of √(1+x)≈1+\frac x2, the circle of Example 6, the two estimates of Example 7, the percentage rule for y=x^n and the first steps of Newton's method.

import math
import sympy as sp

x = sp.symbols('x')


def linearize(f, a):
    """L(x) = f(a) + f'(a) (x - a)"""
    return sp.simplify(f.subs(x, a) + sp.diff(f, x).subs(x, a) * (x - a))


for f, a in ((sp.sqrt(1 + x), 0), (sp.sqrt(1 + x), 3), (sp.cos(x), sp.pi / 2), (sp.exp(x), 0), (sp.log(1 + x), 0), (sp.cbrt(x), 8)):
    print(f, 'at', a, '->', linearize(f, a))
# sqrt(x + 1) at 0 -> x/2 + 1
# sqrt(x + 1) at 3 -> x/4 + 5/4
# cos(x) at pi/2 -> -x + pi/2
# exp(x) at 0 -> x + 1
# log(x + 1) at 0 -> x
# x**(1/3) at 8 -> x/12 + 4/3

# how good is sqrt(1 + x) = 1 + x/2 ?   (x, approximation, true value, error)
for t in (0.005, 0.05, 0.2):
    print(t, 1 + t / 2, round(math.sqrt(1 + t), 6), f'{abs(1 + t / 2 - math.sqrt(1 + t)):.1e}')
# 0.005 1.0025 1.002497 3.1e-06
# 0.05 1.025 1.024695 3.0e-04
# 0.2 1.1 1.095445 4.6e-03

# differentials: the radius of a circle grows from 10 m to 10.1 m
dA = 2 * math.pi * 10 * 0.1
true = math.pi * 10.1**2 - math.pi * 10**2
print(round(dA, 4), round(true, 4), round(true - dA, 4), round(math.pi * 0.01, 4))      # 6.2832 6.3146 0.0314 0.0314

# estimates: the cube root of 7.97 and sin(pi/6 + 0.01)
print(round(2 + (-0.03) / 12, 4), round(7.97 ** (1 / 3), 6))                              # 1.9975 1.997497
print(round(0.5 + math.sqrt(3) / 2 * 0.01, 4), round(math.sin(math.pi / 6 + 0.01), 6))    # 0.5087 0.508635

# a measuring error spreads: y = x^n, p% in x gives about n p% in y   (n, p, estimate, exact)
for n, p in ((2, 2), (3, 0.5), (4, 10)):
    print(n, p, n * p, round(((1 + p / 100) ** n - 1) * 100, 4))
# 2 2 4 4.04
# 3 0.5 1.5 1.5075
# 4 10 40 46.41

# Newton's method is the zero of the linearization, again and again (f(x) = x^2 - 2)
a = 1.5
for step in range(4):
    a = a - (a * a - 2) / (2 * a)
    print(step + 1, a)
# 1 1.4166666666666667
# 2 1.4142156862745099
# 3 1.4142135623746899
# 4 1.4142135623730951

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double PI = acos(-1.0);

int main() {
  // how good is sqrt(1 + x) = 1 + x/2 ?   (x, approximation, true value, error)
  double xs[3] = {0.005, 0.05, 0.2};
  for (int i = 0; i < 3; i++) {
    double x = xs[i], approx = 1 + x / 2, truth = sqrt(1 + x);
    printf("%.3f  %.4f  %.6f  %.1e\n", x, approx, truth, fabs(approx - truth));
  }

  // differentials: the radius of a circle grows from 10 m to 10.1 m
  double dA = 2 * PI * 10 * 0.1;
  double trueChange = PI * 10.1 * 10.1 - PI * 10 * 10;
  printf("%.4f %.4f %.4f %.4f\n", dA, trueChange, trueChange - dA, PI * 0.01);

  // estimates: the cube root of 7.97, and sin(pi/6 + 0.01)
  printf("%.4f %.6f\n", 2 + (-0.03) / 12, pow(7.97, 1.0 / 3));
  printf("%.4f %.6f\n", 0.5 + sqrt(3.0) / 2 * 0.01, sin(PI / 6 + 0.01));

  // a measuring error spreads: y = x^n, p% in x gives about n p% in y   (n, p, estimate, exact)
  int ns[3] = {2, 3, 4};
  double ps[3] = {2, 0.5, 10};
  for (int i = 0; i < 3; i++)
    printf("%d %.1f %.1f %.4f\n", ns[i], ps[i], ns[i] * ps[i], (pow(1 + ps[i] / 100, ns[i]) - 1) * 100);

  // Newton's method is the zero of the linearization, again and again (f(x) = x^2 - 2)
  double a = 1.5;
  for (int step = 1; step <= 4; step++) {
    a = a - (a * a - 2) / (2 * a);
    printf("%d %.13f\n", step, a);
  }
  return 0;
}

/* output:
0.005  1.0025  1.002497  3.1e-06
0.050  1.0250  1.024695  3.0e-04
0.200  1.1000  1.095445  4.6e-03
6.2832 6.3146 0.0314 0.0314
1.9975 1.997497
0.5087 0.508635
2 2.0 4.0 4.0400
3 0.5 1.5 1.5075
4 10.0 40.0 46.4100
1 1.4166666666667
2 1.4142156862745
3 1.4142135623747
4 1.4142135623731
*/

Where do differentials and linear approximations come from?

Differentials began as infinitely small quantities, and linearization was made precise when they were replaced by limits.

• Leibniz. Gottfried Wilhelm Leibniz introduced the symbols dx and dy for infinitely small differences, and published his differential calculus in 1684. For him (dy)/(dx) was a true quotient of two such quantities, and the tangent line was the line that the curve follows over an infinitely small step.
• Newton and the first approximation. Isaac Newton described in 1669 an iteration that improves an approximate root, in which each step solves the linearization of the equation; Joseph Raphson published a simpler form in 1690. Today it is called the Newton-Raphson method.
• Taylor. Brook Taylor, in 1715, gave the general rule that a function can be approximated near a by a polynomial built from its derivatives. The linearization is the first-degree case, and the quadratic approximation, with \frac12f''(a)(x-a)^2 added, is the next.
• Cauchy. In the 1820s Augustin-Louis Cauchy removed the infinitely small. He defined dx as an ordinary finite increment, an independent variable, and the differential as dy=f'(x) dx, which is the definition used here. This is why dy is the change in the tangent line, not an infinitely small change in the curve.

What mistakes do people make with linearization and differentials?

The most common mistakes with linearization and differentials are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: linearization and differentials at a glance

• Linearization: L(x)=f(a)+f'(a)(x-a), the tangent line at the center a; f(x)≈ L(x) near a.
• Accuracy: best near a and worse away from it; the error is about \frac12f''(a)(x-a)^2; choose a center where f(a) and f'(a) are easy.
• Common ones at 0: \sin x≈ x, \cos x≈1, e^x≈1+x, \ln(1+x)≈ x, (1+x)^k≈1+kx.
• Differential: dy=f'(x) dx, the change of the tangent line; \Delta y=dy+\varepsilon dx with \varepsilon→0.
• Estimating: f(a+dx)≈ f(a)+f'(a) dx; relative change (df)/(f), percentage change (df)/(f)×100.
• Measuring errors: for y=cx^n, (dy)/(y)=n(dx)/(x); the estimate is for small changes.
• Best linear approximation: the only line with zero error at a and error negligible next to x-a.
• Code: sp.diff(f, x).subs(x, a) for the slope, and x_new=a-(f(a))/(f'(a)) for Newton's method.

Glossary: key terms

Linearization

The linear function f of a plus f prime of a times x minus a, whose graph is the tangent line to the graph of f at a; it approximates f for x near a.

Standard linear approximation

The approximation of f of x by its linearization L of x at a, valid when x is close to the center a.

Center of the approximation

The point a at which a linearization is built; the approximation is exact there and loses accuracy as x moves away.

Differential

For y equal to f of x, the differential dy is f prime of x times dx, where dx is an independent variable; it is the change in the tangent line when x changes by dx.

Relative and percentage change

The change in a quantity divided by its original value, and the same multiplied by one hundred; for a differential estimate it is df divided by f of a.

Approximation error

The difference between the true change and the differential estimate, equal to epsilon times delta x with epsilon tending to zero, so it is small even compared with delta x.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 3.11: Linearization and Differentials. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 3.10: Linear Approximations and Differentials. Cengage Learning.
3. Brook Taylor (1715). Methodus incrementorum directa et inversa. London.
4. Augustin-Louis Cauchy (1823). Résumé des leçons données à l'École royale polytechnique sur le calcul infinitésimal. Paris.

Related articles

• The Derivative as a Rate of Change: The derivative as a rate of change: velocity, speed, acceleration and jerk, free fall, marginal cost and revenue, sensitivity to change, and rates from data in code.
• Related Rates: Related rates: differentiate an equation in time to find an unknown rate, with the four-step strategy, six worked examples, a sliding ladder and common mistakes.
• Tangent Lines and the Derivative at a Point: Tangent lines and the derivative at a point: secant slopes, the definition of f′(a), tangent and normal lines, rates of change, and where no derivative exists.
• Extreme Values of Functions on Closed Intervals: Extreme values on closed intervals: absolute and local extremes, the Extreme Value Theorem, critical points and the closed interval method, with examples and code.