Numerical derivative calculator

The first or second derivative of any function f(x) at a point, to about 20 significant digits by Richardson extrapolation, with the tangent line drawn.

Drafted with AI writing toolsUpdated Checked examples: 6

Use x as the variable, e.g. x^3 - 2x, exp(x), ln(x). Trig functions use radians here.
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The chart shows x₀ ± this much
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Derivative
Derivative: 2.223244275484
Maximum decimal places: 12; Nearest, ties away from zero
f(x₀)
0.841470984808
Tangent line
y = 2.223244x − 1.381773
Estimated error
9.6 × 10⁻²³

At x = 1 the slope of f is 2.2232443: f rises by about 2.223 × h when x moves by a small step h.

f and its tangent at x = 1

024-10123xy(1, 0.841471)
f(x)Tangent line
Richardson table (central quotient and best estimate) Rows: 6
hDifference quotientBest estimate so far
0.12.219330544382.21933054438
0.052.222266131452.22324466047
0.0252.222999757542.22324427552
0.01252.223183147132.22324427548
0.006252.223228993472.22324427548
0.0031252.223240454982.22324427548
How it's calculated S
  1. Central difference quotient

    D(h)=f(x0+h)−f(x0−h)2h,D(0.1)=2.21933054438038D(h) = \frac{f(x_0 + h) - f(x_0 - h)}{2h},\qquad D(0.1) = 2.21933054438038

    Its error has only even powers of h, which Richardson extrapolation removes one at a time.

  2. Richardson extrapolation

    Ti,j=4j Ti,j−1−Ti−1,j−14j−1,hi=h02i  ⇒  f′(1)≈2.223244275483932730705941T_{i,j} = \frac{4^{j}\,T_{i,j-1} - T_{i-1,j-1}}{4^{j} - 1},\quad h_i = \frac{h_0}{2^{i}} \;\Rightarrow\; f'(1) \approx 2.223244275483932730705941

    Halved h 5 times from h₀ = 0.1; the estimated error is 9.6 × 10⁻²³.

  3. Both sides agree

    D−=2.22324427548393,D+=2.22324427548393D_{-} = 2.22324427548393,\quad D_{+} = 2.22324427548393

    D₋ and D₊ are the left and right estimates. Their agreement supports the numerical derivative, but a finite sampling check is not a proof of differentiability.

  4. Tangent line

    y=f(x0)+f′(x0)(x−x0)=0.841470984807897+2.22324427548393 (x−1)y = f(x_0) + f'(x_0)(x - x_0) = 0.841470984807897 + 2.22324427548393\,(x - 1)

    y = 2.223244x − 1.381773

About the numerical derivative calculator

The derivative f′(x₀) is the slope of f at x₀, the limit of the central difference quotient [f(x₀ + h) − f(x₀ − h)]/2h as h shrinks to 0. The calculator evaluates that quotient for h = 0.1 × max(1, |x₀|) and repeated halvings, then combines the results by Richardson extrapolation (Ridders' method, Numerical Recipes §5.7), which cancels the h², h⁴, … error terms one at a time and reaches about 20 significant digits. The second derivative uses [f(x₀ + h) − 2f(x₀) + f(x₀ − h)]/h² the same way.

Use it to check a derivative worked out by hand, to find a rate of change where no formula is convenient, or to get a tangent line. The default, x² sin x at x = 1, has derivative 2 sin 1 + cos 1 ≈ 2.2232443 and tangent line y = 2.223244x − 1.381773.

Trig functions use radians. When the slopes from the left and right disagree, f has a corner there and the calculator reports that instead of a number.

How to find the derivative of a function at a point

The derivative f′(x₀) is the slope of f at the point x₀: f′(x) = lim(h→0) [f(x+h)−f(x)]/h. For f(x) = x³, the power rule gives 3x², so the slope at x = 2 is 3 × 2² = 12, and the tangent line there is y = 12x − 16.

The answer here is numerical: the value of f′(x₀) or f″(x₀) at one point, to about 20 significant digits. For the formula itself, use the rules below or the table of common derivatives.

Derivative rules: power, product, quotient and chain

  • Power rule. The derivative of xⁿ is nxⁿ⁻¹, so x³ gives 3x². A constant has derivative 0, and x has derivative 1.
  • Product rule. (fg)′ = f′g + fg′. For x² sin x, that is 2x sin x + x² cos x, which at x = 1 equals 2 sin 1 + cos 1 ≈ 2.22324427548.
  • Quotient rule. (f/g)′ = (f′g − fg′)/g², wherever g is not zero.
  • Chain rule. Differentiate the outer function at the inner value, then multiply by the inner derivative: sin(2x) gives 2cos(2x).

The OpenStax differentiation rules work through the power, product and quotient rules with examples.

Common derivatives: trig, exponential and inverse functions

Trig arguments are in radians. The last column shows where each formula holds: tan, sec and cot need a nonzero denominator, and asin and acos have finite derivatives only for −1 < x < 1. atan(x) is the inverse tangent; cot(x) is 1/tan(x).

Input f(x)First derivativeReal-domain requirement
x1All real x
x^33x²All real x
sqrt(x)1/(2√x)x > 0
sin(x)cos(x)All real x; radians
cos(x)−sin(x)All real x; radians
tan(x)sec²(x)cos(x) ≠ 0
sec(x)sec(x) tan(x)cos(x) ≠ 0
cot(x)−csc²(x)sin(x) ≠ 0
sin(2*x)2cos(2x)All real x; radians
cos(2*x)−2sin(2x)All real x; radians
exp(2*x)2e^(2x)All real x
exp(-2*x)−2e^(−2x)All real x
ln(sec(x))tan(x)sec(x) > 0
x^xx^x (ln(x)+1)x > 0
atan(x)1/(1+x²)All real x
asin(x)1/√(1−x²)−1 < x < 1
acos(x)−1/√(1−x²)−1 < x < 1
atan(sqrt(x))1/[2√x(1+x)]x > 0
x*tan(x)tan(x)+x sec²(x)cos(x) ≠ 0
x*sin(x)sin(x)+x cos(x)All real x; radians

In xˣ the exponent varies too, so logarithmic differentiation gives xˣ(ln x + 1). Treating the exponent as a constant gives x · xˣ⁻¹ = xˣ, which misses the ln x term. At x = 2 the slope is 4(ln 2 + 1) ≈ 6.7726.

ln(sec x) has a real logarithm only where sec x > 0. √x has derivative 1/(2√x) for x > 0 and no finite derivative at 0. The OpenStax exponential and logarithm lesson derives these rules.

Derivative values at a point

Order 1 is the slope f′; order 2 is f″, how fast the slope itself changes. For ln x, f″(x) = −1/x², so f″(2) = −0.25. A second derivative is not the square of the first.

Input f(x)Point xOrderDerivative
cos(2*x)0.51-1.68294197
cos(x)11-0.841470985
cot(x)11-1.412282927
x^32112
atan(sqrt(x))410.05
acos(x)0.51-1.154700538
exp(2*x)1114.778112198
exp(-2*x)11-0.270670566
ln(sec(x))0.510.54630249
sqrt(x)410.25
sin(2*x)0.511.080604612
asin(x)0.511.154700538
x211
x*sin(x)111.381773291
x*tan(x)0.511.195525695
x^x216.772588722
sin(x)110.540302306
tan(x)0.511.29844641
sec(x)0.510.62250837
atan(x)110.5
x^32212
ln(x)22-0.25
5110
x^2 * sin(x)112.223244275

The tangent line at x₀ is y = f(x₀)+f′(x₀)(x−x₀). For x³ at 2, f(2) = 8 and f′(2) = 12, so y = 8 + 12(x − 2) = 12x − 16. A second derivative gives no tangent line, because it is not a slope.

How to type the function

  • Use x as the only variable; 2x and 2*x both work.
  • Trig functions use radians, so sin(x) at x = 30 means 30 radians.
  • Write exponent signs in full. A search for "e 2x" can mean exp(2*x), with derivative 2e^(2x), or exp(-2*x), with derivative −2e^(−2x).
  • Bracket the whole argument: sin(2*x) and 2*sin(x) are different functions with different derivatives.
  • Type powers and roots as x^2, sqrt(x) or x^(1/2).

How the numerical derivative is calculated

The first derivative starts from the central difference quotient [f(x₀ + h) − f(x₀ − h)]/2h, with h = 0.1 × max(1, |x₀|) halved repeatedly. Richardson extrapolation (Ridders' method, Numerical Recipes §5.7) combines the quotients.

Each extrapolation step cancels one more of the h², h⁴, … error terms, which takes the result to about 20 significant digits.

The second derivative uses [f(x₀ + h) − 2f(x₀) + f(x₀ − h)]/h² in the same way. The estimated error is the gap between the last two extrapolated estimates: a measure of how well they settled, not a guaranteed bound.

Slopes from the left and right are compared too. abs(x) at 0 has slope −1 from the left and +1 from the right, so it has no derivative there, even though the central quotient averages to 0. 1/x and √x have no derivative at 0 either.

Common mistakes

  • Dropping the chain-rule factor. The derivative of sin(2x) is 2cos(2x); writing cos(2x) gives half the slope.
  • Working in degrees. The derivative of sin x is cos x only when x is in radians.
  • Using the power rule on xˣ. It applies only when the exponent is a constant.
  • Checking a formula at one point only. Matching values at a few points is evidence that a derivative formula is right, not proof.

A derivative is a rate at one point. To add up change over an interval, use the definite integral calculator.

Worked examples

d/dx x³ at x = 2

Function f(x)
x^3
At x =
2
Derivative
First f′(x)
Derivative
12
f(x₀)
8
Tangent line
y = 12x − 16

Checked against: 3x² = 12 at x = 2; tangent 8 + 12(x − 2)

d/dx sin x at 0 (radians)

Function f(x)
sin(x)
At x =
0
Derivative
First f′(x)
Derivative
1

Checked against: cos 0 = 1

d/dx x²·sin x at 1

Function f(x)
x^2 * sin(x)
At x =
1
Derivative
First f′(x)
Derivative
2.223244275484

Checked against: 2 sin 1 + cos 1 with sin/cos by Taylor series in exact decimal arithmetic at 70 digits

d/dx eˣ at 1

Function f(x)
exp(x)
At x =
1
Derivative
First f′(x)
Derivative
2.718281828459

Checked against: Independent calculation

Questions

What is a derivative?

The derivative of f at x₀ is the slope of its graph there: the limit of [f(x₀ + h) − f(x₀)]/h as h approaches 0. For f(x) = x³ at x = 2 the slope is 3 × 2² = 12, so near x = 2 the function rises about 12 units for each unit of x. The line y = 12x − 16, which touches the curve at (2, 8), is the tangent there.

How do you find a derivative numerically?

Evaluate a difference quotient with a small step h. The central quotient [f(x + h) − f(x − h)]/2h beats the one-sided [f(x + h) − f(x)]/h because its error shrinks like h² rather than h: for sin x at 0 with h = 0.1 it gives 0.998334 against the exact 1. A tiny h eventually fails through rounding error, so this calculator extrapolates from moderate steps instead.

What does the second derivative tell you?

The second derivative f″(x) is the rate of change of the slope, so it measures curvature: positive where the graph bends upward, negative where it bends downward. For ln x, f″(x) = −1/x², so f″(2) = −0.25. Numerically it comes from [f(x + h) − 2f(x) + f(x − h)]/h². Where f″ changes sign the graph has an inflection point.

How do you find the equation of a tangent line?

Use y = f(x₀) + f′(x₀)(x − x₀). For f(x) = x³ at x₀ = 2, f(2) = 8 and f′(2) = 12, so y = 8 + 12(x − 2) = 12x − 16. The tangent is also the best straight-line approximation to f near x₀: it estimates 2.1³ as 8 + 12 × 0.1 = 9.2, against the exact 9.261.

Why does a function have no derivative at some points?

The slopes from the left and the right must agree. |x| at 0 has slope −1 from the left and +1 from the right, a corner, so it has no derivative there, even though the central quotient averages to 0. A jump, or a vertical tangent such as the cube root of x at 0, also rules one out. The calculator compares one-sided quotients and reports the corner.

About this calculator

f′(x)=lim⁡h→0f(x+h)−f(x−h)2hf′′(x)=lim⁡h→0δ2fh2δ2f=f(x+h)−2f(x)+f(x−h)\begin{gathered} f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x-h)}{2h} \\[10pt] f''(x) = \lim_{h \to 0}\frac{\delta^2 f}{h^2} \\[4pt] \delta^2 f = f(x+h) - 2f(x) + f(x-h) \end{gathered}

Sources

  1. Press et al., Numerical Recipes (3rd ed.) §5.7 — numerical derivatives (Ridders' method)
  2. Wolfram MathWorld — Richardson Extrapolation

Checked against references

This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.

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