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Kalkulator turunan numerik

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.

Diperbarui Contoh terverifikasi: 6

Use x as the variable, e.g. x^3 - 2x, exp(x), ln(x). Trig functions use radians here.
Opsi lainnya
The chart shows x₀ ± this much
Coba
Turunan
Turunan: 2.223244275484
Tempat desimal maksimum: 12; Terdekat, jika berjarak sama menjauhi nol
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: near that point, 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) Baris: 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
Cara menghitung 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

    left 2.22324427548393,right 2.22324427548393\text{left } 2.22324427548393,\quad \text{right } 2.22324427548393

    One-sided quotients from each side converge to the same value, so f is smooth enough here for the derivative to exist.

  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

Tentang Kalkulator turunan numerik

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.

Contoh penyelesaian

d/dx x³ at x = 2

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

Sumber pemeriksaan: 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
Turunan
First f′(x)
Turunan
1

Sumber pemeriksaan: cos 0 = 1

d/dx x²·sin x at 1

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

Sumber pemeriksaan: 2 sin 1 + cos 1 with sin/cos by Taylor series in Python decimal at 70 digits (hp.py)

d/dx eˣ at 1

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

Sumber pemeriksaan: e (Python Decimal(1).exp())

Pertanyaan

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.

Seberapa akurat “Kalkulator turunan numerik”?

Akurasi bergantung pada masukan dan asumsi metode. Aritmetika desimal memakai 50 digit signifikan, tetapi perkiraan, metode numerik dan data sumber bisa kurang presisi; pembulatan yang ditampilkan tidak menghilangkan batasan itu. Contoh penyelesaian yang diperiksa dengan sumber independen: 6. Misalnya, “d/dx x³ at x = 2” diperiksa dengan 3x² = 12 at x = 2; tangent 8 + 12(x − 2).

Dari mana metode ini berasal?

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

Tentang kalkulator ini

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}

Sumber

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

Diperiksa dengan referensi

Kalkulator ini mencakup 6 contoh perhitungan dengan jawaban dari sumber independen. Contoh tersebut dijalankan dalam rangkaian pengujian dan dapat Anda jalankan di sini juga.

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