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Sayısal türev hesaplayıcı

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.

Güncellendi Doğrulanan örnekler: 6

Use x as the variable, e.g. x^3 - 2x, exp(x), ln(x). Trig functions use radians here.
Diğer seçenekler
The chart shows x₀ ± this much
Dene
Türev
Türev: 2.223244275484
En fazla ondalık basamak: 12; En yakına; eşit uzaklıkta sıfırdan uzağa
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) Satır sayısı: 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
Nasıl hesaplanır 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

Sayısal türev hesaplayıcı hakkında

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.

Çözümlü örnekler

d/dx x³ at x = 2

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

Doğrulama kaynağı: 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
Türev
First f′(x)
Türev
1

Doğrulama kaynağı: cos 0 = 1

d/dx x²·sin x at 1

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

Doğrulama kaynağı: 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
Türev
First f′(x)
Türev
2.718281828459

Doğrulama kaynağı: e (Python Decimal(1).exp())

Sorular

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.

“Sayısal türev hesaplayıcı” ne kadar doğru sonuç verir?

Doğruluk, girdilerinize ve yöntemin varsayımlarına bağlıdır. Ondalık aritmetik 50 anlamlı basamak kullanır; ancak tahminler, sayısal yöntemler ve kaynak veriler daha az hassas olabilir. Gösterilen değerin yuvarlanması bu sınırları ortadan kaldırmaz. Bağımsız kaynaklarla doğrulanan çözümlü örnek sayısı: 6. Örneğin “d/dx x³ at x = 2”, 3x² = 12 at x = 2; tangent 8 + 12(x − 2) ile karşılaştırılarak doğrulanır.

Yöntemin kaynağı nedir?

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

Bu hesaplayıcı hakkında

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}

Kaynaklar

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

Kaynaklarla doğrulandı

Bu hesaplayıcı, yanıtları bağımsız kaynaklardan alınan 6 çözümlü örnek içerir. Bunlar test paketinde çalıştırılır; burada da çalıştırabilirsiniz.

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