Как найти производную от функции в питоне

Derivatives#

To take derivatives, use the diff function.

>>> diff(cos(x), x)
-sin(x)
>>> diff(exp(x**2), x)
     ⎛ 2⎞
     ⎝x ⎠
2⋅x⋅ℯ

diff can take multiple derivatives at once. To take multiple derivatives,
pass the variable as many times as you wish to differentiate, or pass a number
after the variable. For example, both of the following find the third
derivative of (x^4).

>>> diff(x**4, x, x, x)
24⋅x
>>> diff(x**4, x, 3)
24⋅x

You can also take derivatives with respect to many variables at once. Just
pass each derivative in order, using the same syntax as for single variable
derivatives. For example, each of the following will compute
(frac{partial^7}{partial xpartial y^2partial z^4} e^{x y z}).

>>> expr = exp(x*y*z)
>>> diff(expr, x, y, y, z, z, z, z)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ
>>> diff(expr, x, y, 2, z, 4)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ
>>> diff(expr, x, y, y, z, 4)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ

diff can also be called as a method. The two ways of calling diff are
exactly the same, and are provided only for convenience.

>>> expr.diff(x, y, y, z, 4)
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ

To create an unevaluated derivative, use the Derivative class. It has the
same syntax as diff.

>>> deriv = Derivative(expr, x, y, y, z, 4)
>>> deriv
     7
    ∂     ⎛ x⋅y⋅z⎞
──────────⎝ℯ     ⎠
  4   2
∂z  ∂y  ∂x

To evaluate an unevaluated derivative, use the doit method.

>>> deriv.doit()
 3  2 ⎛ 3  3  3       2  2  2                ⎞  x⋅y⋅z
x ⋅y ⋅⎝x ⋅y ⋅z  + 14⋅x ⋅y ⋅z  + 52⋅x⋅y⋅z + 48⎠⋅ℯ

These unevaluated objects are useful for delaying the evaluation of the
derivative, or for printing purposes. They are also used when SymPy does not
know how to compute the derivative of an expression (for example, if it
contains an undefined function, which are described in the Solving
Differential Equations section).

Derivatives of unspecified order can be created using tuple (x, n) where
n is the order of the derivative with respect to x.

>>> m, n, a, b = symbols('m n a b')
>>> expr = (a*x + b)**m
>>> expr.diff((x, n))
  n
 ∂ ⎛         m⎞
───⎝(a⋅x + b) ⎠
  n
∂x

Integrals#

To compute an integral, use the integrate function. There are two kinds
of integrals, definite and indefinite. To compute an indefinite integral,
that is, an antiderivative, or primitive, just pass the variable after the
expression.

>>> integrate(cos(x), x)
sin(x)

Note that SymPy does not include the constant of integration. If you want it,
you can add one yourself, or rephrase your problem as a differential equation
and use dsolve to solve it, which does add the constant (see Solving Differential Equations).

To compute a definite integral, pass the argument (integration_variable,
lower_limit, upper_limit). For example, to compute

we would do

>>> integrate(exp(-x), (x, 0, oo))
1

As with indefinite integrals, you can pass multiple limit tuples to perform a
multiple integral. For example, to compute

do

>>> integrate(exp(-x**2 - y**2), (x, -oo, oo), (y, -oo, oo))
π

If integrate is unable to compute an integral, it returns an unevaluated
Integral object.

>>> expr = integrate(x**x, x)
>>> print(expr)
Integral(x**x, x)
>>> expr
⌠
⎮  x
⎮ x  dx
⌡

As with Derivative, you can create an unevaluated integral using
Integral. To later evaluate this integral, call doit.

>>> expr = Integral(log(x)**2, x)
>>> expr
⌠
⎮    2
⎮ log (x) dx
⌡
>>> expr.doit()
         2
x⋅log (x) - 2⋅x⋅log(x) + 2⋅x

integrate uses powerful algorithms that are always improving to compute
both definite and indefinite integrals, including heuristic pattern matching
type algorithms, a partial implementation of the Risch algorithm, and an algorithm using
Meijer G-functions that is
useful for computing integrals in terms of special functions, especially
definite integrals. Here is a sampling of some of the power of integrate.

>>> integ = Integral((x**4 + x**2*exp(x) - x**2 - 2*x*exp(x) - 2*x -
...     exp(x))*exp(x)/((x - 1)**2*(x + 1)**2*(exp(x) + 1)), x)
>>> integ
⌠
⎮ ⎛ 4    2  x    2        x          x⎞  x
⎮ ⎝x  + x ⋅ℯ  - x  - 2⋅x⋅ℯ  - 2⋅x - ℯ ⎠⋅ℯ
⎮ ──────────────────────────────────────── dx
⎮               2        2 ⎛ x    ⎞
⎮        (x - 1) ⋅(x + 1) ⋅⎝ℯ  + 1⎠
⌡
>>> integ.doit()
                 x
   ⎛ x    ⎞     ℯ
log⎝ℯ  + 1⎠ + ──────
               2
              x  - 1

>>> integ = Integral(sin(x**2), x)
>>> integ
⌠
⎮    ⎛ 2⎞
⎮ sin⎝x ⎠ dx
⌡
>>> integ.doit()
         ⎛√2⋅x⎞
3⋅√2⋅√π⋅S⎜────⎟⋅Γ(3/4)
         ⎝ √π ⎠
──────────────────────
       8⋅Γ(7/4)

>>> integ = Integral(x**y*exp(-x), (x, 0, oo))
>>> integ
∞
⌠
⎮  y  -x
⎮ x ⋅ℯ   dx
⌡
0
>>> integ.doit()
⎧ Γ(y + 1)    for re(y) > -1
⎪
⎪∞
⎪⌠
⎨⎮  y  -x
⎪⎮ x ⋅ℯ   dx    otherwise
⎪⌡
⎪0
⎩

This last example returned a Piecewise expression because the integral
does not converge unless (Re(y) > 1.)

Limits#

SymPy can compute symbolic limits with the limit function. The syntax to compute

is limit(f(x), x, x0).

>>> limit(sin(x)/x, x, 0)
1

limit should be used instead of subs whenever the point of evaluation
is a singularity. Even though SymPy has objects to represent (infty), using
them for evaluation is not reliable because they do not keep track of things
like rate of growth. Also, things like (infty – infty) and
(frac{infty}{infty}) return (mathrm{nan}) (not-a-number). For example

>>> expr = x**2/exp(x)
>>> expr.subs(x, oo)
nan
>>> limit(expr, x, oo)
0

Like Derivative and Integral, limit has an unevaluated
counterpart, Limit. To evaluate it, use doit.

>>> expr = Limit((cos(x) - 1)/x, x, 0)
>>> expr
     ⎛cos(x) - 1⎞
 lim ⎜──────────⎟
x─→0⁺⎝    x     ⎠
>>> expr.doit()
0

To evaluate a limit at one side only, pass '+' or '-' as a fourth
argument to limit. For example, to compute

do

>>> limit(1/x, x, 0, '+')
∞

As opposed to

>>> limit(1/x, x, 0, '-')
-∞

Series Expansion#

SymPy can compute asymptotic series expansions of functions around a point. To
compute the expansion of (f(x)) around the point (x = x_0) terms of order
(x^n), use f(x).series(x, x0, n). x0 and n can be omitted, in
which case the defaults x0=0 and n=6 will be used.

>>> expr = exp(sin(x))
>>> expr.series(x, 0, 4)
         2
        x     ⎛ 4⎞
1 + x + ── + O⎝x ⎠
        2

The (Oleft(x^4right)) term at the end represents the Landau order term at
(x=0) (not to be confused with big O notation used in computer science, which
generally represents the Landau order term at (x) where (x rightarrow infty)). It means that all
x terms with power greater than or equal to (x^4) are omitted. Order terms
can be created and manipulated outside of series. They automatically
absorb higher order terms.

>>> x + x**3 + x**6 + O(x**4)
     3    ⎛ 4⎞
x + x  + O⎝x ⎠
>>> x*O(1)
O(x)

If you do not want the order term, use the removeO method.

>>> expr.series(x, 0, 4).removeO()
 2
x
── + x + 1
2

The O notation supports arbitrary limit points (other than 0):

>>> exp(x - 6).series(x, x0=6)
            2          3          4          5
     (x - 6)    (x - 6)    (x - 6)    (x - 6)         ⎛       6       ⎞
-5 + ──────── + ──────── + ──────── + ──────── + x + O⎝(x - 6) ; x → 6⎠
        2          6          24        120

Finite differences#

So far we have looked at expressions with analytic derivatives
and primitive functions respectively. But what if we want to have an
expression to estimate a derivative of a curve for which we lack a
closed form representation, or for which we don’t know the functional
values for yet. One approach would be to use a finite difference
approach.

The simplest way the differentiate using finite differences is to use
the differentiate_finite function:

>>> f, g = symbols('f g', cls=Function)
>>> differentiate_finite(f(x)*g(x))
-f(x - 1/2)⋅g(x - 1/2) + f(x + 1/2)⋅g(x + 1/2)

If you already have a Derivative instance, you can use the
as_finite_difference method to generate approximations of the
derivative to arbitrary order:

>>> f = Function('f')
>>> dfdx = f(x).diff(x)
>>> dfdx.as_finite_difference()
-f(x - 1/2) + f(x + 1/2)

here the first order derivative was approximated around x using a
minimum number of points (2 for 1st order derivative) evaluated
equidistantly using a step-size of 1. We can use arbitrary steps
(possibly containing symbolic expressions):

>>> f = Function('f')
>>> d2fdx2 = f(x).diff(x, 2)
>>> h = Symbol('h')
>>> d2fdx2.as_finite_difference([-3*h,-h,2*h])
f(-3⋅h)   f(-h)   2⋅f(2⋅h)
─────── - ───── + ────────
     2        2        2
  5⋅h      3⋅h     15⋅h

If you are just interested in evaluating the weights, you can do so
manually:

>>> finite_diff_weights(2, [-3, -1, 2], 0)[-1][-1]
[1/5, -1/3, 2/15]

note that we only need the last element in the last sublist
returned from finite_diff_weights. The reason for this is that
the function also generates weights for lower derivatives and
using fewer points (see the documentation of finite_diff_weights
for more details).

If using finite_diff_weights directly looks complicated, and the
as_finite_difference method of Derivative instances
is not flexible enough, you can use apply_finite_diff which
takes order, x_list, y_list and x0 as parameters:

>>> x_list = [-3, 1, 2]
>>> y_list = symbols('a b c')
>>> apply_finite_diff(1, x_list, y_list, 0)
  3⋅a   b   2⋅c
- ─── - ─ + ───
   20   4    5