util.py 文件源码

def idiff(eq, y, x, n=1):
    """Return ``dy/dx`` assuming that ``eq == 0``.

    Parameters
    ==========

    y : the dependent variable or a list of dependent variables (with y first)
    x : the variable that the derivative is being taken with respect to
    n : the order of the derivative (default is 1)

    Examples
    ========

    >>> from sympy.abc import x, y, a
    >>> from sympy.geometry.util import idiff

    >>> circ = x**2 + y**2 - 4
    >>> idiff(circ, y, x)
    -x/y
    >>> idiff(circ, y, x, 2).simplify()
    -(x**2 + y**2)/y**3

    Here, ``a`` is assumed to be independent of ``x``:

    >>> idiff(x + a + y, y, x)
    -1

    Now the x-dependence of ``a`` is made explicit by listing ``a`` after
    ``y`` in a list.

    >>> idiff(x + a + y, [y, a], x)
    -Derivative(a, x) - 1

    See Also
    ========

    sympy.core.function.Derivative: represents unevaluated derivatives
    sympy.core.function.diff: explicitly differentiates wrt symbols

    """
    if is_sequence(y):
        dep = set(y)
        y = y[0]
    elif isinstance(y, Symbol):
        dep = {y}
    else:
        raise ValueError("expecting x-dependent symbol(s) but got: %s" % y)

    f = dict([(s, Function(
        s.name)(x)) for s in eq.free_symbols if s != x and s in dep])
    dydx = Function(y.name)(x).diff(x)
    eq = eq.subs(f)
    derivs = {}
    for i in range(n):
        yp = solve(eq.diff(x), dydx)[0].subs(derivs)
        if i == n - 1:
            return yp.subs([(v, k) for k, v in f.items()])
        derivs[dydx] = yp
        eq = dydx - yp
        dydx = dydx.diff(x)