scipy.interpolate.LSQUnivariateSpline.derivative

LSQUnivariateSpline.derivative(n=1)[source]

Construct a new spline representing the derivative of this spline.

Parameters
nint, optional

Order of derivative to evaluate. Default: 1

Returns
splineUnivariateSpline

Spline of order k2=k-n representing the derivative of this spline.

See also

splder, antiderivative

Notes

New in version 0.13.0.

Examples

This can be used for finding maxima of a curve:

>>> from scipy.interpolate import UnivariateSpline
>>> x = np.linspace(0, 10, 70)
>>> y = np.sin(x)
>>> spl = UnivariateSpline(x, y, k=4, s=0)

Now, differentiate the spline and find the zeros of the derivative. (NB: sproot only works for order 3 splines, so we fit an order 4 spline):

>>> spl.derivative().roots() / np.pi
array([ 0.50000001,  1.5       ,  2.49999998])

This agrees well with roots \(\pi/2 + n\pi\) of \(\cos(x) = \sin'(x)\).

scipy.interpolate.LSQUnivariateSpline.antiderivative scipy.interpolate.LSQUnivariateSpline.derivatives