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# scipy.interpolate.LSQUnivariateSpline.derivative

LSQUnivariateSpline.derivative(n=1)

  Construct a new spline representing the derivative of this spline.

  Parameters:
    n int, optional

      Order of derivative to evaluate. Default: 1

  Returns:
    spline UnivariateSpline

      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:

      >>> import numpy as np
      >>> 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 \(\cos(x) = \sin'(x)\).