Topic 12.4: Higher-Order Derivatives

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Given three or more equally-spaced points, we can find an interpolating polynomial passing through those points, find the 2nd, 3rd, or even higher derivative of that polynomial, and evaluate the derivative at a point to get an approximation of the derivative at that point.


Useful background for this topic includes:


Interactive Maplet

A Differentiation Formula Generator

To generate the formula for higher derivatives, select 2nd, 3rd, etc. instead of 1st in the third drop-down box.


The theory for finding the nth-order derivatives is the same as that for finding simple derivatives: Given a sequence of equally spaced points (either around or preceding a given point x0), find the interpolating polynomial of the appropriate degree, differentiate the interpolating polynomial n times (instead of once) and evaluate that interpolating polynomial at the point.

For the sake of completeness, we will list the centred and backward divided-difference formulae here. In each case, the format of Richardson extrapolation we found could be used to improve the approximations for the O(h2) formulae on this page.

Centred Divided-Difference Formulae

For each derivative, centred divided-difference formulae are O(h2) and O(h4), respectively.

Second Derivatives (3 and 5 point interpolations)

Third Derivatives (5 and 7 point interpolations)

Fourth Derivatives (5 and 7 point interpolations)

Fifth Derivatives (7 and 9 point interpolations)

Backward Divided-Difference Formulae

For each derivative, backward divided-difference formulae are O(h) and O(h2), respectively. Note that in all cases, the O(h) formula is equivalent to the corresponding O(h2) centred divided-difference formula, only shifted by h.

Second Derivatives (3 and 4 point interpolations)

Third Derivatives (4 and 5 point interpolations)

Fourth Derivatives (5 and 6 point interpolations)

Fifth Derivatives (6 and 7 point interpolations)



Approximate a higher derivative of a univariate function f(x) at a point x0. We will assume that we are given a sequence of points (xi, f(xi)) around the point of interest (either before or around). We will not look at iteration because the process of Richardson extrapolation converges significantly faster.


We need to assume the function has an nth derivative if we are to bound the error on our approximation.


We will use interpolation.


If we are to evaluate the second derivative at the point (xi, f(xi)) and have access to the two surrounding points, (xi − 1, f(xi − 1)) and (xi + 1, f(xi + 1)), then we may find the interpolating polynomial, differentiate it twice, and evaluate that derivative at xi:

This is simply another form of the formula

where h is the distance between the points, that is, h = xi - xi − 1.

If we have access to two points on either side of xi, we can calculate

where h = xi - xi − 1.

This is another form of the formula:

We could perform the same operations with higher derivatives, however, it should be noted that to calculate the nth derivative, we require at least n + 1 points.


Example 1

Given the data points ..., 3.2, 3.3, 3.4, 3.5, 3.6, ... which measure the rotation of a satellite dish at points in time, with angles 1.05837 1.15775 1.25554 1.35078 1.44252, approximate the acceleration of the angle at time 3.4 using both the 2nd-order and 4th-order centred divided-difference formulae for the 2nd derivative.

(1.35078 − 2⋅1.25554 + 1.15775)/0.12 = −0.255 and (-1.44252 + 16⋅1.35078 − 30⋅1.25554 + 16⋅1.15775 − 1.05837)/(12⋅0.12) = −0.2550833333⋅⋅⋅.

Thus, −0.25508333333 is the more accurate answer, but even the 2nd-order formula is reasonably close.


To be completed.


2nd-Order Centred Divided-Difference Formula

To determine the error for the 2nd-order centred divided-difference formula for the second derivative, we need only look at the two Taylor series approximations:


Add these two equations and transfer the 2f(x) to the left-hand side to get

If we divide both sides by h2 and make the approximation that:

then we can rearrange the equation as

Thus, the error is O(h2).

4th-Order Centred Divided-Difference Formula

A similar sum may be used to find the error of the 4th-order divided difference formula. If you add the linear combination -f(x + 2h) + 8 f(x + h) - 8 f(x - h) + f(x - 2h) of the 5th-order Taylor series approximations, then, after dividing by 12h, we are left with the error term:

If we divide through by -1/30 and factor out the h4, we get

Now, examining the contents of the parentheses, we note that the coefficients 2/3 - 1/6 - 1/6 + 2/3 = 1, and therefore, the contents of the parentheses is an approximation of the average of f(5)(x) on the interval [x - 2h, x + 2h], and thus, we may approximate the error by

Thus, the error is O(h4).


Question 1

Approximate the rate-of-change of the current at the midpoint of the incoming data ..., 7.2, 7.3, 7.4, 7.5, 7.6, ... where the charge on a capacitor these times is 0.00242759 0.00241500 0.00240247 0.00239001 0.00237761.

Answer: -0.000007 and -0.0000071667.


In Matlab, you would approximate the derivative numerically:

>> ( sin( 1.7 ) - sin( 1.3 ) )/0.4

which would approximate the actual derivative of cos(1.5).


Maple calculates derivatives symbolically:

> diff( sin(x), x );

For more help on the diff routine, enter:


Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.