As the Dirac delta function is essentially an infinitely high spike at a single point, it may seem odd that its derivatives can be defined. The derivatives are defined using the delta function’s integral property

Consider the integral involving the th derivative and apply integration by parts:

The integrated term is taken to be zero, since the delta function itself is constant (at zero) for all , so all its derivatives are zero except at . Therefore at the limits and . We’re therefore left with

Since this is true for all functions , the integrands must be equal, so we get

A common case is the first derivative, which satisfies

If , we get the relation

By iterating 4, we get

Example 1Suppose . Then

Example 2With we have, using 7

Another use of the derivative of the delta function occurs frequently in quantum mechanics. In this case, we are faced with the integral

where the prime in refers to a derivative with respect to , not . Thus the variable in the derivative is *not* the same as the variable being integrated over, unlike the preceding cases. In this case, since only (and not ) is visible outside the integral, we can move the derivative outside the integral and get

Notice that in this case, there is no minus sign attached to the unlike in 5.

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Xinqiang WangThe equation (7) seems wrong. There should not be the symble of multiple mutiplication. And it shoud be f^(n) (x)

gwrowePost authorFixed now. Thanks.

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