Generalisation of derivative

Sunday 6 June 2004 at 2:25 pm | In Articles | Post Comment

Inspired by a posting on S.O.S. Mathematics CyberBoard

Most students will be familiar with the definition of the derivative of a real-valued function of a real variable defined on some interval (a,b):

    If x \in (a,b) then f is differentiable at x if \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} exists and the limit is denoted f^{\prime}(x)

It is also clear that for this to make sense f must be defined at x (and of course it is a well-known consequence of the definition that f is also continuous at x). But what if f is defined on (a,b) but not at x, can we do anything then? Yes, we can define a pseudo-derivative \Lambda f of f provided f is defined on a neighbourhood of x:

    \displaystyle \Lambda f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}

This pseudo-derivative has similar properties to the derivative and indeed it has the same values where f is differentiable but there are significant differences as the following exercises show:

  1. If f is differentiable at x show that \Lambda f(x)=f^{\prime}x)
  2. If f(x)=|x| show that \Lambda f(0) exists although f^{\prime}(0) does not
  3. If f(x)=\left\{\begin{array}{ll}x & \mbox{ if } x<0\\-x^2 & \mbox{ if } x \geq 0\end{array}\right. show that f has a local maximum at 0 but \Lambda f(0) \neq 0
  4. Suppose f is differentiable on (a,b), except at a point x_0 in (a,b), with f^{\prime}(x)\geq 0 for  x \neq x_0.
    If \Lambda (f)(x_0) exists show that \Lambda (f)(x)\geq0

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