Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiability of piece-wise functions

  1. Feb 18, 2016 #1
    Hello,
    Me and my friend were talking about differentiability of some piece-wise functions, but we thought of a problem that we could were not able to come to an agreement on. If the function is:
    y=sin(x) for x≠0
    and
    y=x^2 for x=0,
    Is this function differentiable? The graph looks like a normal sin graph, but the derivative at x=0 is 0, although on the graph it looks like 1. I thought that because y=x^2 at only 1 point, you cannot get the derivative at x=0 because there wouldn't be a dx or a dy, as there is no change in the function. So I thought that it would default to the derivative of sin(x), making the whole function differentiable. My friend thought that it is not differentiable. What is the correct answer? Thank you.
     
  2. jcsd
  3. Feb 18, 2016 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    In which point(s) has the function y a different value than the sine function?
    The answer to that question should give you a clue.
     
  4. Feb 18, 2016 #3

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There's no such thing as a piecewise function. A function is a function.
     
  5. Feb 18, 2016 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You realize that 0^2= 0, don't you? So this function is no different from a sine function. That derivative at 0 is NOT 0. You cannot take derivatives that way. You can say "since y= x^2, its derivative is 2x" only on the interior of some interval on which y is defined like that. Here, y is not defined as x^2 on any interval, just at a single point. You are just saying "y(0)= 0" which was already true from y= sin(x).

    If you were to define y= sin(x) for x< -a or x> b, y= x^2 for -a< x< b, then the derivative of y would be 2x for -a< x< b and, in particular, y'(0) would be equal to 0. This function would be neither differentiable nor continuous at x= -a nor x= b.

     
  6. Feb 18, 2016 #5

    Mark44

    Staff: Mentor

    But there are piecewise-defined functions, which is what the OP meant, I'm sure.
    $$f(x) = \begin{cases} \sin(x), & x \ne 0 \\ x^2, & x = 0 \end{cases}$$

    For reasons already explained, this definition is silly, as it is no different from the ordinary sine function.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differentiability of piece-wise functions
Loading...