# Differentiability of piece-wise functions

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.

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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.
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.

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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.

There's no such thing as a piecewise function. A function is a function.

Homework Helper
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,
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.

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.

Mentor
There's no such thing as a piecewise function. A function is a function.
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.