# Differentiability of piece-wise functions

In summary: But it is certainly differentiable on its entire domain, as the derivative of sin(x) is cos(x) and the derivative of x^2 is 2x, so it is differentiable at x= 0.
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.

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.

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.

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.

PeroK said:
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.

## 1. What is a piece-wise function?

A piece-wise function is a mathematical function that is defined differently over different intervals on its domain.

## 2. What does it mean for a piece-wise function to be differentiable?

A piece-wise function is differentiable if it has a derivative at every point on its domain. In other words, the function has a well-defined slope at each point.

## 3. How do you determine differentiability of a piece-wise function?

To determine the differentiability of a piece-wise function, you need to check if the function is continuous at each point on its domain and if the derivatives from each of the pieces match at the points where they meet.

## 4. Can a piece-wise function be differentiable at a point but not continuous?

Yes, it is possible for a piece-wise function to be differentiable at a point but not continuous. This can happen if the derivatives from each of the pieces do not match at the point where they meet.

## 5. Are there any special cases of piece-wise functions that are always differentiable?

Yes, a piece-wise function with only one piece is always differentiable. Additionally, a piece-wise function with multiple pieces can also be differentiable if all the pieces have the same derivative at the points where they meet.

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