# Question about domain of derivatives

1. Aug 20, 2010

### Taturana

I was thinking how do I differentiate the domain of functions...

Suppose I have a function:

$$f(x) = \left\{\begin{matrix} x^2 -1, \;\; |x| \leq 1\\ 1 - x^2, \;\; |x| > 1 \end{matrix}\right.$$

And I need to derive it:

$$f'(x) = \left\{\begin{matrix} 2x, \;\; |x| \leq 1\\ -2x, \;\; |x| > 1 \end{matrix}\right.$$

1. What do I do with the conditions of |x| ? Is there a rule to handle this?
2. What will be the domain of f'(x)? Will the domain of f'(x) be the domain of f(x) minus the points where the function is non-differentiable?

Thank you,
Rafael Andreatta

2. Aug 20, 2010

### slider142

With regards to the derivative function for single variable functions, the maximal domain of the function that gives the derivative of f at each point will be the set of points in the domain of f where f is differentiable.
For multivariable and vector-valued functions, a more sophisticated definition of the derivative at a point and the derivative function are necessary.
The derivative at a point is generalized to a linear function over the tangent space to the point in the image of the function, and the derivative function is then a map from points in the domain of f to the space of linear functions over the tangent space to each point.
This simplifies to the single-variable case, as linear functions over the tangent line to the curve are constants. From this perspective, the domain would be isomorphic, but not strictly the same.

Last edited: Aug 20, 2010
3. Aug 20, 2010

In a problem like this one (finding the derivative of a piecewise-defined function) I would state the work steps this way (this is my explanation only, not written to be textbook perfect)

* Use ordinary derivative rules everywhere except for the domain values where the different pieces are put together
* Examine the x-values where the pieces are joined. If the one-sided derivatives there agree, no problem. If the one-sided derivatives do not agree, your original function is not differentiable there.

Note: my original first example had an error, pointed out by l'Hopital below: read my later post for the correction.

However, if I have

$$h(x) = \begin{cases} x^2 + 3x, \quad x \le 2 \\ 5x^2 - 5x, \quad x > 2 \end{cases}$$

I need to examine the two one-sided derivatives at x = 2. The left-hand derivative there
is $7$, while the right-hand derivative is $15$. Since the left- and right-hand derivatives don't agree there, h is not differentiable there . Thus, for my second example (notice that $$\le 2$$ has been replaced by $$< 2$$)

$$h'(x) = \begin{cases} 2x + 3, \quad x < 2 \\ 10x - 5, \quad x > 2 \end{cases}$$

Hope this helps.

Last edited: Aug 20, 2010
4. Aug 20, 2010

### l'Hôpital

I have a question regarding g in statdad's post. It is clear g is not continuous at x = 0. Does it really make sense for g to have a derivative x = 0 ?

5. Aug 20, 2010

Well poop (that's not what I said to myself when I saw your email pop up, but you should get the drift)

Perfect point and I have no excuse for my screwup. Let me use this example.

$$f(x) = \begin{cases} x^2 + 6x - 7, \quad x \le 1 \\ -9 + 10x - x^2, \quad x > 1 \end{cases}$$

A quick check now, using the ideas in my earlier post, gives

$$f'(x) = \begin{cases} 2x + 6, \quad x \le 1\\ 10-2x, \quad x > 1 \end{cases}$$

The idea of my second example still works. Huge apologies for any confusion this may have caused, and a big "Thank-you" to l'Hopital for being a fantastic proof reader.