Differentiability in an Interval

A function \( f(x) \) is said to be differentiable on an interval if it is differentiable at every point in that interval. The nature of differentiability depends on whether the interval is open or closed, and special considerations are required at endpoints.

Differentiability on an Open Interval

A function \( f(x) \) is differentiable on an open interval \( (a, b) \) if it is differentiable at every point \( x \in (a, b) \). That is, for every \( c \in (a, b) \), the limit

\[ f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \]

must exist as a finite real number.

Since \( (a, b) \) does not include the endpoints \( a \) and \( b \), differentiability is not defined at these points.

Differentiability on a Closed Interval

A function \( f(x) \) is differentiable on a closed interval \( [a, b] \) if: 1. It is differentiable at every point \( x \in (a, b) \) (i.e., in the open part of the interval). 2. It is differentiable at the endpoints \( a \) and \( b \) using one-sided derivatives.

At an interior point \( c \in (a, b) \), differentiability means that both the left-hand and right-hand derivatives exist and are equal. However, at the endpoints, only a one-sided derivative can be considered.

Differentiability at Endpoints

At \( x = a \), differentiability means that the right-hand derivative (RHD) exists:

\[ f'(a^+) = \lim_{x \to a^+} \frac{f(x) - f(a)}{x - a}. \]

At \( x = b \), differentiability means that the left-hand derivative (LHD) exists:

\[ f'(b^-) = \lim_{x \to b^-} \frac{f(x) - f(b)}{x - b}. \]

If either of these limits fails to exist or is infinite, the function is not differentiable at that endpoint.

Since differentiation is a local property, a function can be differentiable on an open interval \( (a, b) \) but fail to be differentiable at one or both endpoints of \( [a, b] \), making it differentiable only on \( (a, b] \), \( [a, b) \), or neither.

Example

Consider the function

\[ f(x) = \sqrt{1 - x^2}, \quad x \in [-1,1]. \]

The function \( f \) is differentiable at all points \( x \in (-1,1) \), but at \( x = 1 \) and \( x = -1 \), it is not differentiable. Let us verify at both points:

At \( x = 1 \):

The left-hand derivative (LHD) is given by

\[ \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1}. \]

Since \( f(1) = \sqrt{1 - 1^2} = 0 \),

\[ \lim_{x \to 1^-} \frac{\sqrt{1 - x^2} - 0}{x - 1} = \lim_{x \to 1^-} \frac{\sqrt{1 - x^2}}{x - 1}. \]

\[ = \lim_{x \to 1^-} -\frac{\sqrt{1 - x} \cdot \sqrt{1 + x}}{1 - x}. \]

Canceling \( \sqrt{1 - x} \),

\[ \lim_{x \to 1^-} -\frac{\sqrt{1 + x}}{\sqrt{1 - x}}. \]

As \( x \to 1^- \), \( \sqrt{1 - x} \to 0^+ \) and \( \sqrt{1 + x} \to \sqrt{2} \), so

\[ = -\frac{\sqrt{2}}{0^+} = -\infty. \]

Thus, \( f(x) \) is not differentiable at \( x = 1 \).

At \( x = -1 \):

The right-hand derivative (RHD) is given by

\[ \lim_{x \to -1^+} \frac{f(x) - f(-1)}{x - (-1)}. \]

Since \( f(-1) = \sqrt{1 - (-1)^2} = 0 \),

\[ \lim_{x \to -1^+} \frac{\sqrt{1 - x^2} - 0}{x + 1} = \lim_{x \to -1^+} \frac{\sqrt{1 - x^2}}{x + 1}. \]

\[ = \lim_{x \to -1^+} \frac{\sqrt{1 - x} \cdot \sqrt{1 + x}}{1 + x}. \]

Canceling \( \sqrt{1 + x} \),

\[ \lim_{x \to -1^+} \frac{\sqrt{1 - x}}{\sqrt{1 + x}}. \]

As \( x \to -1^+ \), \( \sqrt{1 - x} \to \sqrt(2)^+ \) and \( \sqrt{1 + x} \to 0^+ \), so

\[ = +\infty. \]

Thus, \( f(x) \) is not differentiable at \( x = -1 \).

Since the left-hand derivative at \( x = 1 \) is \( -\infty \) and the right-hand derivative at \( x = -1 \) is \( +\infty \), we conclude that \( f(x) \) is not differentiable at the endpoints. However, it remains differentiable on the open interval \( (-1,1) \).