Section 3.2 Differentiability
Subsection 3.2.1 Differentiability on an Interval
Functions that are differentiable are intuitively “smooth”. Values of \(x\) in \(D(f)\) where \(f\) is non-differentiable and are not endpoints are called singular points of \(f\text{.}\)
Note that we used an open interval \((a,b)\) and did not include the endpoints. This is because if \(f\) is defined on a closed interval \([a,b]\text{,}\) then it cannot be differentiable at the endpoints \(x = a\) or \(x = b\text{,}\) because the (two-sided) limit cannot exist if the function is only defined on one side.
Subsection 3.2.2 Continuous But Not Differentiable
Example 3.2.2. Derivative of absolute value function.
Consider the derivative of the absolute value function,
\begin{equation*}
f(x) = \abs{x} = \begin{cases} x \amp x \geq 0 \\ -x \amp x \lt 0 \end{cases}
\end{equation*}
For \(x \neq 0\text{,}\) \(f\) is differentiable. In particular, for \(x > 0\text{,}\) \(f'(x) = \frac{d}{dx} x = 1\text{,}\) and for \(x \lt 0\text{,}\) \(f'(x) = \frac{d}{dx} (-x) = -1\text{.}\) However, at \(x = 0\text{,}\)
\begin{align*}
f'(0) \amp = \lim_{h \to 0} \frac{\abs{0 + h} - \abs{0}}{h}\\
\amp = \lim_{h \to 0} \frac{\abs{h}}{h}
\end{align*}
- From the right, as \(h \to 0+\text{,}\) \(\abs{h} = h\text{,}\) and so,\begin{equation*} \lim_{h \to 0+} \frac{\abs{h}}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1 \end{equation*}Thus, from the right, the slopes approach 1 (this can be confirmed graphically).
- However, from the left, as \(h \to 0-\text{,}\) \(\abs{h} = -h\text{,}\) and so,\begin{equation*} \lim_{h \to 0-} \frac{\abs{h}}{h} = \lim_{h \to 0-} \frac{-h}{h} = \lim_{h \to 0} -1 = -1 \end{equation*}From the left, the slopes approach -1.
Since these one sided limits are unequal, \(\lim_{h \to 0} \frac{\abs{h}}{h}\) does not exist, and so \(f\) is not differentiable at \(x = 0\text{.}\) In summary,
\begin{equation*}
\frac{d}{dx} \abs{x} = \begin{cases} 1 \amp x > 0 \\ -1 \amp x \lt 0 \end{cases}
\end{equation*}
This can be summarized as,
\begin{equation*}
\boxed{\frac{d}{dx} \abs{x} = \frac{\abs{x}}{x} \qquad x \neq 0}
\end{equation*}
Subsection 3.2.3 Differentiability and Continuity
If a function \(f\) is continuous at \(x\text{,}\) it may or may not be differentiable at \(x\text{.}\) For example, \(f(x) = \abs{x}\) is continuous but not differentiable at \(x = 0\text{.}\) However, the converse is true, in that differentiability implies continuity.
Theorem 3.2.3.
Let \(f\) be a function. If \(f\) is differentiable at \(x = a\text{,}\) then it is continuous at \(x = a\text{.}\)
Intuitively, this theorem says that if a function is smooth at a point \(a\text{,}\) then it can't have a discontinuity there. Alternatively, it is more intuitive to consider the contrapositive, that if \(x\) is not continuous at \(a\) (i.e. it has a break or jump), then it is not differentiable at \(a\) (the slope of a tangent line cannot be found).
Corollary 3.2.4.
If \(f\) is discontinuous at \(a\text{,}\) then \(f\) is not differentiable at \(a\text{.}\)
