Differentiability of a Function

Differentiability of a Function

Definition: A function $f$ is said to be Differentiable at the value $a$ if $f'(a)$ exists, that is $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ exists. $f$ is said to be differentiable on the interval $I$ if for all values $x \in I$, $f'(x)$ is defined, that is, $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ exists for all $x \in I$.

Once again, consider the function $f(x) = \frac{1}{x}$, and suppose we want to find the slope of the tangent line at $x = 0$. Applying the definition of the derivative, we get:

\begin{align} f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \\ f'(a) = \lim_{x \to 0} \frac{\frac{1}{x} - \frac{1}{0}}{x - 0} \end{align}

Clearly, there is a problem here since $\frac{1}{0}$ is undefined and thus, $f$ is not differentiable at $0$.

We will now look at the three ways in which a function is not differentiable.

Functions Containing Discontinuities

This should be rather obvious, but a function that contains a discontinuity is not differentiable at its discontinuity. Recall that there are three types of discontinuities. The first type of discontinuity is asymptotic discontinuities. If a function has an asymptote at $x = a$, then $f(a)$ itself is not defined and therefore $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ is undefined too.

If a function has a point discontinuity at $x = a$, then $f'(a)$ is undefined since the point $(a, f(a))$ is singled out, so our definition of a limit fails since we can't take arbitrarily close secant lines to hone into a tangent line at $a$. The same intuition holds for jump discontinuities.

We will now look at a very important theorem regarding differentiation and continuity.

Theorem: If $f$ is a function that is differentiable at the value $a$, then $f$ is also continuous at the value $a$. Note that the continuity of a function $f$ at $a$ does not imply $f$ is differentiable at $a$.
  • Proof of Theorem: Since $f$ is differentiable at $a$, it follows that $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$.
  • Now, take $f(x) - f(a)$ and divide/multiply it by $x - a$ and then take the limit of both sides to get:
\begin{align} f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a) \\ \lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \left ( \frac{f(x) - f(a)}{x - a} \cdot (x - a) \right )\\ \end{align}
  • Applying limit laws to the righthand side we get that:
\begin{align} \lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \cdot \lim_{x \to a} (x - a) \\ \lim_{x \to a} [f(x) - f(a)] = f'(a) \cdot 0 \\ \lim_{x \to a} [f(x) - f(a)] = 0 \end{align}
  • Note that we can definitively say $\lim_{x \to a} [f(x) - f(a)] = 0$ since $f'(a)$ exists from earlier. Now, take $f(x)$ and add/subtract $f(a)$ as follows:
\begin{align} f(x) = f(x) + f(a) - f(a) \\ f(x) = f(a) + [f(x) - f(a)] \\ \lim_{x \to a} f(x) = \lim_{x \to a} \left ( f(a) + [f(x) - f(a)] \right ) \\ \lim_{x \to a} f(x) = \lim_{x \to a} f(a) + \lim_{x \to a} [f(x) - f(a)] \\ \lim_{x \to a} f(x) = f(a) + 0 \end{align}
  • So we know that $\lim_{x \to a} f(x) = f(a)$. This implies that $f(a)$ is defined and $\lim_{x \to a} f(x)$ exists. We know all three conditions that need to be satisfied for $f$ to be continuous at $a$. $\blacksquare$

Functions Containing Kinks or Cusps


The diagram above illustrates what exactly kinks and cusps are. A function $f$ containing a kink or cusp at $x = a$ is not differentiable at $x = a$ because $\lim_{x \to a^-} \frac{f(x) - f(a)}{x - a} ≠ \lim_{x \to a^+} \frac{f(x) - f(a)}{x - a}$.

Functions Containing Vertical Asymptotes

Some functions contain vertical asymptotes, that is $x \to a$, $f(x) \to \infty$ without there being an asymptote. One such example is the function $y = x^{1/3}$ as graphed below:


Suppose we calculate $f'(0)$ as follows:

\begin{align} f'(x) = \lim_{h \to 0} \frac{(f(x + h) - f(x)}{h} \\ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \\ f'(0) = \lim_{h \to 0} \frac{h^{1/3}}{h} \\ f'(0) = \lim_{h \to 0} \frac{1}{h^{2/3}} = \infty \end{align}

Therefore, as $x \to 0$, the slope at $0$ approaches $\infty$ and becomes a vertical tangent.

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