Mathematics · Ch 5 — Continuity and Differentiability

Differentiability

5.3

Differentiability

The idea of differentiability builds directly on the derivative. Before we explore when a function fails to be differentiable, recall exactly what it means for a derivative to exist.

The Derivative: A Quick Refresher

Suppose $f$ is a real-valued function and $c$ is a point in its domain. The derivative of $f$ at $c$ is defined by the limit

f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}

provided this limit exists. It is also denoted $\left. \frac{d}{dx} f(x) \right|_{x=c}$ . Considered as a function itself,

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

wherever the limit exists, is called the derivative of $f$ . Other common notations are $\frac{d}{dx} f(x)$ , or, if $y = f(x)$ , $\frac{dy}{dx}$ or $y'$ . The process of finding it is called differentiation.

Algebra of Derivatives

The following rules for combining derivatives form the foundation for all differentiation work:

Sum/Difference Rule: $(u \pm v)' = u' \pm v'$
Product Rule (Leibnitz Rule): $(uv)' = u'v + uv'$
Quotient Rule: $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$ , wherever $v \neq 0$

Derivatives of Standard Functions

These are the building blocks for more complex differentiation:

$f(x)$	$f'(x)$
$x^n$	$n x^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$

When the Derivative Does Not Exist

Every definition above carried the condition "provided the limit exists." If the limit $\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$ does not exist, we say the function $f$ is not differentiable at $c$ . For the limit to exist, the left-hand and right-hand limits must be equal, which leads to a practical criterion.

Important

A function $f$ is differentiable at a point $c$ in its domain if and only if both the left-hand derivative and the right-hand derivative exist, are finite, and are equal.

The left-hand derivative at $c$ is given by:

\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}

and the right-hand derivative at $c$ is given by:

\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}

Differentiability on an Interval

A function is differentiable in an open interval $(a, b)$ if it is differentiable at every point within it. For a closed interval $[a, b]$ , it must be differentiable at every point of $(a, b)$ , and at the endpoints we use the appropriate one-sided derivatives: the right-hand derivative at $x = a$ , and the left-hand derivative at $x = b$ .

Theorem 3: Differentiability Implies Continuity

This is a fundamental result linking the two concepts.

f

Proof:

Since $f$ is differentiable at $c$ , the following limit exists and equals $f'(c)$ :

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

For $x \neq c$ , write the difference as

f(x) - f(c) = \frac{f(x) - f(c)}{x - c} \cdot (x - c)

Using the product rule for limits (both limits exist),

\lim_{x \to c} [f(x) - f(c)] = \left[ \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \right] \cdot \left[ \lim_{x \to c} (x - c) \right] = f'(c) \cdot 0 = 0

Hence $\lim_{x \to c} f(x) = f(c)$ , which is precisely the definition of continuity at $x = c$ .

Note

So every differentiable function is continuous. This is a one-way implication: the converse is not true — a continuous function is not necessarily differentiable.

The Converse is False: A Counterexample …

Theorem 3

Theorem 3 (Differentiability implies Continuity)

If a function $f$ is differentiable at a point $c$ , then $f$ is also continuous at $c$ .

Hypotheses:

$f$ is a real-valued function defined on an interval containing $c$ .
The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ exists (as a finite limit).

Conclusion: $\lim_{x \to c} f(x) = f(c)$ , i.e., $f$ is continuous at $c$ .

›Proof

Since $f$ is differentiable at $c$ , we know

$\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = f'(c).$

For $x \neq c$ , we can write the difference $f(x) - f(c)$ as a product:

$f(x) - f(c) = \frac{f(x) - f(c)}{x - c} \cdot (x - c).$

Now take the limit as $x \to c$ on both sides:

$\lim_{x \to c} \bigl[ f(x) - f(c) \bigr] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \cdot (x - c) \right].$

By the limit product rule (the limit of a product equals the product of the limits, provided each limit exists), we have

$\lim_{x \to c} \bigl[ f(x) - f(c) \bigr] = \left( \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \right) \cdot \left( \lim_{x \to c} (x - c) \right).$

The first factor is $f'(c)$ (by differentiability), and the second factor is $0$ (since $x - c \to 0$ ). Hence

$\lim_{x \to c} \bigl[ f(x) - f(c) \bigr] = f'(c) \cdot 0 = 0.$

This means

$\lim_{x \to c} f(x) - f(c) = 0 \quad \Longrightarrow \quad \lim_{x \to c} f(x) = f(c).$

Therefore $f$ is continuous at $c$ .

Important

The converse is false: continuity does not guarantee differentiability. For example, $f(x) = |x|$ is continuous at $0$ but not differentiable there (left and right derivatives differ).

When is this used? …

Corollary 1

Every differentiable function is necessarily continuous. However, a continuous function need not be differentiable — for example, $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there. This corollary is used to quickly check that if a function has a derivative at a point, it must be continuous at that point, …

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