# 2.7. Differentiable Functions#

We continue our discussion on real functions and focus on a special class of functions which are differentiable.

Definition 2.78 (Differentiable function)

A real function $$f: \RR \to \RR$$ is differentiable at an interior point $$x=a$$ of its domain $$\dom f$$ if the difference quotient

$\frac{f(x) - f(a)}{x - a}, \quad x \neq a$

approaches a limit as $$x$$ approaches $$a$$.

If $$f$$ is differentiable at $$x=a$$, the limit is called the derivative of $$f$$ at $$x=a$$ and is denoted by $$f'(a)$$; thus,

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

An alternative way is to write $$x$$ as $$x = a + h$$ and define $$f'(a)$$ as:

$f'(a) = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}.$

Notes

• The difference quotient is not defined at $$x=a$$. This is okay as computing the limit $$\lim_{x \to a} g(x)$$ doesn’t require $$g$$ to be defined at $$x=a$$.

• The derivative is not defined for the non-interior points of $$\dom f$$. Only one sided limits may be computed at the non-interior points on the difference quotient.

• We can treat $$f'$$ as a function from $$\RR$$ to $$\RR$$ where $$f'$$ is defined only on points at which $$f$$ is differentiable.

• The type signature for $$f'$$ is $$f' : \RR \to \RR$$.

• The domain of $$f'$$ denoted by $$\dom f'$$ is the set of points at which $$f$$ is differentiable.

Remark 2.17 (Domain of the derivative function)

The domain of the derivative of a function $$f$$; i.e., the set of points at which the derivative exists (or is defined) is a subset of the interior of the domain of the function $$f$$ itself.

$\dom f' \subseteq \interior \dom f.$

Definition 2.79 (Differentiable function)

Let $$f$$ be defined on an open set $$A$$. We say that $$f$$ is differentiable on $$A$$ if $$f$$ is differentiable at every point in $$A$$.

If $$f$$ is differentiable on (open) $$A$$, then $$f'$$ is defined on $$A$$. In other words: $$\dom f' = A$$.

Definition 2.80 (Continuously differentiable function)

We say that $$f$$ is continuously differentiable on an open set $$A$$ if $$f$$ is differentiable on $$A$$ and $$f'$$ is continuous on $$A$$.

Definition 2.81 (Second and $$n$$-th derivatives)

If $$f$$ is differentiable on a neighborhood of $$x=a$$ and $$f'$$ is differentiable at $$x=a$$, we denote the derivative of $$f'$$ at $$x=a$$ by $$f''(a)$$ and call it the second derivative of $$f$$ at $$x=a$$. Another notation for the second derivative is $$f^{(2)}(a)$$.

Inductively, if $$f^{(n-1)}$$ is defined on a neighborhood of $$x=a$$ and $$f^{(n-1)}$$ is differentiable at $$x=a$$, then the n-th derivative of $$f$$ at $$x=a$$, denoted by $$f^{(n)}(a)$$, is the derivative of $$f^{(n-1)}$$ at $$x=a$$.

The zeroth derivative of $$f$$ is defined to be $$f$$ itself.

$f^{(0)} = f.$

Another common notation is:

$\frac{d f}{d x} = f' \text{ and } \frac{d^n f}{d x^n} = f^{(n)}.$

Definition 2.82 (Tangent line)

If $$f$$ is differentiable at $$x=a$$, then the tangent to $$f$$ at $$x=a$$ can be given by:

$T(x) = f(a) + f'(a)(x -a).$

It is useful to remove the contribution of the tangent in $$f$$ and study the remaining part of $$f$$.

Lemma 2.1 (Removal of tangent line from function)

If $$f$$ is differentiable at $$x=a$$, then we can write $$f$$ as:

(2.1)#$f(x) = f(a) + [f'(a) + E(x)](x -a)$

where $$E$$ is defined in the neighborhood of $$x=a$$ and

$\lim_{x \to a} E(x) = E(a) = 0.$

In other words, $$E$$ is continuous at $$x=a$$.

Proof. We define $$E$$ as:

$\begin{split} E(x) = \begin{cases} \frac{f(x) - f(a)}{x - a} - f'(a), & x \in \dom f \text{ and } x \neq a\\ 0, & x = a \end{cases}. \end{split}$

This $$E$$ meets the requirements of (2.1). Note that $$E$$ is continuous at $$x=a$$ as $$\lim_{x \to a} E(x) = E(a)= 0$$.

We note that:

$f(x) - T(x) = E(x) (x - a).$

Alternatively

$f(x) = T(x) + E(x) (x - a).$

Remark 2.18 (Difference quotient and derivative)

At $$x \neq a$$, (2.1) can also be written as:

$\frac{f(x) - f(a)}{x -a} = f'(a) + E(x).$

In other words, the difference quotient $$\frac{f(x) - f(a)}{x -a}$$ is the sum of the derivative $$f'(a)$$ and $$E(x)$$.

Theorem 2.48 (Differentiability implies continuity)

If $$f$$ is differentiable at $$x=a$$, then $$f$$ is continuous at $$x=a$$.

Proof. Using Definition 2.82 and Lemma 2.1, we have:

$f(x) = T(x) + E(x) (x - a).$

It is easy to see that $$E(x)$$ and $$T(x)$$ are both continuous at $$x=a$$. Thus, $$f$$ is continuous at $$x=a$$.

Notes:

1. If $$f$$ is not continuous at $$x=a$$ then $$f$$ is not differentiable at $$x=a$$.

2. Continuity is a necessary condition but not sufficient condition for differentiability.

Remark 2.19 (Derivative sign and monotonicity in the neighborhood)

If $$f$$ is differentiable at $$x=a$$, and $$f'(a) \neq 0$$, then there is $$\delta > 0$$ such that if $$f'(a) > 0$$, then

$\sgn (f(x) - f(a)) = \sgn (x - a) \Forall |x - a| < \delta$

and if $$f'(a) < 0$$, then

$\sgn (f(x) - f(a)) = \sgn (a - x) \Forall |x - a| < \delta.$

Proof. We have, from Lemma 2.1, for $$x \neq a$$:

$\frac{f(x) - f(a)}{x -a} = f'(a) + E(x).$

Assume $$f'(a) \neq 0$$. Then $$|f'(a)| > 0$$. Since $$E$$ is continuous at $$a$$, with $$\epsilon = |f'(a)| > 0$$, there exists $$\delta > 0$$ such that

$|E(x) - E(a)| = |E(x)| < \epsilon = |f'(a)| \text{ whenever } |x - a| < \delta.$

Thus,

$\sgn(f'(a) + E(x)) = \sgn (f'(a)) \Forall |x - a| < \delta.$

Thus,

$\sgn \left ( \frac{f(x) - f(a)}{x -a} \right ) = \sgn (f'(a)) \Forall |x - a| < \delta.$

Now, if $$f'(a) > 0$$, then

$\sgn (f(x) - f(a)) = \sgn (x - a) \Forall |x - a| < \delta.$

If $$f'(a) < 0$$, then

$\sgn (f(x) - f(a)) = - \sgn (x - a) = \sgn (a - x) \Forall |x - a| < \delta.$

## 2.7.1. Arithmetic#

Theorem 2.49 (Differentiation and arithmetic )

If $$f$$ and $$g$$ are differentiable at $$x=a$$, then so are $$f+g$$, $$f-g$$, $$fg$$. $$\frac{f}{g}$$ is differentiable at $$x=a$$ if $$g'(a) \neq 0$$. The derivatives are:

1. $$(f + g)'(a) = f'(a) + g'(a)$$.

2. $$(f - g)'(a) = f'(a) - g'(a)$$.

3. $$(f g)'(a) = f'(a) g(a) + f(a) g'(a)$$.

4. $$\left(\frac{f}{g} \right)'(a) = \frac{f'(a) g(a) - f(a) g'(a)}{[g'(a)]^2}$$ provided $$g'(a) \neq 0$$.

## 2.7.2. The Chain Rule#

Theorem 2.50 (The chain rule)

Let $$f$$ be differentiable at $$x=a$$. Assume that $$g$$ is differentiable at $$f(a)$$. Then the composite function given by $$h = g \circ f$$ is differentiable at $$f(a)$$ with

$h'(a) = g'(f(a))f'(a).$

Proof. Let $$b = f(a)$$. Since $$g$$ is differentiable at $$b$$, we can write $$g$$ as (Lemma 2.1):

$g(t) = g(b) + [g'(b) + E(t)] (t - b)$

where $$E$$ is continuous in the neighborhood of $$t=b$$ and $$\lim_{t \to b} E(t) = E(b) = 0$$. Thus,

$g(t) - g(b) = [g'(b) + E(t)] (t - b).$

Putting $$t=f(x)$$, we get:

$g(f(x)) - g(f(a)) = [g'(f(a)) + E(f(x))][f(x) - f(a)].$

Since $$h(x) = g(f(x))$$, we get:

$h(x) - h(a) = [g'(f(a)) + E(f(x))][f(x) - f(a)].$

Dividing both sides by $$(x-a)$$, we get:

$\frac{h(x) - h(a)}{x - a} = [g'(f(a)) + E(f(x))] \frac{f(x) - f(a)}{x-a}.$

Since $$f$$ is continuous at $$x=a$$, $$E$$ is continuous at $$t=b=f(a)$$, and $$b$$ is an interior point of $$\dom E$$, hence $$E\circ f$$ is continuous at $$x=a$$ due to Theorem 2.41. Thus,

$\lim_{x \to a} E(f(x)) = E(f(a)) = E(b) = 0.$

Therefore,

\begin{split} \begin{aligned} h'(a) &= \lim_{x \to a} \frac{h(x) - h(a)}{x - a} \\ &= \lim_{x \to a} [g'(f(a)) + E(f(x))] \lim_{x \to a} \frac{f(x) - f(a)}{x-a}\\ &= [g'(f(a)) + \lim_{x \to a}E(f(x)) ] f'(a) = g'(f(a)) f'(a). \end{aligned} \end{split}

Example 2.12

Let

$f(x) = \frac{1}{x} \text{ and } g(x) = \sin x.$

Then, the composition $$h = g \circ f$$ is given by

$h(x) = (g \circ f)(x) = g (f (x)) = \sin \frac{1}{x}.$

We have:

$f'(x) = -\frac{1}{x^2} \text{ and } g'(x) = \cos x.$

Thus,

$h'(x) = g'(f(x))f'(x) = \left (\cos \frac{1}{x} \right ) \left (-\frac{1}{x^2} \right ).$

## 2.7.3. One Sided Derivatives#

Definition 2.83 (One sided derivatives)

One sided limits of the difference quotient

$\frac{f(x) - f(a)}{x - a}, \quad x \neq a$

are called one-sided derivatives if they exist.

If $$f$$ is defined over $$[a,b)$$, then the right hand derivative is defined as:

$f'_+(a) \triangleq \lim_{x \to a^+} \frac{f(x) - f(a)}{x - a}$

if the limit exists.

If $$f$$ is defined over $$(c,a]$$, then the left hand derivative is defined as:

$f'_-(a) \triangleq \lim_{x \to a^-} \frac{f(x) - f(a)}{x - a}$

if the limit exists.

Remark 2.20 (Differentiability and one-sided derivatives)

A function $$f$$ is differentiable at $$x=a$$ if and only if its left and right hand derivatives exist and are equal. In that case:

$f'_-(a) = f'(a) = f'_+(a).$

This is a direct implication of Theorem 2.37.

Remark 2.21

One sided derivative is not the same thing as one sided limit of a derivative.

1. $$f'_+(a)$$ need not be equal to $$f'(a^+)$$.

2. $$f'_-(a)$$ need not be equal to $$f'(a^-)$$.

## 2.7.4. Closed Intervals#

Definition 2.84 (Differentiability on a closed interval)

We say that $$f$$ is differentiable on the closed interval $$[a,b]$$ if $$f$$ is differentiable on the open interval $$(a,b)$$ and the one sided derivatives $$f'_+(a)$$ and $$f'_-(b)$$ both exist.

We assign $$f'(a) = f'_+(a)$$ and $$f'(b) = f'_-(b)$$ to complete the definition of $$f'$$ over $$[a,b]$$.

Note

While it is possible to use the notion of one sided derivatives to define $$f'$$ on a closed interval, this notion doesn’t generalize to multivariable calculus. The definition of derivative on an open interval (or an open subset of $$\dom f$$) can be easily extended to multivariable calculus.

Definition 2.85 (Continuous differentiability on a closed interval)

We say that $$f$$ is continuously differentiable on the closed interval $$[a,b]$$ if

1. $$f$$ is differentiable on the closed interval $$[a,b]$$

2. $$f'$$ is continuous over the open interval $$(a,b)$$

3. $$f'_+(a) = f'(a^+)$$.

4. $$f'_-(b) = f'(b^-)$$.

## 2.7.5. Extreme Values#

Definition 2.86 (Local extreme value)

We say that $$f(a)$$ is a local extreme value of $$f$$ if there exists $$\delta > 0$$ such that $$f(x) - f(a)$$ doesn’t change sign on

$(a - \delta, a + \delta) \cap \dom f.$

More specifically,

1. $$f(a)$$ is a local maximum value of $$f$$ if for some $$\delta > 0$$:

$f(x) \geq f(a) \Forall x \in (a - \delta, a + \delta) \cap \dom f.$
2. $$f(a)$$ is a local minimum value of $$f$$ if for some $$\delta > 0$$:

$f(x) \geq f(a) \Forall x \in (a - \delta, a + \delta) \cap \dom f.$

The point $$x=a$$ is called a local extreme point of $$f$$ or more specifically, a local maximum or a local minimum point of $$f$$.

Theorem 2.51

If $$f$$ is differentiable at a local extreme point $$a \in \dom f$$, then $$f'(a) = 0$$.

In other words, if the derivative exists at a local extreme point, it vanishes there.

Proof. We show that if $$f'(a) \neq 0$$ then $$a$$ is not a local extreme point. Thus, if $$a$$ is a local extreme point then $$f'(a)$$ must be 0.

Assume $$f'(a) \neq 0$$.

From Lemma 2.1, we have (at $$x \neq a$$):

$\frac{f(x) - f(a)}{x -a} = f'(a) + E(x)$

where $$\lim_{x \to a} E(x) = 0$$ and $$E$$ is continuous at $$x=a$$.

Since $$f'(a) \neq 0$$, hence $$|f'(a)| > 0$$, hence there exists $$\delta > 0$$ such that

$|E(x) | < |f'(a)| \text{ whenever } |x - a | < \delta.$

Thus, in the interval $$|x - a | < \delta$$, the term $$f'(a) + E(x)$$ has the same sign as $$f'(a)$$.

Hence the term $$\frac{f(x) - f(a)}{x -a}$$ must not change sign in $$|x - a | < \delta$$.

But the term $$(x -a)$$ changes sign in $$|x - a | < \delta$$. Hence, $$f(x) - f(a)$$ must also change sign.

Moreover, $$(x -a)$$ changes sign in every neighborhood $$|x - a | < \delta_1$$ with $$\delta_1 > 0$$. Hence $$f(x) - f(a)$$ must also change sign in every neighborhood.

Hence there is no neighborhood of $$a$$ in which $$f(x) - f(a)$$ doesn’t change sign. Hence, $$a$$ is not a local extreme point of $$f$$.

Definition 2.87 (Critical point)

Let $$f : \RR \to \RR$$ be a real function. Let $$a \in \interior \dom f$$. If $$f$$ is not differentiable at $$a$$ or if $$f$$ is differentiable at $$x=a$$ and $$f'(a) = 0$$, then we say that $$a$$ is a critical point of $$f$$.

Definition 2.88 (Stationary point)

If $$f$$ is differentiable at $$x=a$$ and $$f'(a) = 0$$, then we say that $$a$$ is a stationary point of $$f$$.

All stationary points are critical points while all critical points need not be stationary points. If the derivative doesn’t exist at some point $$a \in \interior \dom f$$, it could indicate a potential maximum or minimum.

Remark 2.22

All local extreme points are critical points.

Example 2.13 (A non-extreme critical point)

A critical point need not be a local extreme point. For the function $$f(x) = x^3$$, $$f'(0) = 0$$. Thus, $$x=0$$ is a critical point. But it is not a local extreme point since $$f$$ changes sign around $$x=0$$.

Theorem 2.52 (Rolle’s theorem)

Let $$f$$ be continuous on the closed interval $$[a,b]$$. Assume $$f$$ to be differentiable on the open interval $$(a,b)$$. Further assume that $$f(a) = f(b)$$. Then, $$f'(c) = 0$$ for some $$c \in (a,b)$$.

Proof. Recall that if $$f$$ is continuous on a closed interval, then $$f$$ attains its maximum and minimum value on points in the interval (Theorem 2.43).

Assume

$\alpha = \inf_{a \leq x \leq b} f(x) \text{ and } \beta = \sup_{a \leq x \leq b} f(x).$

If $$\alpha=\beta$$, then $$f$$ is a constant function on $$[a,b]$$. In that case, $$f'(c) = 0$$ for all $$c \in (a,b)$$.

Consider the case where $$\alpha < \beta$$. In that case, either the maximum or the minimum is attained at some point $$c \in (a,b)$$ since $$f(a) = f(b)$$.

1. If $$\alpha = f(a) = f(b)$$, then $$\beta$$ must be attained at some point in $$c \in (a,b)$$.

2. If $$\beta = f(a) = f(b)$$, then $$\alpha$$ must be attained at some point in $$c \in (a,b)$$.

3. If neither of the above hold true, then both $$\alpha$$ and $$\beta$$ are attained at some point in $$(a,b)$$.

Since $$f$$ is differentiable at $$c$$ and $$f(c)$$ is either maximum or minimum (i.e. a local extremum), hence $$f'(c) = 0$$ due to Theorem 2.51.

## 2.7.6. Intermediate Values#

Theorem 2.53 (Intermediate value theorem for derivatives)

Suppose that:

1. $$f$$ is differentiable on an open interval $$I$$.

2. There is a closed interval $$[a,b] \subset I$$.

3. $$f'(a) \neq f'(b)$$.

4. $$\mu$$ is in between $$f'(a)$$ and $$f'(b)$$.

Then $$f'(c) = \mu$$ for some $$c \in (a,b)$$

Note that this result doesn’t require $$f$$ to be continuously differentiable (i.e. $$f'$$ to be continuous).

Proof. Since $$f$$ is differentiable on $$I$$, hence $$f$$ is continuous on $$I$$.

Assume without loss of generality:

$f'(a) < \mu < f'(b).$

Define

$g(x) = f(x) - \mu x.$

Then,

$g'(x) = f'(x) - \mu, \Forall a \leq x \leq b.$

Since $$f'(a) < \mu$$ hence $$g'(a) < 0$$. Similarly, $$g'(b) > 0$$.

Since $$g$$ is continuous on $$[a,b]$$, hence $$g$$ attains a minimum at some point $$c \in [a,b]$$ (Theorem 2.43).

Now, $$g'(a) < 0$$ implies that there exists $$\delta > 0$$ such that:

$g(x) < g(a) \Forall a < x < a + \delta.$

Similarly,

$g(x) < g(b) \Forall b - \delta < x < b.$

Thus, minimum of $$g$$ cannot be at $$a$$ or $$b$$. Hence, $$c \in (a, b)$$. Since $$c$$ is a local extreme point, and $$g$$ is differentiable at $$c$$, Hence $$g'(c) = 0$$ due to Theorem 2.51. This in turn implies that $$f'(c) = \mu$$.

The case of $$f'(a) > \mu > f'(b)$$ can be handled by applying the same argument to $$-f$$.

## 2.7.7. Mean Values#

Theorem 2.54 (Generalized mean value theorem)

If $$f$$ and $$g$$ are continuous on the closed interval $$[a,b]$$ and differentiable on the open interval $$(a,b)$$, then

$[g(b) - g(a)] f'(c) = [f(b) - f(a)]g'(c)$

holds true for some $$c \in (a,b)$$.

Proof. Define the function:

$h(x) = [g(b) - g(a)] f(x) - [f(b) - f(a)]g(x).$
1. Since $$f$$ and $$g$$ are continuous on $$[a,b]$$, so is $$h$$.

2. Since $$f$$ and $$g$$ are differentiable on $$(a,b)$$ so is $$h$$.

3. $$h(a) = h(b) = g(b)f(a) - f(b)g(a)$$.

4. Therefore, by Rolle's theorem, $$h'(c) = 0$$ for some $$c \in (a,b)$$.

5. But $$h'(c) = [g(b) - g(a)] f'(c) - [f(b) - f(a)]g'(c)$$.

6. Hence the result.

Theorem 2.55 (Mean value theorem)

If $$f$$ is continuous on the closed interval $$[a,b]$$ and differentiable on the open interval $$(a,b)$$, then

$f'(c) = \frac{f(b) - f(a)}{b -a}$

for some $$c \in (a,b)$$.

Proof. In Theorem 2.54, let $$g(x) = x$$.

Then,

$(b - a) f'(c) = f(b) - f(a).$

Remark 2.23

Assume $$f$$ to be differentiable on some open interval $$(a,b)$$. Assume $$x_1, x_2 \in (a,b)$$. We haven’t specified whether $$x_1 < x_2$$ or $$x_1 > x_2$$.

1. $$f$$ is continuous on the closed interval with endpoints $$x_1$$ and $$x_2$$.

2. $$f$$ is differentiable on the interior of this closed interval.

3. (2.2)#$f(x_2) - f(x_1) = f'(c) (x_2 - x_1)$

for some $$c$$ in the open interval between $$x_1$$ and $$x_2$$.

Theorem 2.56

If $$f'(x)=0$$ for all $$x \in (a,b)$$, then $$f$$ is constant on $$(a,b)$$.

Proof. For any $$x_1, x_2 \in (a,b)$$, using (2.2), we get:

$f(x_2) - f(x_1) = 0.$

Theorem 2.57 (No change in derivative sign implies monotonicity)

If $$f'$$ exists and does not change sign on $$(a,b)$$, then $$f$$ is monotonic on $$(a,b)$$. In particular:

1. If $$f'(x) > 0$$, then $$f$$ is strictly increasing in $$(a,b)$$.

2. If $$f'(x) \geq 0$$, then $$f$$ is increasing in $$(a,b)$$.

3. If $$f'(x) \leq 0$$, then $$f$$ is decreasing in $$(a,b)$$.

4. If $$f'(x) < 0$$, then $$f$$ is strictly decreasing in $$(a,b)$$.

Proof. Let $$x_1, x_2 \in (a,b)$$ be such that $$x_1 < x_2$$. By mean value theorem, there exists $$c \in (x_1,x_2)$$ such that:

$f(x_2) - f(x_1) = f'(c) (x_2 - x_1).$

Now,

1. If $$f'(x) > 0 \Forall x \in (a,b)$$, then $$f(x_2) - f(x_1) > 0$$.

2. If $$f'(x) \geq 0 \Forall x \in (a,b)$$, then $$f(x_2) - f(x_1) \geq 0$$.

3. If $$f'(x) \leq 0 \Forall x \in (a,b)$$, then $$f(x_2) - f(x_1) \leq 0$$.

4. If $$f'(x) < 0 \Forall x \in (a,b)$$, then $$f(x_2) - f(x_1) < 0$$.

Theorem 2.58 (Bounded derivative implies Lipschitz continuity)

If

$|f'(x)| \leq M, \Forall a < x < b,$

then:

$|f(x) - f(x') | \leq M | x - x'|, \Forall x, x' \in (a,b).$

This is another direct implication of the mean value theorem.