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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 $f:\mathbb{R}\to \mathbb{R}$ is differentiable at an interior point $x=a$ of its domain $\mathrm{dom}f$ if the difference quotient

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

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 $f^{\prime }(a)$; thus,

$$f^{\prime }(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}.$$

An alternative way is to write $x$ as $x=a+h$ and define $f^{\prime }(a)$ as:

$$f^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}.$$

Notes

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

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

• We can treat $f^{\prime }$ as a function from $\mathbb{R}$ to $\mathbb{R}$ where $f^{\prime }$ is defined only on points at which $f$ is differentiable.

• The type signature for $f^{\prime }$ is $f^{\prime }:\mathbb{R}\to \mathbb{R}$.

• The domain of $f^{\prime }$ denoted by $\mathrm{dom}{f}^{\prime }$ 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.

$$\mathrm{dom}{f}^{\prime }\subseteq \mathrm{int}\mathrm{dom}f.$$

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 $f^{\prime }$ is defined on $A$. In other words: $\mathrm{dom}{f}^{\prime }=A$.

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 $f^{\prime }$ is continuous on $A$.

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

If $f$ is differentiable on a neighborhood of $x=a$ and $f^{\prime }$ is differentiable at $x=a$, we denote the derivative of $f^{\prime }$ 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:

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

Definition 2.82 (Tangent line)

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

$$T(x)=f(a)+f^{\prime }(a)(x-a).$$

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)$$f(x)=f(a)+[f^{\prime }(a)+E(x)](x-a)$$

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

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

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

Proof. We define $E$ as:

$$\begin{array}{r}E(x)=\{\begin{array}{ll}\frac{f(x)-f(a)}{x-a}-f^{\prime }(a),& x\in \mathrm{dom}f\text{ and }x\ne a\\ 0,& x=a\end{array}.\end{array}$$

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

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

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

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

Theorem 2.47 (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 $f^{\prime }(a)\ne 0$, then there is $\delta >0$ such that if $f^{\prime }(a)>0$, then

$$\mathrm{sgn}(f(x)-f(a))=\mathrm{sgn}(x-a)\mathrm{\forall }|x-a|<\delta$$

and if $f^{\prime }(a)<0$, then

$$\mathrm{sgn}(f(x)-f(a))=\mathrm{sgn}(a-x)\mathrm{\forall }|x-a|<\delta .$$

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

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

Assume $f^{\prime }(a)\ne 0$. Then $|f^{\prime }(a)|>0$. Since $E$ is continuous at $a$, with $ϵ=|f^{\prime }(a)|>0$, there exists $\delta >0$ such that

$$|E(x)-E(a)|=|E(x)|<ϵ=|f^{\prime }(a)|\text{ whenever }|x-a|<\delta .$$

Thus,

$$\mathrm{sgn}(f^{\prime }(a)+E(x))=\mathrm{sgn}(f^{\prime }(a))\mathrm{\forall }|x-a|<\delta .$$

Thus,

$$\mathrm{sgn}\left(\frac{f(x)-f(a)}{x-a}\right)=\mathrm{sgn}(f^{\prime }(a))\mathrm{\forall }|x-a|<\delta .$$

Now, if $f^{\prime }(a)>0$, then

$$\mathrm{sgn}(f(x)-f(a))=\mathrm{sgn}(x-a)\mathrm{\forall }|x-a|<\delta .$$

If $f^{\prime }(a)<0$, then

$$\mathrm{sgn}(f(x)-f(a))=-\mathrm{sgn}(x-a)=\mathrm{sgn}(a-x)\mathrm{\forall }|x-a|<\delta .$$

2.7.1. Arithmetic

Theorem 2.48 (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 $g^{\prime }(a)\ne 0$. The derivatives are:

1. $(f+g{)}^{\prime }(a)=f^{\prime }(a)+g^{\prime }(a)$.

2. $(f-g{)}^{\prime }(a)=f^{\prime }(a)-g^{\prime }(a)$.

3. $(fg{)}^{\prime }(a)=f^{\prime }(a)g(a)+f(a)g^{\prime }(a)$.

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

2.7.2. The Chain Rule

Theorem 2.49 (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

$$h^{\prime }(a)=g^{\prime }(f(a))f^{\prime }(a).$$

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

$$g(t)=g(b)+[g^{\prime }(b)+E(t)](t-b)$$

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

$$g(t)-g(b)=[g^{\prime }(b)+E(t)](t-b).$$

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

$$g(f(x))-g(f(a))=[g^{\prime }(f(a))+E(f(x))][f(x)-f(a)].$$

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

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

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

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

Since $f$ is continuous at $x=a$, $E$ is continuous at $t=b=f(a)$, and $b$ is an interior point of $\mathrm{dom}E$, hence $E\circ f$ is continuous at $x=a$ due to Theorem 2.40. Thus,

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

Therefore,

$$\begin{array}{r}\begin{array}{rl}{h}^{\prime }(a)& =\underset{x\to a}{lim}\frac{h(x)-h(a)}{x-a}\\ & =\underset{x\to a}{lim}[g^{\prime }(f(a))+E(f(x))]\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}\\ & =[g^{\prime }(f(a))+\underset{x\to a}{lim}E(f(x))]f^{\prime }(a)=g^{\prime }(f(a))f^{\prime }(a).\end{array}\end{array}$$

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:

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

Thus,

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

2.7.3. One Sided Derivatives

Definition 2.83 (One sided derivatives)

One sided limits of the difference quotient

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

are called one-sided derivatives if they exist.

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

$$f_{+}^{\prime }(a)\triangleq \underset{x\to a^{+}}{lim}\frac{f(x)-f(a)}{x-a}$$

if the limit exists.

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

$$f_{-}^{\prime }(a)\triangleq \underset{x\to a^{-}}{lim}\frac{f(x)-f(a)}{x-a}$$

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:

$$f_{-}^{\prime }(a)=f^{\prime }(a)=f_{+}^{\prime }(a).$$

This is a direct implication of Theorem 2.36.

Remark 2.21

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

1. $f_{+}^{\prime }(a)$ need not be equal to $f^{\prime }(a^{+})$.

2. $f_{-}^{\prime }(a)$ need not be equal to $f^{\prime }(a^{-})$.

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 $f_{+}^{\prime }(a)$ and $f_{-}^{\prime }(b)$ both exist.

We assign $f^{\prime }(a)=f_{+}^{\prime }(a)$ and $f^{\prime }(b)=f_{-}^{\prime }(b)$ to complete the definition of $f^{\prime }$ over $[a,b]$.

Note

While it is possible to use the notion of one sided derivatives to define $f^{\prime }$ 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 $\mathrm{dom}f$) 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. $f^{\prime }$ is continuous over the open interval $(a,b)$

3. $f_{+}^{\prime }(a)=f^{\prime }(a^{+})$.

4. $f_{-}^{\prime }(b)=f^{\prime }(b^{-})$.

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

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

More specifically,

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

   $$f(x)\ge f(a)\mathrm{\forall }x\in (a-\delta ,a+\delta )\cap \mathrm{dom}f.$$

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

   $$f(x)\ge f(a)\mathrm{\forall }x\in (a-\delta ,a+\delta )\cap \mathrm{dom}f.$$

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 2.50

If $f$ is differentiable at a local extreme point $a\in \mathrm{dom}f$, then $f^{\prime }(a)=0$.

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

Proof. We show that if $f^{\prime }(a)\ne 0$ then $a$ is not a local extreme point. Thus, if $a$ is a local extreme point then $f^{\prime }(a)$ must be 0.

Assume $f^{\prime }(a)\ne 0$.

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

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

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

Since $f^{\prime }(a)\ne 0$, hence $|f^{\prime }(a)|>0$, hence there exists $\delta >0$ such that

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

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

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 $f:\mathbb{R}\to \mathbb{R}$ be a real function. Let $a\in \mathrm{int}\mathrm{dom}f$. If $f$ is not differentiable at $a$ or if $f$ is differentiable at $x=a$ and $f^{\prime }(a)=0$, then we say that $a$ is a critical point of $f$.

Definition 2.88 (Stationary point)

If $f$ is differentiable at $x=a$ and $f^{\prime }(a)=0$, 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 $a\in \mathrm{int}\mathrm{dom}f$, 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 $f(x)=x^3$, $f^{\prime }(0)=0$. Thus, $x=0$ is a critical point. But it is not a local extreme point since $f$ changes sign around $x=0$.

Theorem 2.51 (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, $f^{\prime }(c)=0$ 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 2.42).

Assume

$$\alpha =\underset{a\le x\le b}{inf}f(x)\text{ and }\beta =\underset{a\le x\le b}{sup}f(x).$$

If $\alpha =\beta$, then $f$ is a constant function on $[a,b]$. In that case, $f^{\prime }(c)=0$ 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 $f^{\prime }(c)=0$ due to Theorem 2.50.

2.7.6. Intermediate Values

Theorem 2.52 (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. $f^{\prime }(a)\ne f^{\prime }(b)$.

4. $\mu$ is in between $f^{\prime }(a)$ and $f^{\prime }(b)$.

Then $f^{\prime }(c)=\mu$ for some $c\in (a,b)$

Note that this result doesn’t require $f$ to be continuously differentiable (i.e. $f^{\prime }$ to be continuous).

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

Assume without loss of generality:

$$f^{\prime }(a)<\mu <f^{\prime }(b).$$

Define

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

Then,

$$g^{\prime }(x)=f^{\prime }(x)-\mu ,\mathrm{\forall }a\le x\le b.$$

Since $f^{\prime }(a)<\mu$ hence $g^{\prime }(a)<0$. Similarly, $g^{\prime }(b)>0$.

Since $g$ is continuous on $[a,b]$, hence $g$ attains a minimum at some point $c\in [a,b]$ (Theorem 2.42).

Now, $g^{\prime }(a)<0$ implies that there exists $\delta >0$ such that:

$$g(x)<g(a)\mathrm{\forall }a<x<a+\delta .$$

Similarly,

$$g(x)<g(b)\mathrm{\forall }b-\delta <x<b.$$

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 $g^{\prime }(c)=0$ due to Theorem 2.50. This in turn implies that $f^{\prime }(c)=\mu$.

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

2.7.7. Mean Values

Theorem 2.53 (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

$$[g(b)-g(a)]f^{\prime }(c)=[f(b)-f(a)]g^{\prime }(c)$$

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, $h^{\prime }(c)=0$ for some $c\in (a,b)$.

5. But $h^{\prime }(c)=[g(b)-g(a)]f^{\prime }(c)-[f(b)-f(a)]g^{\prime }(c)$.

6. Hence the result.

Theorem 2.54 (Mean value theorem)

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

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

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

Proof. In Theorem 2.53, let $g(x)=x$.

Then,

$$(b-a)f^{\prime }(c)=f(b)-f(a).$$

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. By mean value theorem:

   (2.2)$$f(x_2)-f(x_1)=f^{\prime }(c)(x_2-x_1)$$

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

Theorem 2.55

If $f^{\prime }(x)=0$ 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 2.56 (No change in derivative sign implies monotonicity)

If $f^{\prime }$ exists and does not change sign on $(a,b)$, then $f$ is monotonic on $(a,b)$. In particular:

1. If $f^{\prime }(x)>0$, then $f$ is strictly increasing in $(a,b)$.

2. If $f^{\prime }(x)\ge 0$, then $f$ is increasing in $(a,b)$.

3. If $f^{\prime }(x)\le 0$, then $f$ is decreasing in $(a,b)$.

4. If $f^{\prime }(x)<0$, 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:

$$f(x_2)-f(x_1)=f^{\prime }(c)(x_2-x_1).$$

Now,

1. If $f^{\prime }(x)>0\mathrm{\forall }x\in (a,b)$, then $f(x_2)-f(x_1)>0$.

2. If $f^{\prime }(x)\ge 0\mathrm{\forall }x\in (a,b)$, then $f(x_2)-f(x_1)\ge 0$.

3. If $f^{\prime }(x)\le 0\mathrm{\forall }x\in (a,b)$, then $f(x_2)-f(x_1)\le 0$.

4. If $f^{\prime }(x)<0\mathrm{\forall }x\in (a,b)$, then $f(x_2)-f(x_1)<0$.

Theorem 2.57 (Bounded derivative implies Lipschitz continuity)

If

$$|f^{\prime }(x)|\le M,\mathrm{\forall }a<x<b,$$

then:

$$|f(x)-f(x^{\prime })|\le M|x-x^{\prime }|,\mathrm{\forall }x,x^{\prime }\in (a,b).$$

This is another direct implication of the mean value theorem.

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