3.10 Differentiability

Table of Contents

Alternative Definition for the Differentiability of Single-Variable Functions

Show the alternative definition for differentiability of single-variable functions

Let’s start with a function $f$ of single variable $x$ . The derivative of $f$ at the point $x$ is $lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = f^{'} (x)$
if this limit exists. Let’s define $ϕ (h)$ as
$ϕ (h) = \frac{f (x + h) - f (x)}{h} - f^{'} (x) .$
Although $ϕ (h)$ is not defined at $h = 0$ , but^*

$lim_{h \to 0} ϕ (h) = 0.$

^* In general, for a function $g (x)$ suppose $lim_{x \to a} g (x) = L$ . Because $lim_{x \to a} (g (x) - L) = 0$ if we define $ϕ (x)$ as $ϕ (x) = g (x) - L$ , then $lim_{x \to 0} ϕ (x) = 0$ .

We can write (if $h \neq 0$ ):
$ϕ (h) = \frac{f (x + h) - f (x)}{h} - f^{'} (x) \Rightarrow f (x + h) - f (x) = f^{'} (x) h + h ϕ (h)$
or
$\begin{array}{r} f (x + h) = f (x) + f^{'} (x) h + h ϕ (h) \end{array}$

Let’s define $ε (h)$ by:
$ε (h) = {\begin{cases} ϕ (h) & if h > 0 \\ - ϕ (h) & if h < 0 \end{cases}$
Therefore, if $f (x)$ is differentiable, there exists a function $ε (h)$ such that
$f (x + h) = f (x) + f^{'} (x) h + | h | ε (h),$
and $ε (h) \to 0$ as $h \to 0$ .

Recall that $f (x) + f^{'} (x) h$ is the linearization of $f$ at the point $x$ . Therefore $| h | ε (h)$ is the error in this approximation. Eq. (i) means that if $f^{'} (x)$ exists, the function $f$ can be approximated by its linear approximation and the growth of the error is nothing compared to $| h |$ , that is error $| h | \to 0$ as $h \to 0$ [1], where $| h |$ is the distance of the point $(x + h)$ from the point $x$ .

Conversely, suppose there exists a number $a$ and a function $ε (h)$ such that
$\begin{matrix} (i) & f (x + h) = f (x) + a h + | h | ε (h), and lim_{h \to 0} ε (h) = 0. \end{matrix}$
Dividing both sides by $h \neq 0$ :
$\frac{f (x + h) - f (x)}{h} = a + \frac{| h |}{h} ε (h)$
and taking the limit as $h \to 0$ :
$lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = a + \underset{= 0}{\underset{⏟}{lim_{h \to 0} \frac{| h |}{h} ε (h)}} \Rightarrow lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = a .$
This means the derivative of $f$ at the point $x$ exists and is equal to $a$ :
$f^{'} (x) = a .$

Therefore, we can define the differentiability of a function $f$ at the point $x$ as the existence of a number $a$ and a function $ε$ that satisfy Eq. (i). We can extend the definition for functions of two or more variables.

Differentiability of Two-Variable Functions

Let $z = f (x, y)$ . We say $f$ is differentiable at the point $(x, y)$ if there exist two numbers $a$ and $b$ , and a function $ε (h, k)$ such that:
$f (x + h, y + k) = f (x, y) + a h + b k + \underset{= \sqrt{h^{2} + h^{2}}}{\underset{⏟}{| (h, k) |}} ε (h, k)$
and**

$lim_{(h, k) \to (0, 0)} ε (h, k) = 0.$

** Note that the magnitude (or the absolute value) of a vector $(h, k)$ is: $| (h, k) | = \sqrt{h^{2} + k^{2}}$

If such an approximation is valid, let $k = 0$ , divide both sides by $h \neq 0$ and take the limit $h \to 0$ , then we will have:
$\underset{= f_{x} (x, y)}{\underset{⏟}{lim_{h \to 0} \frac{f (x + h, y) - f (x, y)}{h}}} = a + \underset{= 0}{\underset{⏟}{lim_{h \to 0} \frac{\sqrt{h^{2}}}{h} ε (h, 0)}} \Rightarrow f_{x} (x, y) = a .$
Similarly, we can show: $f_{y} (x, y) = b$

Definition 1. We say a function

f

is differentiable at the point

(x, y)

if its partial derivatives

f_{x} (x, y)

and

f_{y} (x, y)

exist and there exists a function

ε

such that:

f (x + h, y + k) = f (x, y) + f_{x} (x, y) h + f_{y} (x, y) k + \sqrt{h^{2} + k^{2}} ε (h, k)

and

lim_{(h, k) \to (0, 0)} ε (h, k) = 0.

The above definition means that if $f (x, y)$ is differentiable, it can be approximated by its linearization and the growth of the error in this approximation, $\sqrt{h^{2} + k^{2}} ε (h, k)$ , is nothing compared to the growth of $ρ$ , where $ρ = \sqrt{h^{2} + k^{2}}$ is the distance of the point $(x + h, y + k)$ from the point $(x, y)$ .

Example 1

Show that

f

is differentiable at each point

(x, y)

f (x, y) = 2 x + 3 y^{2}

Solution

First let’s compute

f_{x} (x, y)

and

f_{y} (x, y)

f_{x} (x, y) = 2, f_{y} (x, y) = 6 y

Now let’s form $f (x + h, y + k)$ :
$\begin{aligned} f (x + h, y + k) & = 2 (x + h) + 3 (y + k)^{2} \\ = 2 x + 2 h + 3 (y^{2} + 2 y k + k^{2}) \\ = \underset{f (x, y)}{\underset{⏟}{2 x + 3 y^{2}}} + \underset{f_{x} (x, y)}{\underset{⏟}{2}} h + \underset{f_{y} (x, y)}{\underset{⏟}{6 y}} k + 3 k^{2} \end{aligned}$
If we call $ε (h, k) = \frac{3 k^{2}}{\sqrt{h^{2} + k^{2}}}$ , then we have shown
$f (x + h, y + k) = f (x, y) + f_{x} (x, y) h + f_{y} (x, y) k + \sqrt{h^{2} + k^{2}} ε (h, k) .$
The last step is to show $lim_{(h, k) \to (0, 0)} ε (h, k) = 0$ . To this end, we can use polar coordinates
$h = r \cos θ, k = r \sin θ$
Now $(h, k) \to (0, 0)$ is equivalent to $r \to 0$ . Therefore:
$\begin{aligned} lim_{(h, k) \to (0, 0)} ε (h, k) & = lim_{(h, k) \to (0, 0)} \frac{3 k^{2}}{\sqrt{h^{2} + k^{2}}} \\ = lim_{r \to 0} \frac{3 r \sin^{2} θ}{r \underset{= 1}{\underset{⏟}{\sqrt{\cos^{2} θ + \sin^{2} θ}}}} \\ = lim_{r \to 0} 3 r \sin θ = 0. \end{aligned}$
[Because $- r \leq r \sin θ \leq r$ and $lim_{r \to 0} r = 0$ , if we use the squeeze theorem, we could conclude $lim_{r \to 0} 3 r \sin θ = 0$ .]

Therefore, we have proved that $f (x, y)$ is differentiable at each $(x, y)$ .

From the definition of differentiability it is clear^† that if a function is differentiable at $(x, y)$ then it is continuous there. Therefore, if a function is not continuous, it cannot be differentiable.

^†Take the limit $(h, k) \to (0, 0)$ from both sides, and remember that $f_{x} (x, y)$ and $f_{y} (x, y)$ are the values of $f_{x}$ and $f_{y}$ at $(x, y)$ not two functions of $h$ or $k$ , so they are just two constants.

differentiability $\Rightarrow$ continuity

Also according to Definition 1, if a function is differentiable, its first partial derivatives exist. Therefore, if any of the first partial derivatives of a function does not exist at a point, the function is not differentiable at that point.

differentiability $\Rightarrow$ existence of first partial derivatives

Example 2

Show that $f (x, y)$ is not differentiable at $(0, 0)$ if $f (x, y) = {\begin{cases} \frac{x y}{x^{2} + y^{2}} & if (x, y) \neq (0, 0) \\ 0 & if (x, y) = (0, 0) \end{cases}$

Solution

In Example 5 of the Section on Limits and Continuity, we showed that $lim_{(x, y) \to (0, 0)} f (x, y)$ does not exist. Therefore, $f$ is discontinuous at $(0, 0)$ . A discontinuous function cannot be differentiable.

Also in Example 2 of the Section on Linear Approximation, we showed that $f_{x} (0, 0) = f_{y} (0, 0) = 0$ and the linearization of $f$ does not provide a good approximation for $f$ at $(0, 0)$ . So the existence of the partial derivatives does not guarantee that the function is differentiable or even continuous.

Using Definition 1 to verify whether a function is differentiable is often hard. Here we introduce a theorem that can be applied to most functions to show that they are differentiable.

Theorem 1. If the first partial derivatives of a function $f$ exist in some neighborhood of $x_{0}$ and are continuous at $x_{0}$ , then $f$ is differentiable at $x_{0}$ .

Example 3

Use Theorem 1 to show that $f$ is differentiable everywhere if $f (x, y) = {\begin{cases} \frac{x^{2} y^{2}}{x^{2} + y^{2}} & if (x, y) \neq (0, 0) \\ 0 & if (x, y) = (0, 0) \end{cases}$

Figure 1. Graph of

z = \frac{x^{2} y^{2}}{x^{2} + y^{2}}

Solution

To find $f_{x}$ and $f_{y}$ , we have to consider two cases: (1) $(x, y) = (0, 0)$ and (2) $(x, y) \neq (0, 0)$ .

Case (1): To calculate $f_{x} (0, 0)$ and $f_{y} (0, 0)$ , we need to use the definition of partial derivatives (similar to Example 2 of the Section on Linear Approximation):
$f_{x} (0, 0) = lim_{h \to 0} \frac{f (0 + h, 0) - f (0, 0)}{h} = lim_{h \to 0} \frac{0 - 0}{h} = 0,$
$f_{y} (0, 0) = lim_{k \to 0} \frac{f (0, 0 + k) - f (0, 0)}{k} = lim_{k \to 0} \frac{0 - 0}{k} = 0,$

Case (2): If $(x, y) \neq (0, 0)$ , we can differentiate $f$ with respect to $x$ while treating $y$ as a constant to find $f_{x} (x, y)$ and differentiate $f (x, y)$ with respect to $y$ , while treating $x$ as a constant to find $f_{y} (x, y)$ (although because of symmetry of $x$ and $y$ , to find $f_{y} (x, y)$ , we just need to replace $x$ by $y$ and $y$ by $x$ in the formula of $f_{x} (x, y)$ ). We have
$\begin{aligned} f_{x} (x, y) & = \frac{2 x y^{2} (x^{2} + y^{2}) - 2 x (x^{2} y^{2})}{(x^{2} + y^{2})^{2}} = \frac{2 x y^{4}}{(x^{2} + y^{2})^{2}}, \\ f_{y} (x, y) & = \frac{2 y x^{4}}{(x^{2} + y^{2})^{2}} . \end{aligned}$
Therefore:
$f_{x} (x, y) = {\begin{cases} \frac{2 x y^{4}}{(x^{2} + y^{2})^{2}} & if (x, y) \neq (0, 0) \\ 0 & if (x, y) = (0, 0) \end{cases}$
and
$f_{y} (x, y) = {\begin{cases} \frac{2 x^{4} y}{(x^{2} + y^{2})^{2}} & if (x, y) \neq (0, 0) \\ 0 & if (x, y) = (0, 0) \end{cases}$
It is clear that when $(x, y) \neq (0, 0)$ , $f_{x} (x, y)$ and $f_{y} (x, y)$ are continuous, because the numerators and denominators are continuous functions and the the ratio of two continuous functions is a continuous function if the denominator is nonzero (recall Theorem 3 in the Section on Limits and Continuity). We just need to show that $f_{x}$ and $f_{y}$ are continuous at $(0, 0)$ . To this end, again we can use polar coordinates:
$x = r \cos θ, y = r \sin θ .$
In polar coordinates $(x, y) \to (0, 0)$ is equivalent to $r \to 0$ :
$lim_{(x, y) \to (0, 0)} \frac{2 x y^{4}}{(x^{2} + y^{2})^{2}} = lim_{r \to 0} \frac{2 r^{5} \cos θ \sin^{4} θ}{r^{4} \underset{= 1}{\underset{⏟}{(\cos^{2} θ + \sin^{2} θ)^{2}}}} = lim_{r \to 0} 2 r \cos θ \sin^{4} θ = 0.$

So we have shown $lim_{(x, y) \to (0, 0)} f_{x} (x, y) = f_{x} (0, 0) = 0$ , that is $f_{x}$ is continuous at $(0, 0)$ . In a similar way, you can show that $lim_{(x, y) \to (0, 0)} f_{y} (x, y) = f_{y} (0, 0) = 0$ . Therefore $f$ has continuous first partial derivatives at each $(x, y)$ . So according to Theorem 1, $f$ is differentiable everywhere.

Differentiability of Functions in n-Space

At the end of this section, we can extend the concept of differentiability of functions of several variables:

Definition 2. We say a function $f : U \subseteq R^{n} \to R$ is differentiable at $(x_{1}, \dots, x_{n})$ , if its first partial derivatives $\frac{\partial f}{\partial x_{i}} (x_{1}, \dots, x_{n})$ (for $i = 1, \dots, n$ ) exist and there exists a function $ε (h_{1}, \dots, h_{n})$ such that:
$\begin{aligned} f (x_{1} + h_{1}, \dots, x_{n} + h_{n}) = f & (x_{1}, \dots, x_{n}) + \frac{\partial f}{\partial x_{1}} \cdot h_{1} + \dots + \frac{\partial f}{\partial x_{n}} \cdot h_{n} \\ (i) & + \sqrt{h_{1}^{2} + \dots + h_{n}^{2}} ε (h_{1}, \dots, h_{n}) \end{aligned}$
and
$lim_{(h_{1}, \dots, h_{n}) \to (0, \dots, 0)} ε (h_{1}, \dots, h_{n}) = 0.$
The partial derivatives are evaluated at $(x_{1}, \dots, x_{n})$ .

Equation (i) can also be written as $f (x + h) = f (x) + \sum_{i = 1}^{n} \frac{\partial f}{\partial x_{i}} h_{i} + | h | ε (h),$ with $x = (x_{1}, \dots, x_{n})$ , $h = (h_{1}, \dots, h_{n})$ , and $lim_{h \to 0} ε (h) = 0.$

Continuous Differentiability and Functions of Class $C^{p}$

A function that has continuous first partial derivatives is called continuously differentiable or a function of class $C^{1}$ . A function that not only its first partial derivatives are continuous but also all its second partial derivatives are continuous is called twice continuously differentiable or a function of class $C^{2}$ . In the same manner we can define function of class $C^{3}$ , $C^{4}$ , and so on. A function that has continuous partial derivatives of all orders is called a $C^{\infty}$ function. A function that is continuous is referred to a function of class $C^{0}$ .

Definition 3.Let

f

be defined on an open set

U \subseteq R^{n}

. We say

f

is of class

C^{p}

U

if all the partial derivatives of

f

of order

\leq p

are continuous everywhere in

U

[1] Mathematically, this is often written as $e r r o r = o (| h |)$

Alternative Definition for the Differentiability of Single-Variable Functions

Show the alternative definition for differentiability of single-variable functions

Hide the alternative definition

Differentiability of Two-Variable Functions

Differentiability of Functions in n-Space

Continuous Differentiability and Functions of Class Cp

Continuous Differentiability and Functions of Class $C^{p}$