Basic Definitions

Section 4.1 Basic Definitions

Consider a function

f

with domain

D \subset R^{N}

with values in

R .

For every

i = 1, \dots, n

we can then define a function of one variable,

t,

by setting

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g_{i} (t) = f (a_{1}, \dots, a_{i - 1}, a_{i} + t, a_{i + 1}, \dots, a_{n}) .

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N = 2

then the graph of this function is a cross-section of the graph of

f

in the

i

-th coordinate direction as seen in Figure 4.1.

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Figure 4.1. Cross-sections of the graph of a function through a given point
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a

is an interior point of

D

then

g_{i}

is defined on a small interval

I = (- ε, ε),

and zero is an interior point. From the calculus of one variables we know that

g_{i}^{'} (0)

is the slope of the graph of the function

g_{i}

t = 0 .

Hence

g_{i}^{'} (0)

is the slope of the cross-section of the graph of

f

through

a

in the

x_{i}

-direction. This motivates the following definition.

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Definition 4.2. Partial derivative.

Suppose that

a

is an interior point of

D \subset R^{N},

and that

f

is a function on

D

with values in

R .

We define the partial derivative of

f

with respect to

x_{i}

a

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\begin{aligned} \frac{\partial}{\partial x_{i}} f (a) & := g_{i}^{'} (0) \\ = lim_{t \to 0} \frac{1}{t} (f (a_{1}, \dots, a_{i} + t, \dots, a_{N}) - f (a_{1}, \dots, a_{i}, \dots, a_{N})) \end{aligned}

whenever the limit exists.

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Warning 4.3. Notation for derivatives.

There are other possible notations for partial derivatives. We also write

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\begin{aligned} \frac{\partial}{\partial x_{i}} f (a) & = \frac{\partial f}{\partial x_{i}} (a) \\ = f_{x_{i}} (a) \\ = \partial_{x_{i}} f (a) \\ = \partial_{i} f (a) . \end{aligned}

Look at the symbol for the partial derivative carefully, and practise its writing a little bit.

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Write \frac{\partial}{\partial x_{i}} and NOT \frac{δ}{δ x_{i}} or \frac{d}{d x_{i}}

as it is often seen among beginning students. The last two symbols have a meaning on their own right, the first being the ‘variation’, and the other being the ‘total derivative’ of a function. An example showing the difference between partial and total derivatives will be shown in Example 4.23.

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Given a function of

N

variables we can form a vector having as its components the partial derivatives of that function.

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Definition 4.4. Gradient.

Suppose

f

is a function defined on

D \subset R^{N}

having partial derivatives with respect to all

N

variables at the interior point

a .

Then the vector

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\begin{aligned} grad f (a) & = \nabla f (a) \\ = (\frac{\partial f}{\partial x_{1}} (a), \dots, \frac{\partial f}{\partial x_{N}} (a)) \end{aligned}

is called the gradient of

f

a

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We conclude this section by some examples.

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Example 4.5.

Consider

f (x, y) := x^{2} - y^{2}

defined on

R^{2} .

To compute its derivatives we treat one variable as a constant, and differentiate with respect to the other:

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\frac{\partial}{\partial x} f (x, y) = 2 x, \frac{\partial}{\partial y} f (x, y) = - 2 y .

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The gradient of the function is

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\nabla f (x, y) = (2 x, - 2 y) = 2 (x, - y) .

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Example 4.6.

Compute the partial derivatives and the gradient of the function

f

given by

f (x, y, z, t) := (x^{2} + y^{2} + z^{2}) e^{- t} .

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Solution.

As in the previous example we treat all variables as constants except for one. Hence the partial derivatives are

\begin{aligned} \frac{\partial f}{\partial x} & = 2 x e^{- t}, & \frac{\partial f}{\partial y} & = 2 y e^{- t}, & \frac{\partial f}{\partial z} & = 2 z e^{- t}, & \frac{\partial f}{\partial t} & = - (x^{2} + y^{2} + z^{2}) e^{- t} . \end{aligned}

The gradient is

\begin{aligned} grad f (x, y, z, t) & = (2 x e^{- t}, 2 y e^{- t}, 2 z e^{- t}, - (x^{2} + y^{2} + z^{2}) e^{- t}) \\ = (2 x, 2 y, 2 z, - (x^{2} + y^{2} + z^{2})) e^{- t} . \end{aligned}

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Example 4.7.

Compute the partial derivatives of

f (x, y) := x / y

on its domain of definition.

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Solution.

We saw in Example 3.27 that

R^{2}

minus the

x

-axis is the domain of the function under consideration. Hence, if

x \neq 0

then

\begin{aligned} \frac{\partial}{\partial x} f (x, y) & = \frac{1}{y}, & \frac{\partial}{\partial y} f (x, y) & = - \frac{x}{y^{2}} . \end{aligned}

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Example 4.8.

Compute the gradient of

f (x, t) := x^{t}

defined for

x > 0

and

t \in R .

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Solution.

We first compute the partial derivatives:

\begin{aligned} \frac{\partial}{\partial x} f (x, t) & = t x^{t - 1} \\ \frac{\partial}{\partial t} f (x, t) & = \frac{\partial}{\partial t} e^{t \ln x} = e^{t \ln x} \ln x = x^{t} \ln x . \end{aligned}

Hence the gradient of

f

grad f (x, t) = (t x^{t - 1}, x^{t} \ln x) = x^{t} (t x^{- 1}, \ln x) .

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Example 4.9.

Define the function

f

R^{2}

by setting

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f (s, t) := {\begin{cases} \frac{s t^{2}}{s^{2} + t^{2}} & if (s, t) \neq (0, 0), \\ 0 & if (s, t) = (0, 0) \end{cases}

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Determine whether

f

has partial derivatives at

(0, 0) .

Determine the gradient at

(0, 0)

if it exists.

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Solution.

By definition of the partial derivative with respect to

s

we have

\frac{\partial f}{\partial s} f (0, 0) = lim_{h \to 0} \frac{1}{h} (\frac{(0 + h) 0^{2}}{(0 + h)^{2} + 0^{2}} - 0) = lim_{h \to 0} 0 = 0

Similarly we have

\frac{\partial f}{\partial t} f (0, 0) = lim_{h \to 0} \frac{1}{h} (\frac{0 (0 + h)^{2}}{0^{2} + (0 + h)^{2}} - 0) = lim_{h \to 0} 0 = 0 .

Hence both partial derivatives exist, and thus

grad f (0, 0) = (0, 0) .

Look at the cross-sections through the graph of

f

shown in Figure 3.29 along the coordinate axis to visualise the result.

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Example 4.10.

Consider the function

f

R^{2}

defined by

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f (x, y) := {\begin{cases} \frac{x y^{2}}{x^{2} + y^{4}} & if (x, y) \neq (0, 0), \\ 0 & if (x, y) = (0, 0) . \end{cases}

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Determine whether

f

has partial derivatives at

(0, 0),

and determine them if possible.

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Solution.

By the definition of partial derivatives we have

\frac{\partial f}{\partial x} f (0, 0) = lim_{t \to 0} \frac{1}{t} (\frac{t 0^{2}}{t^{2} + 0^{4}} - 0) = lim_{t \to 0} 0 = 0

and

\frac{\partial f}{\partial y} f (0, 0) = lim_{t \to 0} \frac{1}{t} (\frac{0 t^{2}}{0^{2} + t^{4}} - 0) = lim_{t \to 0} 0 = 0 .

Hence both partial derivatives exist and are zero.

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Remark 4.11.

For functions of one variable we know that differentiability at a point implies continuity of the function at that point. Note however, that the existence of all partial derivatives of a function of several variables does not imply the continuity of the function at the corresponding point: The function considered in the previous example has partial derivatives at

(0, 0),

but is discontinuous at

(0, 0)

as shown in Example 3.30.

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