**Example.** Let
Let

Then

We may express these relations with the following matrix equation:

**Example.** Show that

satisfies the partial differential equation

where
is any smooth function.

Proof. Let

with

Then

Therefore

Let
Show that

Proof. From

we have

Therefore

Some surfaces in 3-space are described by Cartesian equations of the form

An equation like this is said to provide an *implicit representation*
of the surface. For example, the equation
represents the surface of a unit sphere with center at the origin. Sometimes
it is possible to solve the equation
for one of the variables in terms of the other two, say for
in terms of
and
.
This leads to one or more equations of the form

For the sphere we have two solutions,

one representing the upper hemisphere, the other the lower hemisphere.

In the general case it may not be an easy matter to obtain an explicit formula for in terms of and . For example, there is no easy method for solving for in the equation Nevertheless, a judicious use of the chain rule makes it possible to deduce various properties of the partial derivatives and without an explicit knowledge of . The procedure is described in this section.

We assume that there is a function
such that

all
in
some open set
,
although we may not have explicit formulas for calculating
.
We describe this by saying that the equation
defines
*implicitly* as a function of
and
,
and we write

Now we introduce an auxiliary function
defined on
as follows:

Equation
states that
on
;
hence the partial derivatives
and
are also
on
.
But we can also compute these partial derivatives by the chain rule. To do
this we write

where
,
,
and
.
The chain rule gives us the formulas

and

where each partial derivative
is to be evaluated at
Since we have

the first of the foregoing equations becomes

Solving this for
we obtain

at those points at which
By a similar argument we obtain a corresponding formula for
:

at those points at which
These formulas are usually written more briefly as follows:

**EXAMPLE.** Assume that the equation
defines
as a function of
and
,
say
Find a value of the constant
such that
,
and compute the partial derivatives
and
at the point

*Solution.* When
,
and
,
the equation becomes
,
and this is satisfied by
.
Let
From

and

we have

When
,
and
we find
and
Note that we were able to compute the partial derivatives
and
using only the value of
at the single point

The foregoing discussion can be extended to functions of more than two variables.

**THEOREM.** *Let
be a scalar field differentiable on an open set
in
Assume that the equation
defines
implicitly as a differentiable function of
,
say
for all points
in some open set
in
Then for each
the partial derivative
is given by the formula
at those points at which
.
The partial derivatives
and
which appear in this equation are to be evaluated at the point
*

The discussion can be generalized in another way. Suppose we have two surfaces
with following implicit representations:

If these surfaces intersect along a curve
,
it may be possible to obtain a parametric representation of
by solving the two equations simultaneously for two of the variables in terms
of the third, say for
and
in terms of
.
Let us suppose that it is possible to solve for
and
and that solutions are given by the equations

for all
in some open interval
.
Then when
and
are replaced by
and
,
respectively, the two equations in this equation are identically satisfied.
That is, we can write
and
for all
in
Again, by using the chain rule, we can compute the derivatives
and
without an explicit knowledge of
and
.
To do this we introduce new functions
and
by means of the equations

Then
for every
in
and hence the derivatives
and
are also zero on
.
By the chain rule these derivatives are given by the formula

Since
and
are both zero we can determine
and
by solving the following pair of simultaneous *linear* equations:

At those points at which the determinant of the system is not zero, these
equations have a unique solution which can be expressed as follows, using
Cramer's rule:

The determinants which appear in this equation are determinants of Jacobian
matrices and are called *Jacobian determinants*. A special notation is
often used to denote Jacobian determinants. We write

In this notation, the formulas for
and
can be expressed more briefly in the form

(The minus sign has been incorporated into the numerators by interchanging the
columns.)

The method can be extended to treat more general situations in which equations in variables are given, where and we solve for of the variables in terms of the remaining variables. The partial derivatives of the new functions so defined can be expressed as quotients of Jacobian determinants.

**EXAMPLE 1.** Assume that the equation
determines
as a differentiable function of
,
say
for all
in some open interval
.
Express the derivative
in terms of the partial derivatives of

Solution. Let
for
in
.
Then the equation
implies
in
.
By the chain rule we have

from which we obtain

at those points
in
at which
The partial derivatives
and
are given by the formulas
and
.

**EXAMPLE 2** When
is eliminated from the two equations
and
,
the result can be expressed in the form
.
Express the derivative
in terms of the partial derivatives of
and
.

Solution. Let us assume that the equation
may be solved for
in terms of
and that a solution is given by
for all
in some open interval
.
Then the function
is given by the formula

Applying the chain rule we have

Using the equation of Example 1 we obtain the formula

The partial derivatives on the right are to be evaluated at the point
.
Note that the numerator can also be expressed as a Jacobian determinant,
giving us

**EXAMPLE 3.** The two equations
and
define
and
as functions of
and
,.
Find formulas for
.

Solution. If we hold
,
fixed and differentiate the two equations in question with respect to
,
remembering that
and
are functions of
and
,
we obtain

Solving these simultaneously for
and
we find

On the other hand, if we hold
fixed and differentiate the two given equations with respect to
we obtain the equations

Solving these simultaneously we find

**EXAMPLE 4.** Let
be defined as a function of
and
by means of the equation

Find
and
in terms of the partial derivatives of
.

Solution. Suppose that
for all
in some open set
.
Substituting
for
in the original equation we must have

where
and
Now we hold
fixed and differentiate both sides with respect to
,
using the chain rule on the right, to obtain

But
and
Hence the equation becomes

Solving this equation for
(and writing
for
)
we obtain

In a similar way we find

This leads to the equation

The partial derivatives
and
are to be evaluated at the point

**EXAMPLE 5.** When
is eliminated from the two equations
and
,
we get an equation of the form
which defines
implicitly as a function of
and
,
say
Prove that

and find a similar formula for

*Solution.* Eliminating
from the two given equations, we obtain the relation

Let
be the function defined by the equation

The discussion in the previous section is now applicable and we can write

But
and
Hence the equations

become

and

**EXAMPLE 6.** The equation
defines
implicitly as a function of
and
,
say
.
Assuming that
show that

where the partial derivatives on the right are to be evaluated at

Solution. We have

We must remember that this quotient really means

Let us introduce
and
.
Our object is to evaluate the partial derivative with respect to
of the quotient

holding
fixed. The rule for differentiating quotients gives us

Since
and
are composite functions, we use the chain rule to compute the partial
derivatives
and
.
For
we have

Similarly, we find

Substituting these into

and replacing
by

we obtain the desired formula.