# Chain Rule: Further Examples

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

# Derivatives of functions defined implicitly

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.

# Worked Examples

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

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.

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