Download 7.2 Partial Derivatives

158
CHAPTER 7. FUNCTIONS FROM EUCLIDEAN PLANE TO R
Basically, the contributions to the change of f made by the change of x and by
the change of y add together to make the total change of f .
7.2
Partial Derivatives
We develop now a way to study local properties of functions of two variables. Let
f (x, y) be a function of two variables and let (a, b) be a point in the domain of
f . How does the value of the function change (in comparison with f (a, b)) when
we evaluate it at some point (a + 4x, b + 4y)? We will answer this question
completely in the next section but we provide a partial answer here (which will
be of major importance later).
Assume that 4y = 0. We consider the limit
f (a + 4x, b) − f (a, b)
4x→0
4x
lim
(analogous to the derivative of a function of one variable) and we denote this
limit by ∂f
∂x (a, b) or fx (a, b). This limit is called the partial derivative of f
with respect to x at the point (a, b). The meaning of fx (a, b) is simply
the rate of change of the values of f (in comparison with f (a, b)) as only the
x-coordinate of the argument is allowed to change. Similarly, if we allow only
y-coordinate of the argument to change (i.e. we keep 4x = 0), then the limit
(a,b)
lim4y→0 f (a,b+4y)−f
is called the partial derivative of f with respect
4y
to y at the point (a, b) and is usually denoted by ∂f
∂y (a, b) or fy (a, b).
If the function f (x, y) is given by a formula, then the values fx (a, b) and
fy (a, b) are very easy to compute. In order to find fx (a, b) we consider the
function f (x, b) of ONE variable x and simply differentiate this function as we
did in Calculus I. The value of that derivative at x = a is equal to fx (a, b) –
just compare the definitions. Similarly, fy (a, b) is equal to the value at y = b
of the usual derivative of the function f (a, y) of ONE variable y. Recall from
Calculus I that in order to compute the value of the derivative of a function g (t)
of one variable at certain point t = 4 one usually computes the FUNCTION
g 0 (t) and then evaluates this function at t = 4. We are going to follow the same
procedure for partial derivatives.
We see now that there is a certain value fx (a, b) associated to every point
(a, b) of the domain of f . That means we can consider a new function fx
of two variables. If we use notation fx (x, y) notice the different meanings of
the two appearances of x: fx or ∂f
∂x is a notation for the function and should
be considered as one symbol while x in the brackets denotes an independent
variable for the function. Similarly we can consider a new function fy of two
variables.
The function fx can be computed as follows: consider the formula for f (x, y)
and treat it as a function of one variable x considering y as a constant; differentiate the formula with respect to x as we usually did in Calculus I; treat
the resulting formula as a function of two variables x and y – this is exactly
7.2. PARTIAL DERIVATIVES
159
fx (x, y). The function fy can be computed as follows: consider the formula for
f (x, y) and treat it as a function of one variable y considering x as a constant;
differentiate the formula with respect to y as we usually did in Calculus I; treat
the resulting formula as a function of two variables x and y – this is exactly
fy (x, y).
Problem. Partial derivative xy
Find the first partial derivatives of the function f (x, y) = xy .
Generalization. Find the first partial derivatives of the function f (x, y).
Strategy.
1. Treat y as a constant and consider f (x, y) as a function of one
variable x. Then differentiate with respect to x to find fx (x, y).
2. Treat x as a constant and consider f (x, y) as a function of one variable y.
Then differentiate with respect to y to find fy (x, y).
Computation.
1. fx (x, y) = yxy−1
2. fy (x, y) = xy ln x
Problem. Partial derivative
Rb
a
g (x) dx
Find the first partial derivatives of the function f (a, b) =
Rb
a
g (x) dx.
Generalization. Find the first partial derivatives of the function f (a, b).
Strategy.
1. Treat b as a constant and consider f (a, b) as a function of one
variable a. Then differentiate with respect to a (using the Fundamental
Theorem of Calculus) to find fa (a, b).
2. Treat a as a constant and consider f (a, b) as a function of one variable b.
Then differentiate with respect to b (using the Fundamental Theorem of
Calculus) to find fb (a, b).
Computation.
1. f (a, b) =
Rb
a
g (x) dx = −
Ra
b
g (x) dx
fa (a, b) = −g (a)
2. fb (a, b) = g (b)
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7.2.1
CHAPTER 7. FUNCTIONS FROM EUCLIDEAN PLANE TO R
Implicit differentiation
Sometimes the dependence of the function z = f (x, y) on its variables x and
y is expressed implicitly and cannot be written using a simple explicit formula
(for example, x + z = eyz ). One can still compute the partial derivatives fx
and fy using the method of implicit differentiation (similar to the one used in
Calculus I). Namely, differentiate the given equation with respect to x keeping
in mind that x and y are independent variables (therefore y itself differentiates
to zero) but z is a function of both x and y which differentiates to zx . Then
solve for zx = fx in the new equation.
Does this method work for any equation? Explain your answer.
Problem. Implicit differentiation x + z = eyz
Use implicit differentiation to find zx : x + z = eyz
Generalization. Given an equation on x, y, z, where z is considered a function
of two independent variables x and y, find zx .
Strategy.
1. Differentiate the given equation with respect to x keeping in
mind that x and y are independent variables (therefore y itself differentiates to zero) but z is a function of x which differentiates to zx .
2. Solve for zx = fx in the new equation.
Computation.
2. zx =
1. 1 + zx = eyz · yzx
1
yeyz −1
Problem. Implicit differentiation x4 + y 6 + z 9 = 9xyz
Use implicit differentiation to find zy (2, 0): x4 + y 6 + z 9 = 9xyz.
Generalization. Given an equation on x, y, z, where z is considered a function
of two independent variables x and y, find zy at a given point (x0 , y0 ).
Strategy.
1. Differentiate the given equation with respect to y keeping in
mind that x and y are independent variables (therefore x itself differentiates to zero) but z is a function of y which differentiates to zy .
2. Solve for zy in the new equation.
3. Solve for z0 = z (x0 , y0 ) in the given equation.
4. Substitute the values x0 , y0 , z0 into the formula for zy .
Computation.
2. zy =
1. 0 + 6y 5 + 9z 8 zy = 9x (z + yzy )
6y 5 −9xz
9xy−9z 8
3. 24 + 0 + z0 9 = 0
z0 = −24/9
7.3. TAYLOR SERIES
4. zy (2, 0) =
7.2.2
161
0−18(−24/9 )
0−9(−24/9 )
8
4
32
19
= −21+ 9 − 9 = −2− 9
Higher order partial derivatives
Similarly to the higher order derivatives considered in Calculus I, we consider
higher order partial derivatives as results of iterative applications of partial
derivatives to the given function f (x, y). For example, if we want to differentiate
f (x, y) twice: first with respect to x, then with respect to y, the resulting second
∂2f
or fxy . Higher order partial derivatives will
partial derivative is denoted ∂x∂y
be used in Taylor approximations. Here we simply practice computing these
derivatives.
There is one theorem that might be useful in computations: for any function
we consider (any smooth function) one has fyx = fxy . More generally: the order
of finding partial derivatives is not important, for instance fyxyxx = fxxxyy . You
may simplify your computation by changing the order of partial differentiation.
Problem. Higher order partial derivative 8xy 7 + x9 y 5
Find the partial derivative fxxy of the function f (x, y) = 8xy 7 + x9 y 5 .
Generalization. Find a higher order partial derivative of a given function.
Strategy. Compute partial derivatives successively as indicated.
Computation.
fx (x, y) = 8y 7 + 9x8 y 5
fxx (x, y) = 0 + 72x7 y 5
fxxy (x, y) = 360x7 y 4
7.3
Taylor series
Given two independent variables x and y, a function f (x, y) of two variables is a
rule of assigning a numerical value corresponding to any pair of values of x and
y. The simplest function is a constant function like f (x, y) = 5 when the value
assigned to a pair (x, y) does not depend on either x or y. A more interesting
class of functions of two variables is the class of linear functions. Any linear
function of two variables can be written as f (x, y) = a + bx + cy where a, b, c
are constants.
Our goal here is to find Taylor series for a function of two variables f (x, y).
First of all, what should be the terms of the Taylor series? In order to simplify
our discussion, let us talk about Taylor series centered at the point (0, 0). Since
we have two variables now, we expect all powers of each variable to be present.
Can we express any function in the following form?
f (x, y) = constant + (sum of powers of x) + (sum of powers of y)