To understand the topic of differentiability, we will start with the idea of increment. Consider the function of a single variable. If y = f (x) is differentiable at x = x ₀, then the change in the values of f that results from changing x from x ₀ to x ₀ + D x is D y = f ' (x ₀) D x + e D x in which e ® 0 as D x ® 0. For functions of two variables, the analogous property becomes the definition of differentiability.

THEOREM:

Suppose that the first partial derivatives of f (x, y) are defined throughout an open region R containing the point (x ₀, y ₀) and that f _x and f _y are continuous at (x ₀, y ₀). Then the change D z = f (x ₀ + D x, y ₀ + D y) - f (x ₀, y ₀) in the value of f that results from moving from (x₀, y ₀) to another point (x ₀ + D x, y ₀ + D y) in R satisfies an equation of the form

D z = f _x (x ₀, y ₀)D x + f _y (x ₀, y ₀)D y + e ₁ D x + e ₂ D y (§ )

in which e ₁, e ₂ ® 0 as D x, D y ® 0.

DEFINITION:

A function f (x, y) is differentiable at (x ₀, y ₀) if f _x (x ₀, y ₀) and f _y (x ₀, y ₀) exists and equation (§ ) holds for f at (x ₀, y ₀). We call f differentiable if it is differentiable at every point in the domain.

COROLLARY:

If the partial derivatives f _x and f _y of a function f (x, y) are continuous throughout an open region R, then f is differentiable at every point of R.

Now, if we replace D z with f (x, y) - f (x ₀, y ₀), we will have the following.

(§ § ) f (x, y) = f (x ₀, y ₀) + f _x (x ₀, y ₀)D x + f _y (x ₀, y ₀)D y + e ₁ D x + e ₂ D y

THEOREM:

If a function f (x, y) is differentiable at (x ₀, y ₀), then f is continuous at (x ₀, y ₀).

Functions of two variables can be complicated to work with, so we would like to replace them with simpler ones with a given accuracy. Suppose z = f (x, y), we want a replacement to be effective near a point (x ₀, y ₀) at which we know the values of f, f _x, f _y, and at which f is differentiable.

Since f is differentiable, then equation (§ § ) holds for f at (x ₀, y ₀). Therefore, is we move from (x ₀, y ₀) to any point (x, y) by increments D x = x - x ₀ and D y = y - y ₀, the new value of f will be

f (x, y) = f (x ₀, y ₀) + f _x (x ₀, y ₀)(x - x ₀) + f _y (x ₀, y ₀)(y - y ₀) + e ₁ D x + e ₂ D y

where e ₁, e ₂ ® 0 as D x, D y ® 0.

If the increments D x and D y are small, then the products e ₁ D x and e ₂ D y will eventually be smaller still and

f (x, y) » f (x ₀, y ₀) + f _x (x ₀, y ₀)(x - x ₀) + f _y (x ₀, y ₀)(y - y ₀) = L (x, y).

This is the linearization of a function f (x, y) at a point (x ₀, y ₀).

EXAMPLE 1:

Find the linearization L (x, y) of the function f (x, y) = (x + y + 2) ² at the point (1, 2).

SOLUTION:

f (1, 2) = (1 + 2 + 2) ² = 25

f _x = 2(x + y + 2)

f _y = 2(x + y + 2)

f _x (1, 2) = 2(1 + 2 + 2) = 10

f _y (1, 2) = 2(1 + 2 + 2) = 10

L (x, y) = 25 + 10(x - 1) + 10(y - 2) = 25 + 10x - 10 + 10y - 20

=10x + 10y - 5

EXAMPLE 2:

Find the linearization L (x, y) of the function f (x, y) = x ² + 3y ² at the point (1, 2).

SOLUTION:

f (1, 2) = 1 + 3(4) = 13

f _x = 2x

f _y = 6y

f _x (1, 2) = 2

f _y (1, 2) = 12

L (x, y) = 13 + 2(x - 1) + 12(y -2) = 13 + 2x - 2 + 12y - 24

= 2x + 12y -13

EXAMPLE 3:

Find the linearization L (x, y) of the function f (x, y) = e ^{2 y - x} at the point (0, 0).

SOLUTION:

f (0, 0) = e ⁰ = 1

f _x = -e ^{2 y - x}

f _y = 2e ^{2 y - x}

f _x (0, 0) = -e ⁰ = -1

f _y (0, 0) = 2e ⁰ = 2

L (x, y) = 1 - 1(x - 0) + 2(y - 0) = 1 - x + 2y

Suppose we know the value of a differential function f (x, y) and its partial derivatives at a point (x ₀, y ₀), and we want to predict how much the value of f will change if we move to a point (x ₀ + D x, y ₀ + D y). If D x and D y are small, then f and its linearization at (x ₀, y ₀) will change by nearly the same amount. Therefore L will a practical estimate of the change in f.

D f = f (x ₀ + D x, y ₀ + D y) - f (x ₀, y ₀)

D L = L (x ₀ + D x, y ₀ + D y) - L (x ₀, y ₀) = f _x (x ₀, y ₀) D x + f _y (x ₀, y ₀) D y

The formula for D f is hard to work with, but D L is easier. D L is usually described in simpler terms.

DEFINITION:

If we move from (x ₀, y ₀) to a point (x ₀ + dx, y ₀ + dy) nearby. The resulting differential in f is

df = f _x (x ₀, y ₀)dx + f _y (x ₀, y ₀)dy.

The change in the linearization of f is called the total differential of f.

EXAMPLE 4a:

Around the point (1, 0) is f (x, y) = x ² (y + 1) more sensitive to changes in x, or to changes in y? Give reasons for your answer.

SOLUTION:

f _x = 2x(y + 1)

f _x (1, 0) = 2(1) = 2

f _y = x ²

f _y (1, 0) = 1

df = 2dx + dy

df will be more sensitive to changes in x since the coefficient of the dx term is 2, whereas the coefficient of dy is 1.

EXAMPLE 4b:

What ratio of dx to dy will make df equal to zero at (1, 0)?

SOLUTION:

To determine the ratio of dx to dy that will make df equal zero at the given point, I will set df = 0 and solve for dx/dy.

When we move from (x ₀, y ₀) to a point nearby, we can describe the corresponding change in the values of a function f (x, y) in three different ways.

TRUE

ESTIMATE

ABSOLUTE CHANGE

D f

df

RELATIVE CHANGE

PERCENTAGE CHANGE:

EXAMPLE 5:

About how accurately may V = p r ² h be calculated from measurements of r and h that are in error by 1%?

SOLUTION:

First of all, an error by 1% is a percentage change, so

V_r = 2p r h

V_h = p r ²

dV = 2p r h dr + p r ² dh

I want to find the percentage change of V.

EXAMPLE 6:

To estimate the volume of a cylinder of radius about 2 m and height about 3 m, about how accurately should the radius and height be measured so that the error in the volume estimate will not exceed 0.1 m ³. Assume that the possible error dr in measuring r is equal to the possible error dh in measuring h.

SOLUTION:

V = p r ² h

V_r = 2p r h

V_h = p r ²

dV = V_r dr + V_hdh = 2p r h dr + p r ² dr = (2p r h + p r ²) dr

We are given that dV £ 0.1 m ³ and r = 2 and h = 3.

(2p (2)(3) + p (2) ²) dr £ 0.1

16p dr £ 0.1

dr £ 0.1/(16p ) » 0.002 m or | dr | = | dh | £ 0.002 m

FACT:

The linearization of f (x, y, z) at a point P₀ (x ₀, y ₀, z ₀) is

L (x, y, z) = f (P₀) + f _x (P₀)(x - x ₀) + f _y (P₀)(y - y ₀) + f _z (P₀)(z - z ₀).

FACT:

The total differential is df = f _x (P₀)dx + f _y (P₀)dy + f _z (P₀)dz.

EXAMPLE 7:

Find the linearization L (x, y, z) of the function f (x, y, z) = tan ^-1 (xyz) at (1, 1, 1).

SOLUTION:

f (1, 1, 1) = tan ^-1 1 = p /4.

L (x, y, z) = p /4 + (1/2)(x - 1) + (1/2)(y - 1) + (1/2)(z - 1)

= p /4 -1.5 + 0.5x + 0.5y + 0.5z

EXAMPLE 8:

Find the linearization L (x, y, z) of the function

f (x, y, z) = e ^x + cos (y + z)

at (0, p /2, 0).

SOLUTION:

f (0, p /2, 0) = e ⁰ + cos (p /2 + 0) = 1

f _x = e ^x

f _y = -sin (y + z)

f _z = -sin (y + z)

f _x (0, p /2, 0) = 1

f _y (0, p /2, 0) = -sin (p /2 + 0) = -1

f _z (0, p /2, 0) = -sin (p /2 + 0) = -1

L (x, y, z) = 1 + 1(x - 0) - 1(y - p /2) - 1(z - 0)

= 1 + x - y + p /2 - z = 1 + p /2 + x - y - z