Calculus II, Spring 2023 (http://www.math.nthu.edu.tw/˜ wangwc/)

Remark on Definition of Differentiability

The following definitions are equivalent (i.e. the same, even though they look different) Definition 1: f(x, y) is differentiable at (x_{0}, y_{0}) if

∂_{x}f(x_{0}, y_{0}) and ∂_{y}f(x_{0}, y_{0}) exist and

f(x, y) = L(x, y) + ε_{1} · (x − x_{0}) + ε_{2} · (y − y_{0})  (1)

where

L(x, y) = f(x_{0}, y_{0}) + ∂_{x}f(x_{0}, y_{0}) · (x − x_{0}) + ∂_{y}f(x_{0}, y_{0}) · (y − y_{0})  (2)

and

lim_{(x,y)→(x_{0},y_{0})} ε_{1} = lim_{(x,y)→(x_{0},y_{0})} ε_{2} = 0

Definition 2: f(x, y) is differentiable at (x_{0}, y_{0}) if

There exists a, b, c ∈ ℝ such that the linear function

L(x, y) = a · (x − x_{0}) + b · (y − y_{0}) + c

satisfies

f(x, y) = L(x, y) + ε_{1} · (x − x_{0}) + ε_{2} · (y − y_{0})  (3)

where

lim_{(x,y)→(x_{0},y_{0})} ε_{1} = lim_{(x,y)→(x_{0},y_{0})} ε_{2} = 0

Moreover, if a, b, c ∈ ℝ do exist, we have

∂_{x}f(x_{0}, y_{0}) and ∂_{y}f(x_{0}, y_{0}) exist, and a = ∂_{x}f(x_{0}, y_{0}), b = ∂_{y}f(x_{0}, y_{0}), c = f(x_{0}, y_{0}). (4)

The proof of (4) is identical to the one-variable case. See page 6 of Lecture 14.

Other equivalent definitions:

In view of the identity (see homework 07, problem 2 for hint of proof)

ε_{1} · (x − x_{0}) + ε_{2} · (y − y_{0}) = ε · \sqrt{(x − x_{0})^{2}(y − y_{0})^{2}},  (5)

we can further modify definition 1 and definition 2 into following equivalent definitions: Definition 1’: f(x, y) is differentiable at (x_{0}, y_{0}) if

∂_{x}f(x_{0}, y_{0}) and ∂_{y}f(x_{0}, y_{0}) exist and

f(x, y) = L(x, y) + ε · \sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}}  (6)

where

L(x, y) = f(x_{0}, y_{0}) + ∂_{x}f(x_{0}, y_{0}) · (x − x_{0}) + ∂_{y}f(x_{0}, y_{0}) · (y − y_{0})  (7)

and

lim_{(x,y)→(x_{0},y_{0})} ε = 0  (8)

Definition 2’: f(x, y) is differentiable at (x_{0}, y_{0}) if

There exists a, b, c ∈ ℝ such that the linear function

L(x, y) = a · (x − x_{0}) + b · (y − y_{0}) + c

satisfies

f(x, y) = L(x, y) + ε · \sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}}  (9)

where

lim_{(x,y)→(x_{0},y_{0})} ε = 0

Moreover, if a, b, c ∈ ℝ do exist, we have

∂_{x}f(x_{0}, y_{0}) and ∂_{y}f(x_{0}, y_{0}) exist, and a = ∂_{x}f(x_{0}, y_{0}), b = ∂_{y}f(x_{0}, y_{0}), c = f(x_{0}, y_{0}). (10)

Remark: It is straight forward to generalize to higher dimensional case. For example Definition 1’-3D: f(x, y, z) is differentiable at (x_{0}, y_{0}, z_{0}) if

∂_{x}f(x_{0}, y_{0}, z_{0}), ∂_{y}f(x_{0}, y_{0}, z_{0}) and ∂_{z}f(x_{0}, y_{0}, z_{0}) exist and

f(x, y, z) = L(x, y, z) + ε · \sqrt{(x − x_{0})^{2} + (y − y_{0})^{2} + (z − z_{0})^{2}}  (11)

where

L(x, y, z) = f(x_{0}, y_{0}, z_{0})+∂_{x}f(x_{0}, y_{0}, z_{0})·(x−x_{0})+∂_{y}f(x_{0}, y_{0}, z_{0})·(y−y_{0})+∂_{z}f(x_{0}, y_{0}, z_{0})·(z−z_{0})

(12)

and

lim_{(x,y,z)→(x_{0},y_{0},z_{0})} ε = 0

Definitions 1-3D, 2-3D and 2’-3D are similar.

Question: How do we verify whether f(x, y) is differentiable at (x_{0}, y_{0})? (For example, Homework 07 problem 3) Answer:

Step 1: Find ∂_{x}f(x_{0}, y_{0}) and ∂_{y}f(x_{0}, y_{0}). If they don’t exist, then f(x, y) is NOT differentiable at (x_{0}, y_{0}).

Step 2: If ∂_{x}f(x_{0}, y_{0}) and ∂_{y}f(x_{0}, y_{0}) exist, we still need to check whether L(x, y) given by (2) satisfies (1) or (6). In most cases, it is easier to check (6). In other words, to check whether the following is true:

lim_{(x,y)→(x_{0},y_{0})} \frac{f(x, y) − L(x, y)}{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}}} = 0 ?  (13)

If (13) is true, then we can write

ε =^{def} \frac{f(x, y) − L(x, y)}{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}}} , lim_{(x,y)→(x_{0},y_{0})} ε = 0.  (14)

which is the same as (6) and (8), thus we conclude that f(x, y) is differentiable at (x_{0}, y_{0}) if (13) holds. One can follow the procedures above to solve problem 3 of Homework 07 for (x_{0}, y_{0}) = (0, 0).

On the other hand, Definition 1 is more convenient for the proof of Theorem 2 (page 830) given in Appendix 9. Tangent plane interpretation:

In Definition 1’: (6) and (8) leads to

lim_{(x,y)→(x_{0},y_{0})} \frac{f(x, y) − L(x, y)}{\sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}}} = 0,  (15)

meaning that f(x, y) and L(x, y) are very close, their difference is even much smaller than \sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} when \sqrt{(x − x_{0})^{2} + (y − y_{0})^{2}} → 0. This is the definition for the two surfaces z = f(x, y) and z = L(x, y) to be tangent at (x_{0}, y_{0}, f(x_{0}, y_{0})). Linearization (Linear Approximation):

When f(x, y) is differentiable at (x_{0}, y_{0}) (and therefore f(x, y) and L(x, y) are very close near (x_{0}, y_{0})), we call z = L(x, y) the linearization or linear approximation (since L is linear) of z = f(x, y) at (x_{0}, y_{0}). Remark on the Definition of tangent plane on page 853:

When f(x, y) is differentiable at (x_{0}, y_{0}), the function F (x, y, z) = z − f(x, y) is also differentiable at (x_{0}, y_{0}, z_{0}) for any z_{0}. In particular, the surface z = f(x, y) passing through (x_{0}, y_{0}, f(x_{0}, y_{0})) is the same as the level surface S_{0} = {(x, y, z), F (x, y, z) = 0}. Therefore, the tangent plane of S_{0} at (x_{0}, y_{0}, z_{0} = f(x_{0}, y_{0})) can be obtained from

∇F (x_{0}, y_{0}, z_{0}) · (x − x_{0}, y − y_{0}, z − z_{0}) = 0,  z_{0} = f(x_{0}, y_{0}).  (16)

It is easy to see that (16) leads to z = L(x, y).