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Continuity of first partials implies differentiability

Let z = f ( x , y ) be a function of two variables with ( x 0 , y 0 ) in the domain of f . If f ( x , y ) , f x ( x , y ) , and f y ( x , y ) all exist in a neighborhood of ( x 0 , y 0 ) and are continuous at ( x 0 , y 0 ) , then f ( x , y ) is differentiable there.

Recall that earlier we showed that the function

f ( x , y ) = { x y x 2 + y 2 ( x , y ) ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 )

was not differentiable at the origin. Let’s calculate the partial derivatives f x and f y :

f x = y 3 ( x 2 + y 2 ) 3 / 2 and f y = x 3 ( x 2 + y 2 ) 3 / 2 .

The contrapositive of the preceding theorem states that if a function is not differentiable, then at least one of the hypotheses must be false. Let’s explore the condition that f x ( 0 , 0 ) must be continuous. For this to be true, it must be true that lim ( x , y ) ( 0 , 0 ) f x ( 0 , 0 ) = f x ( 0 , 0 ) :

lim ( x , y ) ( 0 , 0 ) f x ( x , y ) = lim ( x , y ) ( 0 , 0 ) y 3 ( x 2 + y 2 ) 3 / 2 .

Let x = k y . Then

lim ( x , y ) ( 0 , 0 ) y 3 ( x 2 + y 2 ) 3 / 2 = lim y 0 y 3 ( ( k y ) 2 + y 2 ) 3 / 2 = lim y 0 y 3 ( k 2 y 2 + y 2 ) 3 / 2 = lim y 0 y 3 | y | 3 ( k 2 + 1 ) 3 / 2 = 1 ( k 2 + 1 ) 3 / 2 lim y 0 | y | y .

If y > 0 , then this expression equals 1 / ( k 2 + 1 ) 3 / 2 ; if y < 0 , then it equals ( 1 / ( k 2 + 1 ) 3 / 2 ) . In either case, the value depends on k , so the limit fails to exist.

Differentials

In Linear Approximations and Differentials we first studied the concept of differentials. The differential of y , written d y , is defined as f ( x ) d x . The differential is used to approximate Δ y = f ( x + Δ x ) f ( x ) , where Δ x = d x . Extending this idea to the linear approximation of a function of two variables at the point ( x 0 , y 0 ) yields the formula for the total differential for a function of two variables.

Definition

Let z = f ( x , y ) be a function of two variables with ( x 0 , y 0 ) in the domain of f , and let Δ x and Δ y be chosen so that ( x 0 + Δ x , y 0 + Δ y ) is also in the domain of f . If f is differentiable at the point ( x 0 , y 0 ) , then the differentials d x and d y are defined as

d x = d x and d y = Δ y .

The differential d z , also called the total differential    of z = f ( x , y ) at ( x 0 , y 0 ) , is defined as

d z = f x ( x 0 , y 0 ) d x + f y ( x 0 , y 0 ) d y .

Notice that the symbol is not used to denote the total differential; rather, d appears in front of z . Now, let’s define Δ z = f ( x + Δ x , y + Δ y ) f ( x , y ) . We use d z to approximate Δ z , so

Δ z d z = f x ( x 0 , y 0 ) d x + f y ( x 0 , y 0 ) d y .

Therefore, the differential is used to approximate the change in the function z = f ( x 0 , y 0 ) at the point ( x 0 , y 0 ) for given values of Δ x and Δ y . Since Δ z = f ( x + Δ x , y + Δ y ) f ( x , y ) , this can be used further to approximate f ( x + Δ x , y + Δ y ) :

f ( x + Δ x , y + Δ y ) = f ( x , y ) + Δ z f ( x , y ) + f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y .

See the following figure.

A surface f in the xyz plane, with a tangent plane at the point (x, y, f(x, y)). On the (x, y) plane, there is a point marked (x + Δx, y + Δy). There is a dashed line to the corresponding point on the graph of f and the line then continues to the tangent plane; the distance to the graph of f is marked f(x + + Δx, y + Δy), and the distance to the tangent plane is marked as the linear approximation.
The linear approximation is calculated via the formula f ( x + Δ x , y + Δ y ) f ( x , y ) + f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y .

One such application of this idea is to determine error propagation. For example, if we are manufacturing a gadget and are off by a certain amount in measuring a given quantity, the differential can be used to estimate the error in the total volume of the gadget.

Approximation by differentials

Find the differential d z of the function f ( x , y ) = 3 x 2 2 x y + y 2 and use it to approximate Δ z at point ( 2 , −3 ) . Use Δ x = 0.1 and Δ y = −0.05 . What is the exact value of Δ z ?

First, we must calculate f ( x 0 , y 0 ) , f x ( x 0 , y 0 ) , and f y ( x 0 , y 0 ) using x 0 = 2 and y 0 = −3 :

f ( x 0 , y 0 ) = f ( 2 , −3 ) = 3 ( 2 ) 2 2 ( 2 ) ( −3 ) + ( −3 ) 2 = 12 + 12 + 9 = 33 f x ( x , y ) = 6 x 2 y f y ( x , y ) = −2 x + 2 y f x ( x 0 , y 0 ) = f x ( 2 , −3 ) = 6 ( 2 ) 2 ( −3 ) = 12 + 6 = 18 f y ( x 0 , y 0 ) = f y ( 2 , −3 ) = −2 ( 2 ) + 2 ( −3 ) = −4 6 = −10.

Then, we substitute these quantities into [link] :

d z = f x ( x 0 , y 0 ) d x + f y ( x 0 , y 0 ) d y d z = 18 ( 0.1 ) 10 ( −0.05 ) = 1.8 + 0.5 = 2.3.

This is the approximation to Δ z = f ( x 0 + Δ x , y 0 + Δ y ) f ( x 0 , y 0 ) . The exact value of Δ z is given by

Δ z = f ( x 0 + Δ x , y 0 + Δ y ) f ( x 0 , y 0 ) = f ( 2 + 0.1 , −3 0.05 ) f ( 2 , −3 ) = f ( 2.1 , −3.05 ) f ( 2 , −3 ) = 2.3425.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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