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The last term in [link] is referred to as the error term and it represents how closely the tangent plane comes to the surface in a small neighborhood ( δ disk) of point P . For the function f to be differentiable at P , the function must be smooth—that is, the graph of f must be close to the tangent plane for points near P .

Demonstrating differentiability

Show that the function f ( x , y ) = 2 x 2 4 y is differentiable at point ( 2 , −3 ) .

First, we 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 , then we use [link] :

f ( 2 , −3 ) = 2 ( 2 ) 2 4 ( −3 ) = 8 + 12 = 20 f x ( 2 , −3 ) = 4 ( 2 ) = 8 f y ( 2 , −3 ) = −4.

Therefore m 1 = 8 and m 2 = −4 , and [link] becomes

f ( x , y ) = f ( 2 , −3 ) + f x ( 2 , −3 ) ( x 2 ) + f y ( 2 , −3 ) ( y + 3 ) + E ( x , y ) 2 x 2 4 y = 20 + 8 ( x 2 ) 4 ( y + 3 ) + E ( x , y ) 2 x 2 4 y = 20 + 8 x 16 4 y 12 + E ( x , y ) 2 x 2 4 y = 8 x 4 y 8 + E ( x , y ) E ( x , y ) = 2 x 2 8 x + 8.

Next, we calculate lim ( x , y ) ( x 0 , y 0 ) E ( x , y ) ( x x 0 ) 2 + ( y y 0 ) 2 :

lim ( x , y ) ( x 0 , y 0 ) E ( x , y ) ( x x 0 ) 2 + ( y y 0 ) 2 = lim ( x , y ) ( 2 , −3 ) 2 x 2 8 x + 8 ( x 2 ) 2 + ( y + 3 ) 2 = lim ( x , y ) ( 2 , −3 ) 2 ( x 2 4 x + 4 ) ( x 2 ) 2 + ( y + 3 ) 2 = lim ( x , y ) ( 2 , −3 ) 2 ( x 2 ) 2 ( x 2 ) 2 + ( y + 3 ) 2 lim ( x , y ) ( 2 , −3 ) 2 ( ( x 2 ) 2 + ( y + 3 ) 2 ) ( x 2 ) 2 + ( y + 3 ) 2 = lim ( x , y ) ( 2 , −3 ) 2 ( x 2 ) 2 + ( y + 3 ) 2 = 0.

Since E ( x , y ) 0 for any value of x or y , the original limit must be equal to zero. Therefore, f ( x , y ) = 2 x 2 4 y is differentiable at point ( 2 , −3 ) .

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Show that the function f ( x , y ) = 3 x 4 y 2 is differentiable at point ( −1 , 2 ) .

f ( −1 , 2 ) = −19 , f x ( −1 , 2 ) = 3 , f y ( −1 , 2 ) = −16 , E ( x , y ) = −4 ( y 2 ) 2 .

lim ( x , y ) ( x 0 , y 0 ) E ( x , y ) ( x x 0 ) 2 + ( y y 0 ) 2 = lim ( x , y ) ( −1 , 2 ) −4 ( y 2 ) 2 ( x + 1 ) 2 + ( y 2 ) 2 lim ( x , y ) ( −1 , 2 ) −4 ( ( x + 1 ) 2 + ( y 2 ) 2 ) ( x + 1 ) 2 + ( y 2 ) 2 = lim ( x , y ) ( 2 , −3 ) 4 ( x + 1 ) 2 + ( y 2 ) 2 = 0.

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The function f ( x , y ) = { x y x 2 + y 2 ( x , y ) ( 0 , 0 ) 0 ( x , y ) = ( 0 , 0 ) is not differentiable at the origin. We can see this by calculating the partial derivatives. This function appeared earlier in the section, where we showed that f x ( 0 , 0 ) = f y ( 0 , 0 ) = 0 . Substituting this information into [link] using x 0 = 0 and y 0 = 0 , we get

f ( x , y ) = f ( 0 , 0 ) + f x ( 0 , 0 ) ( x 0 ) + f y ( 0 , 0 ) ( y 0 ) + E ( x , y ) E ( x , y ) = x y x 2 + y 2 .

Calculating lim ( x , y ) ( x 0 , y 0 ) E ( x , y ) ( x x 0 ) 2 + ( y y 0 ) 2 gives

lim ( x , y ) ( x 0 , y 0 ) E ( x , y ) ( x x 0 ) 2 + ( y y 0 ) 2 = lim ( x , y ) ( 0 , 0 ) x y x 2 + y 2 x 2 + y 2 = lim ( x , y ) ( 0 , 0 ) x y x 2 + y 2 .

Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and the function f is not differentiable at the origin as shown in the following figure.

A curved surface in xyz space that remains constant along the positive x axis and curves downward along the line y = –x in the second quadrant.
This function f ( x , y ) is not differentiable at the origin.

Differentiability and continuity for functions of two or more variables are connected, the same as for functions of one variable. In fact, with some adjustments of notation, the basic theorem is the same.

Differentiability implies continuity

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

[link] shows that if a function is differentiable at a point, then it is continuous there. However, if a function is continuous at a point, then it is not necessarily differentiable at that point. For example,

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

is continuous at the origin, but it is not differentiable at the origin. This observation is also similar to the situation in single-variable calculus.

[link] further explores the connection between continuity and differentiability at a point. This theorem says that if the function and its partial derivatives are continuous at a point, the function is differentiable.

Practice Key Terms 4

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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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