Second-Order Partial Derivatives Previously we have taken the partial derivative of a function f(x, y). But those partial derivatives were themselves functions and so we can take their partial derivatives.

The Second-Order Partial Derivatives of z = f(x, y)   z   2 z  ( f x ) x  f xx   x  x  x 2

  z   2 z  ( f y ) x  f yx   x  y  xy

  z   2 z  ( f x ) y  f xy   y  x  yx

  z   2 z  ( f y ) y  f yy   y  y  y 2

Example 1: Compute the second-order partial derivatives of f(x, y) = x2y + 5xsin(y). Solution: Notice that fx(x, y) = 2xy + 5sin(y) and fy(x, y) = x2 + 5xcos(y). Thus, we have that fxx(x, y) = 2y fxy(x, y) = 2x + 5cos(y) fyx(x, y) = 2x + 5cos(y) fyy(x, y) = –5xsin(y). Notice that in Example 1 above, we have that fxy(x, y) = fyx(x, y). Indeed, this is typically always the case. Thus, it does not matter if we take the partial derivative with respect to x first or with respect to y first.

The Equality of Mixed Partial Derivatives If fxy(x, y) and fyx(x, y) are continuous at (a, b), an interior point of their domain, then f xy (a, b)  f yx (a, b) .

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We can visualize this statement in the following diagram.

z = f x,y ∑z ∑y

∑z ∑x

fy x,y

fx x,y ∑ fx ∑x

∑ fx ∑y

∑ fy

∑ fy

∑x

∑y

fxxx,y fxyx,y=fyxx,y fyyx,y Figure 1: Equality of Mixed Partial Derivatives

Example 2:

Find all second-order partial derivatives of f(x, y) = ln(3x + 5y). Solution:

Notice that f x ( x, y ) 

f xx ( x, y ) 

  3   9 1 2     3(3 x  5 y )   9(3x  5 y )  (3 x  5 y ) 2 x  3x  5 y  x

f xy ( x, y )  f yx ( x, y )  f yy ( x, y ) 

3 5 and f y ( x, y )  . Thus, we have that 3x  5 y 3x  5 y

  3    15 1 2 .     3(3 x  5 y )   15(3 x  5 y )  (3 x  5 y ) 2 y  3x  5 y  y

  5   25 1 2 .     5(3x  5 y )   25(3 x  5 y )  (3x  5 y ) 2 y  3x  5 y  y

At this point, the reader might be wondering why we care about these second-order partial derivatives. We shall see many applications of this in the near future as we seek to determine maximum and minimum values, but for the time being, we can think of the second-order partial derivative as telling us the rate of change of the rate of change of the function. (No, there is a not a typo in that sentence.)

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If we think back to a one-variable example, if s(t) measured the position of a particle at time t, then we have that s(t )  v(t ) measured the particle’s rate of change (velocity), and s(t )  v(t )  a (t ) measured the rate of change of the velocity (acceleration). The same idea applies to a function of two variables, but now we can ask how does the function change as the rate of change (in a particular direction, say x or y) changes in another direction (either x or y).

Recall, we introduced Taylor polynomials as a way to better fit a polynomial to a particular function at a point. The Degree 1 Taylor Polynomial approximation was just the tangent line approximation that we saw quite a while ago. That is, f ( x)  f (a )  f (a)( x  a)

for x near a

We saw that a Degree 2 Taylor Polynomial approximation did a better job approximating the function. That is, f ( x)  f (a)  f (a )( x  a ) 

f (a ) ( x  a) 2 2

for x near a

Earlier in the chapter, we defined the local linearization for a function of two variables, f(x, y), near the point (a, b) as

f ( x, y )  f (a, b)  f x (a, b)( x  a)  f y (a, b)( y  b) Using the same notation as we did when discussing Taylor Polynomials, we have the following result:

Degree 1 Taylor Polynomial approximation of f(x, y) for (x, y) near (a, b)

If f(x, y) has continuous first-order partial derivatives, fx(x, y) and fy(x, y), then f ( x, y )  f (a, b)  f x (a, b)( x  a)  f y (a, b)( y  b)

As one might infer from the discussion above, we get a better approximation to the function f(x, y) if instead we use a quadratic polynomial. In the same manner that we constructed the Degree 2 Taylor Polynomial approximation for f(x) for x near a, we have the following result:

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Degree 2 Taylor Polynomial approximation of f(x, y) for (x, y) near (a, b)

If f(x, y) has continuous second-order partial derivatives, then f ( x, y )  f (a, b)  f x (a, b)( x  a )  f y ( a, b)( y  b) 

f ( a, b) f xx (a, b) ( x  a) 2  f xy (a, b)( x  a)( y  b)  yy ( y  b) 2 2 2

One can check that the first- and second-order partial derivatives of both sides are equal, ensuring that our polynomial is a close approximation to the true function.

Example 3:

Find the Degree 1 and Degree 2 Taylor Polynomials for f(x, y) = x2 + xy + y2 at the point (1, 2). What do you notice about the Degree 2 approximation? Solution:

It will probably be easiest to put all of this information in a table. Function f ( x, y ) f x ( x, y ) f y ( x, y )

Equation x 2  xy  y 2 2x  y x  2y

Value at (1, 2) 7 4 5

Function f xx ( x, y ) f xy ( x, y ) f yy ( x, y )

Equation 2 1

Value at (1, 2) 2 1

2

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Using the above table, we see that the Degree 1 Taylor Polynomial approximation for f(x, y) near (1, 2) is given by 7 + 4(x – 1) + 5(y – 2) = –7 + 4x + 5y. The Degree 2 Taylor Polynomial approximation for f(x, y) near (1, 2) is given by 2 2 7  4( x  1)  5( y  2)  ( x  1) 2  1( x  1)( y  2)  ( y  2) 2 2 2 2  7  4 x  4  5 y  10  ( x  1)  ( x  1)( y  2)  ( y  2) 2  7  4 x  4  5 y  10  x 2  2 x  1  xy  2 x  y  2  y 2  4 y  4  x 2  xy  y 2 We remark that the Degree 2 Taylor Polynomial approximation for f(x, y) is precisely the function f(x, y). This is because f(x, y) does not contain any powers greater than 2.

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Example 4:

Find the Degree 2 Taylor Polynomial for the function f ( x, y ) 

1 at the point (2, 3). xy

Solution:

Again, we construct a table similar to that in Example 3. Function f ( x, y ) f x ( x, y ) f y ( x, y )

Equation 1 ( xy ) 1 ( x 2 y )  1 ( xy 2 )

Value at (2,3) 16  1 12 1 18

Function f xx ( x, y ) f xy ( x, y ) f yy ( x, y )

Equation 2 ( x3 y) 1 ( x2 y2 ) 2 ( xy 3 )

Value at (2,3) 1 12 1 36 1 27

Thus, the Degree 2 Taylor Polynomial approximation for f(x, y) near (1, 2) is given by

1 1 1 1 1 1  ( x  2)  ( y  3)  ( x  2) 2  ( x  2)( y  3)  ( y  3) 2 . 6 12 18 24 36 54

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