Advanced Calculus Chapter 3 Applications of partial differentiation                    37
                        3 Applications of partial differentiation
                        3.1       Stationary points
                        Higher derivatives
                                        2
                        Let U ⊆ R and f : U → R. The partial derivatives f and f
                                                                                                      x         y
                        are functions of x and y and so we can find their partial deriva-
                        tives. We write fxy to denote fy differentiated with respect to x.
                        Similarly fxx denotes fx differentiated with respect to x, and fyx
                        and fyy denote fx and fy, respectively, differentiated with respect
                        to y. The functions f , f , f                  and f      are the second partial
                                                      xx    xy     yx         yy
                        derivatives of f. For the other notation for partial derivatives, we
                        write ∂ µ∂f¶ or just ∂2f for fxx, ∂ µ∂f¶ or just ∂2f for
                                ∂x ∂x                     ∂x2               ∂x     ∂y                ∂x∂y
                        f , ∂ µ∂f¶ or just ∂2f for f                     and ∂ µ∂f¶ or just ∂2f for
                         xy   ∂y     ∂x                ∂y∂x          yx        ∂y     ∂y                ∂y2
                        fyy.
                        We can also differentiate the second partial derivatives to get the
                        third partial derivatives, and so on. For example, f                     , or   ∂3f , is
                                                                                             xyy       ∂x∂y2
                        the third partial derivative obtained from differentiating fyy with
                        respect to x.
                        Example 1 Let f(x,y) = 3x3y +2xy2 −4x2y. Then
                                                f (x,y) = 9x2y+2y2−8xy;
                                                  x
                                                f (x,y) = 3x3+4xy−4x2;
                                                  y
                                               fxx(x,y) = 18xy−8y;
                                               f (x,y) = 9x2+4y−8x;
                                                 yx
                                               fyy(x,y) = 4x;
                                               f (x,y) = 9x2+4y−8x.
                                                 xy
                        Notethat, in Example 1, fyx = fxy. This is true for all well-behaved
                        functions. In particular, if f            and f      are both continuous on R2
                                                              xy          yx
                                                           2                              2
                        (or an open subset U of R ) then fyx = fxy on R (U).
                        Advanced Calculus Chapter 3 Applications of partial differentiation                    38
                        Example 2 Let z = xcos2y. Then
                                                         ∂z = cos2y,
                                                         ∂x
                                                         ∂z = −2xsin2y,
                                                         ∂y
                                                        ∂2z
                                                        ∂x2 = 0,
                                                       ∂2z
                                                      ∂y∂x = −2sin2y,
                                                       ∂2z
                                                      ∂x∂y = −2sin2y,
                                                        ∂2z
                                                        ∂y2     = −4xcos2y.
                                                      xe2x
                        Example 3 Let z = yn . Find all the possible values of n given
                        that
                                                        ∂2z           ∂2z
                                                     3x       −xy2          =12z.
                                                        ∂x2           ∂y2
                                   xe2x
                        For z = yn we have
                                                             2x         2x
                                                ∂z = e +2xe ,
                                                ∂x                yn
                                                            e2x(1 + 2x)
                                                      =           yn       ,
                                                ∂z            nxe2x
                                                ∂y = −yn+1 ,
                                                 2             2x                2x
                                               ∂ z    = 2e (1+2x)+e ×2,
                                               ∂x2                       yn
                                                            4e2x(x+1)
                                                      =           yn       ,
                                               ∂2z          n(n+1)xe2x
                                               ∂y2    =          yn+2        .
                        Thus
                             2            2             2x                          2 2x         2x
                        3x∂ z−xy2∂ z = 12xe (x+1)−n(n+1)x e                               =xe (12(x+1)−n(n+1)x),
                               2           2               n                    n                n
                           ∂x           ∂y               y                    y                y
                        Advanced Calculus Chapter 3 Applications of partial differentiation                    39
                        and so
                             2             2                       2x                                            2x
                        3x∂ z −xy2∂ z = 12z ⇔ xe (12(x+1)−n(n+1)x)= 12xe ,
                               2             2                     n                                           n
                           ∂x            ∂y                      y                                            y
                                                          ⇔ 12(x+1)−n(n+1)x=12,
                                                                (Since the result is true for all x and y.)
                                                          ⇔ 12x+12−n(n+1)x=12,
                                                          ⇔ 12−n(n+1)=0,
                                                                (Since the result is true for all x.)
                                                          ⇔ n2+n−12=0,
                                                          ⇔ (n+4)(n−3)=0,
                                                          ⇔ n=−4orn=3.
                        Stationary points
                        Let U ⊆ R2, and let f : U → R. Then (a,b) is a stationary point
                        of f if the tangent plane at (a,b) is horizontal. That is, the tangent
                        plane is parallel to the (x,y)-plane, and so has the form z = c for
                        some constant c.
                        Sincethetangentplaneat(a,b)passesthroughthepoint(a,b,f(a,b)),
                        then if it is horizontal it has the form
                                                             z = f(a,b).
                        Recall from Chapter 1 that the equation of the tangent plane at
                        (a,b) is
                                    f(a,b)−z+fx(a,b)(x−a)+fy(a,b)(y−b)=0.
                        Now, if (a,b) is a stationary point then f(a,b) − z = 0, and so
                                             f (a,b)(x −a)+f (a,b)(y −b) = 0.
                                              x                      y
                        The last equation is true for all x and y. Putting x = a and y any
                        value other than b into this equation gives fy(a,b) = 0. Similarly,
                        putting y = b and x any value other than a into this equation
                        gives fx(a,b) = 0. Hence, if (a,b) is a stationary point of f then
                        fx(a,b) = 0 and fy(a,b) = 0. It is straightforward to establish the
                        converse of this result. Thus, to find the stationary points of f, we
                        have to find the solutions of the equations
                                                           f (a,b) = 0,
                                                            x
                                                           fy(a,b) = 0.
                        Example 4 Find all of the stationary points of
                                                                 2           2
                                                  f(x,y) = x y +3xy −3xy.
                        Advanced Calculus Chapter 3 Applications of partial differentiation                    40
                        The partial derivative of f are
                                     f (x,y) = 2xy+3y2−3y =y(2x+3y−3);
                                       x
                                                         2
                                     fy(x,y) = x +6xy−3x=x(x+6y−3).
                        Putting fx(x,y) = fy(x,y) = 0 gives
                                                      y(2x+3y−3) = 0,                                         (1)
                                                       x(x+6y−3) = 0.                                         (2)
                        From equation (1) either y = 0 or 2x + 3y = 3. If y = 0 then
                        equation 2 gives x(x−3) = 0, and so x = 0,3. If 2x+3y = 3 then
                        6y = 6−4x. Thus, in this case, equation 2 gives x(3−3x) = 0, and
                        so x = 0,1 and y = 1, 1, respectively. Thus the stationary points
                                                       3               1
                        of f are (0,0), (0,3), (0,1) and (1, 3).
                        Types of stationary points
                        There are three main types of stationary point: a local minimum,
                        a local maximum and a saddle point. The following diagrams show
                        a typical graph and contour-plot of these three types.
                                   Alocal minimum
                                                                                                    2
                                                                                                    1
                                                 4
                                                 3
                                                                                                    0 y
                                                 2                             -2        -1        0        1
                                                                                                  x
                                                 1 -2
                                                  -1                                                -1
                                                 0
                                    2     1     00 y
                                                 1    -1     -2
                                                x
                                                2                                                   -2
                                   Alocal maximum
                                                                                                    2
                                          -2          -2
                                             -1 xy -1
                                                00                                                  1
                                                0
                                             1      1
                                         2     -1       2
                                                                                                  x
                                               -2                               -2       -1        0        1
                                               -3                                                   0 y
                                               -4
                                               -5                                                   -1
                                                                                                    -2