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section 14 4 3 23 08 chain rules with two variables overview in this section we discuss procedures for dierentiating composite functions with two vari ables then we consider second ...

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               Section 14.4                                                                           (3/23/08)
               Chain Rules with two variables
               Overview: In this section we discuss procedures for differentiating composite functions with two vari-
               ables. Then we consider second-order and higher-order derivatives of such functions.
               Topics:
                   • Using the Chain Rule for one variable
                   • The general Chain Rule with two variables
                   • Higher order partial derivatives
               Using the Chain Rule for one variable
               Partial derivatives of composite functions of the forms z = F (g(x;y)) can be found directly with the
               Chain Rule for one variable, as is illustrated in the following three examples.
               Example 1       Find the x-and y-derivatives of z = (x2y3 +sinx)10.
               Solution        Tofindthex-derivative, we consider y to be constant and apply the one-variable Chain
                               Rule formula d (f10) = 10f9 df from Section 2.8. We obtain
                                            dx             dx
                                     ∂ [(x2y3 +sinx)10] = 10(x2y3 +sinx)9 ∂ (x2y3 +sinx)
                                    ∂x                                   ∂x
                                                       =10(x2y3 +sinx)9(2xy3 +cosx):
                                     Similarly, we find the y-derivative by treating x as a constant and using the same
                               one-variable Chain Rule formula with y as variable:
                                     ∂ [(x2y3 +sinx)10] = 10(x2y3 +sinx)9 ∂ (x2y3 +sinx)
                                    ∂y                                   ∂y
                                                       =10(x2y3 +sinx)9(3x2y2): 
               Example 2       The radius (meters) of a spherical balloon is given as a function r = r(P;T) of the
                               atmospheric pressure P (atmospheres) and the temperature T (degrees Celsius). At
                               one moment the radius is ten meters, the rate of change of the radius with respect to
                               atmospheric pressure is −0:01 meters per atmosphere, and the rate of change of the
                               radius with respect to the temperature is 0.002 meter per degree. What are the rates of
                               change of the volume V = 4πr3 of the balloon with respect to P and T at that time?
                                                        3
               Solution        We first take the P-derivative with T constant and then take the T-derivative with
                               P constant, using the Chain Rule for one variable in each case to differentiate r3. We
                               obtain
                                                ∂V = ∂ 4πr3= 1πr2 ∂r
                                                ∂P    ∂P 3         3    ∂P
                                                ∂V = ∂ 4πr3= 1πr2 ∂r:
                                                ∂P    ∂T 3         3    ∂T
                               Setting r = 10;∂r=∂P = −0:01, and ∂r=∂T = 0:002 then gives
                                      ∂V    1    2            1   :       cubic meters
                                      ∂P = 3π(10 )(−0:01) = −3π = −1:05 atmosphere
                                      ∂V    1    2           1   :     cubic meters
                                      ∂T = 3π(10 )(0:002) = 15π = 0:21    degree   : 
                                                            317
                  p. 318 (3/23/08)                                                        Section 14.4, Chain Rules with two variables
                  Example 3         What are the x- and y-derivatives of z = F(g(x;y)) at x = 5;y = 6 if
                                    g(5;6) = 10;F′(10) = −7;g (5;6) = 3, and g (5;6) = 11?
                                                                 x                 y
                  Solution          By the Chain Rule formula d [F (u(t))] = F′(u(t)) u′(t) for one variable with first x
                                                                  dt
                                    and then y in place of t, we obtain
                                      h ∂               i             ′
                                           {F(g(x;y))}            =F (g(5;6))g (5;6)
                                        ∂x                                       x
                                                         x=5;y=6
                                                                  =F′(10)g (5;6) = (−7)(3) = −21
                                                                             x
                                      h ∂               i             ′
                                           {F(g(x;y))}            =F (g(5;6))g (5;6)
                                        ∂y                                       y
                                                         x=5;y=6
                                                                  =F′(10)g (5;6) = (−7)(11) = −77: 
                                                                             y
                        Partial derivatives of composite functions of the forms F(t) = f (x(t);y(t)) and F(s;t)
                  =f(x(s;t);y(s;t)) can be found directly with the Chain Rule for one variable if the “outside” function
                  z = f(x;y) is given in terms of power functions, exponential functions, logarithms, trigonometric
                  functions, and inverse trigonometric functions rather than just by a letter name. This is illustrated
                  in the following example.
                                                                                                         5 6            t
                  Example 4         Find the t-derivative of z = f (x(t);y(t)), where f(x;y) = x y ;x(t) = e , and
                                    y(t) = √t.
                  Solution          Because f(x;y) is a product of powers of x and y, the composite function f (x(t);y(t))
                                    can be rewritten as a function of t. We obtain
                                                                   5     6      t 5 1=2 6     5t 3
                                             f (x(t);y(t)) = [x(t)] [y(t)] = (e ) (t    ) =e t :
                                    Then the Product and Chain Rules for one variable give
                                         d                   d   5t 3      5t d  3     3 d   5t
                                         dt[f (x(t);y(t))] = dt(e t ) = e    dt(t ) +t dt(e )
                                                               2 5t    3 5t d          2 5t     3 5t
                                                          =3t e +t e dt(5t)=3t e +5t e :
                  The general Chain Rule with two variables
                  Wethe following general Chain Rule is needed to find derivatives of composite functions in the form
                  z = f(x(t);y(t)) or z = f (x(s;t);y(s;t)) in cases where the outer function f has only a letter name. We
                  begin with functions of the first type.
                  Theorem 1 (The Chain Rule)           The t-derivative of the composite function z = f (x(t);y(t)) is
                                          d [f (x(t);y(t))] = f (x(t);y(t))x′(t) + f (x(t);y(t))y′(t):                      (1)
                                          dt                   x                     y
                        We assume in this theorem and its applications that x = x(t) and y = y(t) have first derivatives
                  at t and that z = f(x;y) has continuous first-order derivatives in an open circle centered at (x(t);y(t)).
                        Learn equation (1) as the following statement: the t-derivative of the composite function equals
                  the x-derivative of the outer function z = f(x;y) at the point (x(t);y(t)) multiplied by the t-derivative
                  of the inner function x = x(t), plus the y-derivative of the outer function at (x(t);y(t)) multipled by the
                  t-derivative of the inner function y = y(t).
             Section 14.4, Chain Rules with two variables                       p. 319 (3/23/08)
            Proof of Theorem 1: We fix t and set (x;y) = (x(t);y(t)). We consider nonzero ∆t so small that
            (x(t+∆t);y(t+∆t))is in the circle where f has continuous first derivatives and set ∆x = x(t+∆t)−x(t)
            and ∆y = y(t+∆t)−y(t). Then, by the definition of the derivative,
                            d [f(x(t);y(t))] = lim f(x(t+∆t);y(t+∆t))−f(x(t);y(t))
                            dt            ∆t→0             ∆t                          (2)
                                        = lim f(x+∆x;y+∆y)−f(x;y):
                                          ∆t→0          ∆t
                 We express the change f(x + ∆x;y + ∆y) − f(x;y) in the value of z = f(x;y) from (x;y) to
            (x + ∆x;y + ∆y) as the change in the x-direction from (x;y) to (x + ∆x;y) plus the change in the
            y-direction from (x+∆x;y) to (x+∆x;y+∆y), as indicated in Figure 1:
                f(x+∆x;y+∆y)−f(x;y)=[f(x+∆x;y)−f(x;y)]+[f(x+∆x;y+∆y)−f(x+∆x;y)]: (3)
            (Notice that the terms f(x + ∆x;y) and −f(x + ∆x;y) on the right side of (3) cancel to give the left
            side.)
                                      (x+∆x;y+∆y)                        (x+∆x;y+∆y)
                                                                                (x+∆x;c2)
                    (x;y)                    (x+∆x;y) (x;y)      (c1;y)         (x+∆x;y)
                           FIGURE1                             FIGURE2
                 WecanapplytheMeanValue Theoremfrom Section 3.3 to the expression in the first set of square
            brackets on the right of (3) where y is constant and to the expression in the second set of square brackets
            where x is constant. We conclude that there is a number c1 between x and x + ∆x and a number c2
            between y and y +∆y (see Figure 2) such that
                                       f(x+∆x;y)−f(x;y)=f (c ;y)∆x
                                                           x 1
                                                                                       (4)
                               f(x+∆x;y+∆y)−f(x+∆x;y)=f (x+∆x;c )∆y:
                                                           y       2
                 Wecombine equations (3) and (4) and divide by ∆t to obtain
                        f(x+∆x;y+∆y)−f(x;y) =f (c ;y )∆x+f (x+∆x;c )∆y:            (5)
                                  ∆t            x 1 0   ∆t    y        2  ∆t
                 The functions x = x(t) and x = y(t) are continuous at t because they have derivatives at that
            point. Consequently, as ∆t → 0, the numbers ∆x and ∆y both tend to zero and the triangle in Figure 2
            collapses to the point (x;y). Because the partial derivatives of f are continuous, the term fx(c1;y +∆y)
            in (5) tends to f (x;y) and the term f (x;c ) tends to f (x;y) as ∆t → 0. Moreover ∆x=∆t → x′(t) and
                        x               y   2         y
            ∆y=∆t→y′(t) as ∆t →0, so equation (5) with (2) gives
                              d [f(x(t);y(t))] = f (x(t);y(t))x′(t)+f (x(t);y(t))y′(t)
                              dt            x               y
            to establish the theorem. QED
                   p. 320 (3/23/08)                                                                Section 14.4, Chain Rules with two variables
                   Example 5            What is the t-derivative of z = f (x(t);y(t)) at t = 1 if x(1) = 2;y(1) = 3;
                                        x′(1) = −4, y′(1) = 5, f (2;3) = −6, and f (2;3) = 7?
                                                                   x                     y
                   Solution             By formula (1) with t = 1,
                                         h d                  i                         ′                        ′
                                             {f (x(t);y(t))}        =f (x(1);y(1))x (1)+f (x(1);y(1))y (1)
                                           dt                           x                        y
                                                               t=1
                                                                    =f (2;3)x′(1)+f (2;3)y′(1)
                                                                        x                 y
                                                                    =(−6)(−4)+(7)(5)=59: 
                                                ′                          2 3
                   Example 6            Find G (2) where G(t) = h(t ;t ) and h = h(x;y) is such that hx(4;8) = 10 and
                                        hy(4;8) = −20.
                   Solution             Formula (1) gives
                                                ′       d      2 3            2 3 d 2              2 3 d 3
                                              G(t)= dt[h(t ;t )] = hx(t ;t )dt(t )+hy(t ;t )dt(t )
                                                     =2th (t2;t3)+3t2h (t2;t3):
                                                           x                 y
                                        Therefore,
                                              ′                 2   3        2      2   3
                                            G(2)=2(2)hx(2 ;2 )+3(2 )hy(2 ;2 )
                                                   =4hx(4;8)+12hy(4;8)=4(10)+12(−20) =−200:
                           In applications it often helps to interpret the Chain Rule formula (1) in terms of rates of change.
                   Wewrite it in the form
                                                                  dF = ∂F dx + ∂F dy                                                    (6)
                                                                   dt     ∂x dt      ∂y dt
                   without reference to where the derivatives are evaluated. Equation (6) states that the rate of change of
                   F with respect to t equals the rate of change of F with respect to x multiplied by the rate of change of
                   x with respect to t, plus the rate of change of F with respect to y multiplied by the rate of change of y
                   with respect to t.
                   Example 7            Asmall plane uses gasoline at the rate of r = r(h;v) gallons per hour when it is flying
                                        at an elevation of h feet above the ground and its air speed is v knots (nautical miles per
                                        hour). Atamomentwhentheplanehasanaltitudeof8000feetandaspeedof120knots,
                                        its height is increasing 500 feet per minute and it is accelerating 3 knots per minute.
                                        At what rate is its gasoline consumption increasing or decreasing at that moment if at
                                        h=8000andv=120thefunctionr and its derivatives have the values r = 7:2 gallons
                                                                       −4
                                        per hour, ∂r=∂h = −2×10            gallons per hour per foot, and ∂r=∂v = 0:13 gallons per
                                        hour per knot?(1)
                           (1)Data adapted from Cessna 172N Information Manual , Wichita Kansas: Cessna Aircraft Company, 1978, p.5-16.
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