Section 14.4 (3/23/08) Chain Rules with two variables Overview: In this section we discuss procedures for differentiating composite functions with two variables. 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 = (x 2 y 3 + sin x) 10 . Solution To find the x-derivative, we consider y to be constant and apply the one-variable Chain Rule formula d DX(f 10) = 10 f 9 df dx from Section 2.8. We obtain ∂ ∂x[(x 2 y 3 + sin x) 10] = 10(x 2 y 3 + sin x) 9 ∂ ∂x(x 2 y 3 + sin x) = 10(x 2 y 3 + sin x) 9 (2xy 3 + cos x). 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: ∂ ∂y [(x 2 y 3 + sin x) 10] = 10(x 2 y 3 + sin x) 9 ∂ ∂y (x 2 y 3 + sin x) = 10(x 2 y 3 + sin x) 9 (3x 2 y 2 ). 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 = 4 3 πr3 of the balloon with respect to P and T at that time? 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 r 3 . We obtain ∂V ∂P = ∂ ∂P 4 3 πr 3 = 1 3 πr 2 ∂r ∂P ∂V ∂P = ∂ ∂T 4 3 πr 3 = 1 3 πr 2 ∂r ∂T . Setting r = 10, ∂r/∂P = −0.01, and ∂r/∂T = 0.002 then gives ∂V ∂P = 1 3 π(102 )(−0.01) = − 1 3 π .= −1.05 cubic meters atmosphere ∂V ∂T = 1 3 π(102 )(0.002) = 1 15π .= 0.21 cubic meters degree . 3 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, F0 (10) = −7, gx(5, 6) = 3, and gy(5, 6) = 11? Solution By the Chain Rule formula d dt[F (u(t))] = F 0 (u(t)) u 0 (t) for one variable with first x and then y in place of t, we obtain h ∂ ∂x{F(g(x, y))} i x=5,y=6 = F 0 (g(5, 6)) gx(5, 6) = F 0 (10) gx(5, 6) = (−7)(3) = −21 h ∂ ∂y {F(g(x, y))} i x=5,y=6 = F 0 (g(5, 6)) gy(5, 6) = F 0 (10) gy(5, 6) = (−7)(11) = −77. 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. Example 4 Find the t-derivative of z = f (x(t), y(t)), where f(x, y) = x 5 y 6 , x(t) = e t , 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 f (x(t), y(t)) = [x(t)]5 [y(t)]6 = (e t ) 5 (t 1/2 ) 6 = e 5t t 3 . Then the Product and Chain Rules for one variable GIVE d dt[f (x(t), y(t))] = d dt (e 5t t 3 ) = e 5t d dt (t 3 ) + t 3 d dt (e 5t ) = 3t 2 e 5t + t 3 e 5t d dt (5t) = 3t 2 e 5t + 5t 3 e 5t . The general Chain Rule with two variables We the 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 dt[f (x(t), y(t))] = fx (x(t), y(t)) x 0 (t) + fy (x(t), y(t)) y 0 (t). (1) 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(Explanation: