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Unformatted text preview: (3/23/08) Section 14.4 Chain Rules with two variables Overview: In this section we discuss procedures for differentiating composite functions with two vari ables. Then we consider secondorder and higherorder 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 xand yderivatives of z = ( x 2 y 3 + sin x ) 10 . Solution To find the xderivative, we consider y to be constant and apply the onevariable 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 (2 xy 3 + cos x ) . Similarly, we find the yderivative by treating x as a constant and using the same onevariable 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 (3 x 2 y 2 ) . square 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 . 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 πr 3 of the balloon with respect to P and T at that time? Solution We first take the Pderivative with T constant and then take the Tderivative 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 = . 01, and ∂r/∂T = 0 . 002 then gives ∂V ∂P = 1 3 π (10 2 )( . 01) = 1 3 π . = 1 . 05 cubic meters atmosphere ∂V ∂T = 1 3 π (10 2 )(0 . 002) = 1 15 π . = 0 . 21 cubic meters degree . square 317 p. 318 (3/23/08) Section 14.4, Chain Rules with two variables Example 3 What are the x and yderivatives of z = F ( g ( x,y )) at x = 5 ,y = 6 if g (5 , 6) = 10 ,F prime (10) = 7 ,g x (5 , 6) = 3, and g y (5 , 6) = 11? Solution By the Chain Rule formula d dt [ F ( u ( t ))] = F prime ( u ( t )) u prime ( t ) for one variable with first x and then y in place of t , we obtain bracketleftBig ∂ ∂x { F ( g ( x,y )) } bracketrightBig x =5 ,y =6 = F prime ( g (5 , 6)) g x (5 , 6) = F prime (10) g x (5 , 6) = ( 7)(3) = 21 bracketleftBig ∂ ∂y { F ( g ( x,y )) } bracketrightBig x =5 ,y =6 = F prime ( g (5 , 6)) g y (5 , 6) = F prime (10) g y (5 , 6) = ( 7)(11) = 77 . square...
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 Summer '08
 Eggers
 Math, Chain Rule, Derivative, Composite Functions, Xu, yu, chain rules

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