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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 r 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 r 3 ) = 1 3 r 2 r P V P = T ( 4 3 r 3 ) = 1 3 r 2 r T . Setting r = 10 ,r/P = . 01, and r/T = 0 . 002 then gives V P = 1 3 (10 2 )( . 01) = 1 3 . = 1 . 05 cubic meters atmosphere V T = 1 3 (10 2 )(0 . 002) = 1 15 . = 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

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