{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Section14_4

# Section14_4 - Section 14.4 Chain Rules with two variables...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 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 (2 xy 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 (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 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 =- . 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 y-derivatives 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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 10

Section14_4 - Section 14.4 Chain Rules with two variables...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online