Explanation Given that the function is F ( x , y ) = y ln ( x + 2 y ) . Here, the dependent variable is z and the independent variables are x and y . Differentiate F ( x , y ) = y ln ( x + 2 y ) partially with respect to x as follows. ∂ F ∂ x = ∂ ∂ x ( y ln ( x + 2 y ) ) = y ∂ ∂ x ( ln ( x + 2 y ) ) = y ⋅ 1 x + 2 y ⋅ ∂ ∂ x ( x + 2 y ) = y x + 2 y ⋅ ( 1 ) That is, ∂ F ∂ x = y x + 2 y . Differentiate ∂ F ∂ x = y x + 2 y partially with respect to x as follows. ∂ 2 F ∂ x 2 = ∂ ∂ x ( y x + 2 y ) = ( x + 2 y ) ∂ ∂ x ( y ) − y ∂ ∂ x ( x + 2 y ) ( x + 2 y ) 2 = ( x + 2 y ) ( 0 ) − y ( 1 ) ( x + 2 y ) 2 = − y ( x + 2 y ) 2 That is, ∂ 2 F ∂ x 2 = − y ( x + 2 y ) 2 . Differentiate ∂ F ∂ x = y x + 2 y partially with respect to y as follows. ∂ 2 F ∂ y ∂ x = ∂ ∂ y ( y x + 2 y ) = x + 2 y ∂ ∂ y ( y ) − y ∂ ∂ y ( x + 2 y ) ( x + 2 y ) 2 = x + 2 y ( 1 ) − y ( 2 ) ( x + 2 y ) 2 = x ( x + 2 y ) 2 Differentiate F ( x , y ) = y ln ( x + 2 y ) partially with respect to y as follows. ∂ F ∂ y = ∂ ∂ y ( y ln ( x + 2 y ) ) = y ∂ ∂ y ( ln ( x + 2 y ) ) + ln ( x + 2 y ) ∂ ∂ y ( y ) = y ⋅ 1 x + 2 y ∂ ∂ y ( x + 2 y ) + ln ( x + 2 y ) ( 1 ) = y x + 2 y ( 2 ) + ln ( x + 2 y ) That is, ∂ F ∂ y = 2 y x + 2 y + ln ( x + 2 y ) . Differentiate ∂ F ∂ y = 2 y x + 2 y + ln ( x + 2 y ) partially with respect to y again as follows

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Chapter 29.3, Problem 32E

To determine

All the second partial derivatives of the function, ∂2F∂x2,∂2F∂y2 and ∂2F∂x∂y.

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Chapter 29.3, Problem 31E

Chapter 29.3, Problem 33E

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Chapter 29 Solutions

Basic Technical Mathematics

Ch. 29.1 - Practice Exercise If f(x, y) = 4xy2 – 3x2y, find...Ch. 29.1 - Prob. 1E Ch. 29.1 - Prob. 2E Ch. 29.1 - Prob. 3E Ch. 29.1 - Prob. 4E Ch. 29.1 - Prob. 5E Ch. 29.1 - Prob. 6E Ch. 29.1 - Prob. 7E Ch. 29.1 - Prob. 8E Ch. 29.1 - Prob. 9E

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Knowledge Booster

All the second partial derivatives of the function, ∂ 2 F ∂ x 2 , ∂ 2 F ∂ y 2 and ∂ 2 F ∂ x ∂ y .

All the second partial derivatives of the function, ∂2F∂x2,∂2F∂y2 and ∂2F∂x∂y.

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Chapter 29 Solutions