Explanation It is given that the function is f ( x , y ) = 2 sin 3 2 x 1 − 3 y . Here, the dependent variable is f and the independent variables are x and y . Differentiate f ( x , y ) = 2 sin 3 2 x 1 − 3 y partially with respect to x as follows. ∂ f ∂ x = ∂ ∂ x ( 2 sin 3 2 x 1 − 3 y ) = 2 1 − 3 y ∂ ∂ x ( sin 3 2 x ) = 6 ( sin 2 2 x ) 1 − 3 y ∂ ∂ x ( sin 2 x ) ( use chain rule ) = 6 ( sin 2 2 x ) cos 2 x 1 − 3 y ∂ ∂ x ( 2 x ) On further simplifications, ∂ f ∂ x = 6 ( sin 2 2 x ) cos 2 x 1 − 3 y ( 2 ) = 12 sin 2 2 x cos 2 x 1 − 3 y Thus, the partial derivative of f ( x , y ) = 2 sin 3 2 x 1 − 3 y with respect to x is 12 sin 2 2 x cos 2 x 1 − 3 y . Differentiate f ( x , y ) = 2 sin 3 2 x 1 − 3 y partially with respect to y as follows

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Chapter 29.3, Problem 21E

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The partial derivatives ∂f∂x and ∂f∂y of the function, f(x,y)=2sin32x1−3y.

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Chapter 29.3, Problem 20E

Chapter 29.3, Problem 22E

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Chapter 29 Solutions

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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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The partial derivatives ∂ f ∂ x and ∂ f ∂ y of the function, f ( x , y ) = 2 sin 3 2 x 1 − 3 y .

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The partial derivatives ∂f∂x and ∂f∂y of the function, f(x,y)=2sin32x1−3y.

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Chapter 29 Solutions