Explanation It is given that the function is w = u v 2 ( 1 − u 3 ) 1 3 . Here, the dependent variable is w and the independent variables are u and v . Differentiate w = u v 2 ( 1 − u 3 ) 1 3 partially with respect to u as follows. ∂ w ∂ u = ∂ ∂ u ( u v 2 ( 1 − u 3 ) 1 3 ) = v 2 ∂ ∂ u ( u ( 1 − u 3 ) 1 3 ) = v 2 ( 1 − u 3 ) 1 3 ∂ ∂ u ( u ) + u v 2 ∂ ∂ u ( ( 1 − u 3 ) 1 3 ) = v 2 ( 1 − u 3 ) 1 3 ( 1 ) + u v 2 ( 1 3 ( 1 − u 3 ) 1 3 − 1 ) ∂ ∂ u ( 1 − u 3 ) On further simplification, ∂ w ∂ u = v 2 ( 1 − u 3 ) 1 3 + u v 2 3 ( 1 − u 3 ) − 2 3 ( − 3 u 2 ) = v 2 ( 1 − u 3 ) 1 3 − u 3 v 2 ( 1 − u 3 ) − 2 3 = v 2 ( 1 − u 3 ) 1 3 ( 1 − u 3 ) 2 3 − u<

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Chapter 29.3, Problem 14E

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The partial derivatives ∂w∂u and ∂w∂v of the function w=uv2(1−u3)13.

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Chapter 29.3, Problem 13E

Chapter 29.3, Problem 15E

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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 ∂ w ∂ u and ∂ w ∂ v of the function w = u v 2 ( 1 − u 3 ) 1 3 .

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The partial derivatives ∂w∂u and ∂w∂v of the function w=uv2(1−u3)13.

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Chapter 29 Solutions