Explanation It is given that the function is z = y e x 2 y sin ( 2 x − y ) . Here, the dependent variable is z and the independent variables are x and y . Differentiate z = y e x 2 y sin ( 2 x − y ) partially with respect to x as follows. ∂ z ∂ x = ∂ ∂ x ( y e x 2 y sin ( 2 x − y ) ) = y e x 2 y ∂ ∂ x ( sin ( 2 x − y ) ) + sin ( 2 x − y ) ∂ ∂ x ( y e x 2 y ) = y e x 2 y cos ( 2 x − y ) ∂ ∂ x ( 2 x − y ) + sin ( 2 x − y ) [ e x 2 y ∂ ∂ x ( y ) + y ∂ ∂ x ( e x 2 y ) ] = y e x 2 y cos ( 2 x − y ) ( 2 ) + sin ( 2 x − y ) [ ( 0 ) + y e x 2 y ∂ ∂ x ( x 2 y ) ] That is, ∂ z ∂ x = 2 y e x 2 y cos ( 2 x − y ) + sin ( 2 x − y ) [ y e x 2 y ( 2 x y ) ] = 2 y e x 2 y ( cos ( 2 x − y ) + x y sin ( 2 x − y ) ) Thus, the partial derivative of z = y e x 2 y sin ( 2 x − y ) with respect to x is 2 y e x 2 y ( cos ( 2 x − y ) + x y sin ( 2 x − y ) ) . Differentiate z = y e x 2 y sin ( 2 x − y ) partially with respect to y as follows

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Chapter 29, Problem 22RE

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The partial derivatives of the function, ∂z∂x and ∂z∂y.

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Chapter 29, Problem 21RE

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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 of the function, ∂ z ∂ x and ∂ z ∂ y .

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Chapter 29 Solutions