Chapter 12: Q. 32 (page 944)

Find the first-order partial derivatives for the functions in Exercises 27–36.
$g (x, y) = x \cos y$

Short Answer

Expert verified

The first-order partial derivatives are $\frac{\partial g}{\partial x} = \cos y and \frac{\partial g}{\partial y} = - x \sin y$

Step by step solution

Given

The given function is: $g (x, y) = x \cos y$

To find

We have to find the first-order partial derivatives.

Calculation

$g (x, y) = x \cos y \frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (x \cos y) \frac{\partial g}{\partial x} = \cos y g (x, y) = x \cos y \frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (x \cos y) \frac{\partial g}{\partial y} = x (- \sin y) \frac{\partial g}{\partial y} = - x \sin y Hence, \frac{\partial g}{\partial x} = \cos y and \frac{\partial g}{\partial y} = - x \sin y$

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Find the first-order partial derivatives for the functions in Exercises 27–36.
$g (x, y) = x \cos y$

Short Answer

Step by step solution

Given

To find

Calculation

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Find the first-order partial derivatives for the functions in Exercises 27–36. gx,y=xcosy

Short Answer

Step by step solution

Given

To find

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Geometry

Discrete Mathematics

Mechanics Maths

Applied Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Find the first-order partial derivatives for the functions in Exercises 27–36.
$g (x, y) = x \cos y$