Chapter 12: Q. 26 (page 944)

Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23–26.
$\frac{\partial g}{\partial x} and \frac{\partial g}{\partial y} when g (x, y) = \frac{\sqrt{y}}{x + 1}$ .

Short Answer

Expert verified

Required partial derivatives are $\frac{\partial g}{\partial x} = - \frac{\sqrt{y}}{{(x + 1)}^{2}} and \frac{\partial g}{\partial y} = \frac{1}{2 (x + 1) \sqrt{y}} .$

Step by step solution

Given

The given function is: $g (x, y) = \frac{\sqrt{y}}{x + 1}$ .

To find

We have to find $\frac{\partial g}{\partial x} and \frac{\partial g}{\partial y} .$

Calculation

Differentiate both sides with respect to x and y respectively.

$g (x, y) = \frac{\sqrt{y}}{x + 1} \frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (\frac{\sqrt{y}}{x + 1}) \frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (\sqrt{y} {(x + 1)}^{- 1}) \frac{\partial g}{\partial x} = \sqrt{y} (- 1) {(x + 1)}^{- 2} \frac{\partial}{\partial x} (x + 1) \frac{\partial g}{\partial x} = \sqrt{y} (- 1) {(x + 1)}^{- 2} (1) \frac{\partial g}{\partial x} = - \frac{\sqrt{y}}{{(x + 1)}^{2}} g (x, y) = \frac{\sqrt{y}}{x + 1} \frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (\frac{\sqrt{y}}{x + 1}) \frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (\frac{y^{\frac{1}{2}}}{x + 1}) \frac{\partial g}{\partial y} = \frac{1}{(x + 1)} \frac{\partial}{\partial y} (y^{\frac{1}{2}}) \frac{\partial g}{\partial y} = \frac{1}{(x + 1)} (\frac{1}{2}) (y^{- \frac{1}{2}}) \frac{\partial g}{\partial y} = \frac{1}{2 (x + 1) \sqrt{y}} Hence, \frac{\partial g}{\partial x} = - \frac{\sqrt{y}}{{(x + 1)}^{2}} and \frac{\partial g}{\partial y} = \frac{1}{2 (x + 1) \sqrt{y}} .$

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Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23–26.
$\frac{\partial g}{\partial x} and \frac{\partial g}{\partial y} when g (x, y) = \frac{\sqrt{y}}{x + 1}$ .

Short Answer

Step by step solution

Given

To find

Calculation

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Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23–26. ∂g∂xand∂g∂ywhengx,y=yx+1.

Short Answer

Step by step solution

Given

To find

Calculation

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Logic and Functions

Calculus

Discrete Mathematics

Probability and Statistics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23–26.
$\frac{\partial g}{\partial x} and \frac{\partial g}{\partial y} when g (x, y) = \frac{\sqrt{y}}{x + 1}$ .