Q. 26 Use the definition of the partia... [FREE SOLUTION]

Chapter 12: Q. 26 (page 944)

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

Short Answer

Expert verified

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

Step by step solution

Given

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

To find

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

Calculation

Differentiate both sides with respect to x and y respectively.

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

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

Short Answer

Step by step solution

Given

To find

Calculation

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91影视

Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23鈥�26. 鈭�g鈭�xand鈭�g鈭�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

Mechanics Maths

Pure Maths

Decision Maths

Applied Mathematics

Discrete Mathematics

Geometry

Study anywhere. Anytime. Across all devices.

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