Q. 30 Find the first-order partial der... [FREE SOLUTION]

Chapter 12: Q. 30 (page 944)

Find the first-order partial derivatives for the functions in Exercises 27鈥�36.
$g (x, y) = \frac{\ln (y^{2} + 1)}{x}$

Short Answer

Expert verified

The first-order partial derivatives are $\frac{鈭� g}{鈭� x} = - \frac{\ln (y^{2} + 1)}{x^{2}} and \frac{鈭� g}{鈭� y} = \frac{2 y}{x (y^{2} + 1)}$

Step by step solution

Given

The given function is: $g (x, y) = \frac{\ln (y^{2} + 1)}{x}$

To find

We have to find the first-order partial derivatives.

Calculation

$g (x, y) = \frac{\ln (y^{2} + 1)}{x} \frac{鈭� g}{鈭� x} = \frac{鈭�}{鈭� x} [\frac{\ln (y^{2} + 1)}{x}] \frac{鈭� g}{鈭� x} = \frac{鈭�}{鈭� x} [\ln (y^{2} + 1) 路 x^{- 1}] \frac{鈭� g}{鈭� x} = \ln (y^{2} + 1) \frac{鈭�}{鈭� x} [x^{- 1}] \frac{鈭� g}{鈭� x} = \ln (y^{2} + 1) (- 1) (x^{- 2}) \frac{鈭� g}{鈭� x} = - \frac{\ln (y^{2} + 1)}{x^{2}} g (x, y) = \frac{\ln (y^{2} + 1)}{x} \frac{鈭� g}{鈭� y} = \frac{鈭�}{鈭� y} [\frac{\ln (y^{2} + 1)}{x}] \frac{鈭� g}{鈭� y} = \frac{1}{x} 路 \frac{鈭�}{鈭� y} [\ln (y^{2} + 1)] \frac{鈭� g}{鈭� y} = \frac{1}{x} 路 \frac{1}{(y^{2} + 1)} \frac{鈭�}{鈭� y} (y^{2} + 1) \frac{鈭� g}{鈭� y} = \frac{1}{x} 路 \frac{1}{(y^{2} + 1)} (2 y) \frac{鈭� g}{鈭� y} = \frac{2 y}{x (y^{2} + 1)} Hence, \frac{鈭� g}{鈭� x} = - \frac{\ln (y^{2} + 1)}{x^{2}} and \frac{鈭� g}{鈭� y} = \frac{2 y}{x (y^{2} + 1)}$

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Find the first-order partial derivatives for the functions in Exercises 27鈥�36.
$g (x, y) = \frac{\ln (y^{2} + 1)}{x}$

Short Answer

Step by step solution

Given

To find

Calculation

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91影视

Find the first-order partial derivatives for the functions in Exercises 27鈥�36. gx,y=lny2+1x

Short Answer

Step by step solution

Given

To find

Calculation

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Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

Find the first-order partial derivatives for the functions in Exercises 27鈥�36.
$g (x, y) = \frac{\ln (y^{2} + 1)}{x}$