Chapter 12: Q. 30 (page 944)

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

Short Answer

Expert verified

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

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

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

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

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=lny2+1x

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Step by step solution

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Calculation

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