Chapter 12: Q. 28 (page 944)

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

Short Answer

Expert verified

The first-order partial derivatives are

$\frac{\partial g}{\partial x} = \frac{y^{2}}{1 + x^{2} y^{4}} and \frac{\partial g}{\partial y} = \frac{2 x y}{1 + x^{2} y^{4}}$

Step by step solution

Given

The given function is: $g (x, y) = \tan^{- 1} (x y^{2})$ .

To find

We have to find the first-order partial derivatives.

Calculation

$g (x, y) = \tan^{- 1} (x y^{2}) \frac{\partial g}{\partial x} = \frac{\partial}{\partial x} (\tan^{- 1} (x y^{2})) \frac{\partial g}{\partial x} = \frac{1}{1 + {(x y^{2})}^{2}} \frac{\partial}{\partial x} (x y^{2}) \frac{\partial g}{\partial x} = \frac{1}{1 + x^{2} y^{4}} (y^{2}) \frac{\partial g}{\partial x} = \frac{y^{2}}{1 + x^{2} y^{4}} g (x, y) = \tan^{- 1} (x y^{2}) \frac{\partial g}{\partial y} = \frac{\partial}{\partial y} (\tan^{- 1} (x y^{2})) \frac{\partial g}{\partial y} = \frac{1}{1 + {(x y^{2})}^{2}} \frac{\partial}{\partial y} (x y^{2}) \frac{\partial g}{\partial y} = \frac{1}{1 + x^{2} y^{4}} (2 x y) \frac{\partial g}{\partial y} = \frac{2 x y}{1 + x^{2} y^{4}} Hence, \frac{\partial g}{\partial x} = \frac{y^{2}}{1 + x^{2} y^{4}} and \frac{\partial g}{\partial y} = \frac{2 x y}{1 + x^{2} y^{4}}$

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

Short Answer

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To find

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Find the first-order partial derivatives for the functions in Exercises 27–36.gx,y=tan-1xy2

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