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Chapter 12: Q. 62. (page 945)

For each pair of functions in Exercises 59鈥�62, use Theorem
12.24 to show that there is a function of two variables,
$f (x, y)$ such that role="math" localid="1653976646361" $\frac{d F}{d x} = g (x, y)$ and $\frac{d F}{d y} = h (x, y)$ Then find $F$ .
$g = \frac{2 x}{y}, h (x, y) = - \frac{x^{2}}{y^{2}}$

Short Answer

Expert verified

The required answer is $F (x, y) = \frac{x^{2}}{y} + C$

Step by step solution

01

Given information

Think about, $g = \frac{2 x}{y}, h = - \frac{x^{2}}{y^{2}}$

Then,

$g_{y} = - \frac{2 x}{y^{2}} = h_{x}$

There is a function $F (x, y)$ based on Theorem $12.24$

02

The objective is to find Fintegrate, g with respect to x

Think about,

$鈭� (\frac{2 x}{y}) d x = \frac{2}{y} \frac{x^{2}}{2} + q (y) = \frac{x^{2}}{y} + q (y) = F (x, y)$

Think about,

$\frac{d}{d y} (F (x, y)) = \frac{d}{d y} (\frac{x^{2}}{y} + q (y)) = - \frac{x^{2}}{y^{2}} + q^{'} (y)$

Suppose,

$\frac{d}{d y} (F (x, y)) = h 鈬� - \frac{x^{2}}{y^{2}} + q^{'} (y) = - \frac{x^{2}}{y^{2}} 鈬� q^{'} (y) = 0 q (y) = C$

Hence, $F (x, y) = \frac{x^{2}}{y} + C$

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Most popular questions from this chapter

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.

$f (x, y) = \frac{y}{x}, P = (4, 3), v = 鉄� 3, 鈭� 2 鉄�$

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable.

$\frac{d z}{d t} w h e n z = x^{3} e^{y}, x = \sin t, a n d y = \cos t$

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$f (x, y) = \frac{y}{x}, P = (4, 3)$

$A chain rule for functions of two variables : Let f (x, y) = x^{2} y^{3}, x = s \cos t, and y = s \sin t . Find \frac{鈭� f}{鈭� s} and \frac{鈭� f}{鈭� t} using the preceding results .$

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