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Chapter 12: Q. 24 (page 944)

Use the definition of the partial derivative to find the partial derivatives specified in Exercises 23鈥�26.
$f_{x} (0, - 3) and f_{y} (0, - 3) when f (x, y) = \frac{3 x}{y^{2}}$

Short Answer

Expert verified

Required partial derivatives are $f_{x} (0, - 3) = \frac{1}{3} and f_{y} (0, - 3) = 0$

Step by step solution

01

Given

The given function is: $f (x, y) = \frac{3 x}{y^{2}}$ .

02

To find

We have to find $f_{x} (0, - 3) and f_{y} (0, - 3)$ .

03

Calculation

Differentiate both sides with respect to x and y respectively and then plug $x = 0$ and $y = - 3$ .

$f (x, y) = \frac{3 x}{y^{2}} f_{x} (x, y) = \frac{鈭�}{鈭� x} (\frac{3 x}{y^{2}}) f_{x} (x, y) = \frac{3}{y^{2}} f_{x} (0, - 3) = \frac{3}{{(- 3)}^{2}} f_{x} (0, - 3) = \frac{3}{9} f_{x} (0, - 3) = \frac{1}{3} f (x, y) = \frac{3 x}{y^{2}} f_{y} (x, y) = \frac{鈭�}{鈭� y} (\frac{3 x}{y^{2}}) f_{y} (x, y) = \frac{鈭�}{鈭� y} (3 x y^{- 2}) f_{y} (x, y) = 3 x (- 2) y^{- 3} f_{y} (x, y) = - \frac{6 x}{y^{3}} f_{y} (0, - 3) = - \frac{6 (0)}{{(- 3)}^{3}} f_{y} (0, - 3) = 0 Hence, f_{x} (0, - 3) = \frac{1}{3} and f_{y} (0, - 3) = 0$

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Most popular questions from this chapter

Consider the function f(x, y) = 2x + 3y.

(a) Why is the graph of f a plane?

(b) In what direction is f increasing most rapidly at the

point (鈭�1, 4)?

(c) In what direction is f increasing most rapidly at the

point (x 0, y 0)?

(d) Why are your answers to parts (b) and (c) the same?

Evaluate the following limits, or explain why the limit does not exist.

$\lim_{(x, y, z) 鈫� (0, 0, 0)} \frac{x^{2} + y^{2}}{x^{2} + y^{2} + z^{2}}$

Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P.

$f (x, y) = x \sin 鈦� y, P = (3, \frac{蟺}{2})$

Partial derivatives: Find all first- and second-order partial derivatives for the following functions:

$f (x, y, z) = x z^{2} e^{y}$

In Exercises $37 - 42$ , use the partial derivatives of $g (x, y) = x \cos y$ and the point $(1, 蟺)$ specified to

$(a)$ find the equation of the line tangent to the surface defined by the function in the $x$ direction,

$(b)$ find the equation of the line tangent to the surface defined by the function in the $y$ direction, and

$(c)$ find the equation of the plane containing the lines you found in parts $(a)$ and $(b)$ .

See all solutions

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