Answered: f(x, y) : ;; P(4, 1); {-1,2) х — у

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter2: Linear Equations

Section: Chapter Questions

Problem 71SGR

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Question

Computing directional derivatives with the gradient Compute the directional derivative of the following functions at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector.

Expert Solution

Step 1

Given:

$f (x, y) = \frac{x}{x - y}; P (4, 1); <- 1, 2>$

To find:

Directional derivative of the given function.

Step 2

To find the directional derivative of function f at some point in direction of the vector u,

we can use the formula,

$D_{u} f (x, y) = \nabla f (x, y) \cdot u$

where u must be unit vector.

That is u must be vector which has a magnitude of 1.

Let,

$\nabla f (x, y) = <f_{x} (x, y), f_{y} (x, y)>$

Here,

$f (x, y) = \frac{x}{x - y} \Rightarrow f_{x} (x, y) = \frac{(x - y) \cdot 1 - x (1)}{{(x - y)}^{2}} = \frac{x - y - x}{{(x - y)}^{2}} = - \frac{y}{{(x - y)}^{2}}$

$\Rightarrow f_{y} (x, y) = \frac{(x - y) \cdot 0 - x (- 1)}{{(x - y)}^{2}} = \frac{x}{{(x - y)}^{2}}$

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