# Find the local linearization of the function f (x, y) = x^3 y at the point (2, 1)

## Question:

Find the local linearization of the function {eq}f (x, y) = x^3 y{/eq} at the point {eq}(2, 1){/eq}

## Linearization of Function:

Let us consider a function of two variables {eq}f(x,y) {/eq}.

The linearization of the function at the point {eq}(x_0,y_0) {/eq} can be found using the following formula

{eq}L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) {/eq}

where {eq}f_x,\; f_y {/eq} are the partial derivatives of the function.

## Answer and Explanation: 1

We are given the function

{eq}f(x, y) = x^3 y {/eq}

having its first partial derivatives equal to

{eq}f_x(x, y) = 3x^2 y \\ f_y(x, y) = x^3. {/eq}

The local linearization of the function at the point {eq}(2, 1){/eq} is found as

{eq}\displaystyle f(2,1) = 8 \\ \displaystyle f_x(2,1) = 12 \\ \displaystyle f_y(2,1) = 8 \\ \displaystyle L(x,y) = f(2,1) + f_x(2,1)(x-2) + f_y(2,1)(y-1) \\ \displaystyle = 8 + 12(x-2) + 8(y-1) \\ \displaystyle = 12x+8y-24. {/eq}

#### Learn more about this topic: Linearization of Functions

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Chapter 10 / Lesson 1
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Over the river and through the woods to Grandmother's house we go ... Are we there yet? In this lesson, apply linearization to estimate when we will finally get to Grandma's house!