# Let f(x, y) = x^{4} y^{-3}. Estimate the change \Delta f = f(3.03, 4.9) - f(3, 5) using the...

## Question:

Let {eq}f(x, y) = x^{4} y^{-3}{/eq}.

Estimate the change

{eq}\Delta f = f(3.03, 4.9) - f(3, 5){/eq} using the equation below. {eq}\Delta f \approx f_x (a, b) \Delta x + f_y (a, b) \Delta y{/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} has the following mathematical expression

{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:

We are given the function

{eq}\displaystyle f(x, y) = x^{4} y^{-3} {/eq}.

The change of the function

{eq}\Delta f = f(3.03, 4.9) - f(3, 5) {/eq}

is estimated by using the linearization of the function at the point (3,5) that is

{eq}\displaystyle f(3,5) = 3^{4} \times 5^{-3} = \:0.648 \\ \displaystyle f_x(x,y) = 4x^{3} y^{-3} \Rightarrow f_x(3,5)=\:0.864 \\ \displaystyle f_y(x,y) = -3x^{4} y^{-4} \Rightarrow f_y(3,5)=\:-0.3888 \\ \displaystyle \Delta f \approx f_x(3,5) \Delta x + f_y (3, 5) \Delta y \\ = 0.864(3.03-3) -0.3888(4.9-5) \\ = 0.0648. {/eq}

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