# Find the linearization, L(x,y) of the function f(x,y) = tan^(-1) (x + 8y) at the point (9,-1).

## Question:

Find the linearization, {eq}L(x,y) {/eq} of the function

{eq}f(x,y) = tan^{-1}(x + 8y) {/eq} at the point (9,-1).

## Linearization:

The linear approximation, linearization, of a function of two variables at one point is defined by three elements:

the value of the function and the value of the two partial derivatives of the function at the point.

First, calculating the value of the function and the partial derivatives, we have:

{eq}f(x,y) = ta{n^{ - 1}}(x + 8y) \to f\left( {9, - 1} \right) = \frac{\pi }{4}\\ \left\{ {\begin{array}{*{20}{c}} {\frac{{\partial f}}{{\partial x}}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {x + 8y} \right)}^2}}} \cdot 1 \to \frac{{\partial f}}{{\partial x}}\left( {9, - 1} \right) = \frac{1}{2}}\\ {\frac{{\partial f}}{{\partial y}}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {x + 8y} \right)}^2}}} \cdot 8 \to \frac{{\partial f}}{{\partial y}}\left( {9, - 1} \right) = 4} \end{array}} \right. {/eq}

So, the linearization of the function can be written as:

{eq}L\left( {x,y} \right) = f\left( {a,b} \right) + \frac{{\partial f}}{{\partial x}}\left( {a,b} \right)\left( {x - a} \right) + \frac{{\partial f}}{{\partial y}}\left( {a,b} \right)\left( {y - b} \right)\\ L\left( {x,y} \right) = f\left( {9, - 1} \right) + \frac{{\partial f}}{{\partial x}}\left( {9, - 1} \right)\left( {x - 9} \right) + \frac{{\partial f}}{{\partial y}}\left( {9, - 1} \right)\left( {y + 1} \right)\\ L\left( {x,y} \right) = \frac{\pi }{4} + \frac{1}{2} \cdot \left( {x - 9} \right) + 4\left( {y + 1} \right) {/eq} 