Find the linearization of the function at the given point.
-$f ( x , y , z ) = \tan ^ { - 1 } x y z$ at $( 9,9,9 )$
A) $L ( x , y , z ) = \frac { 81 } { 6562 } x + \frac { 81 } { 6562 } y + \frac { 81 } { 6562 } z + \tan ^ { - 1 } 81 - \frac { 243 } { 6562 }$
B) $L ( x , y , z ) = \frac { 81 } { 6562 } x + \frac { 81 } { 6562 } y + \frac { 81 } { 6562 } z + \tan ^ { - 1 } 729 - \frac { 243 } { 6562 }$
C) $L ( x , y , z ) = \frac { 81 } { 531,442 } x + \frac { 81 } { 531,442 } y + \frac { 81 } { 531,442 } z + \tan ^ { - 1 } 729 - \frac { 2187 } { 531,442 }$
D) $\mathrm { L } ( \mathrm { x } , \mathrm { y } , \mathrm { z } ) = \frac { 81 } { 531,442 } \mathrm { x } + \frac { 81 } { 531,442 } \mathrm { y } + \frac { 81 } { 531,442 } \mathrm { z } + \tan ^ { - 1 } 81 - \frac { 2187 } { 531,442 }$

Question

Quizplus · Accepted Answer

The linearization of a function $f(x, y, z)$ at a point $(a, b, c)$ is given by $L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c)$, where $f_x$, $f_y$, and $f_z$ are the partial derivatives of $f$ with respect to $x$, $y$, and $z$, respectively. For $f(x, y, z) = \tan^{-1}(xyz)$ at $(9, 9, 9)$, the correct linearization must account for the value of the function and its derivatives at this point. Choice C correctly calculates these values, including the derivative terms and the constant term, which is the function value at the point plus the necessary adjustments from the linearization formula.

Find the Linearization of the Function at the Given Point f(x,y,z)=tan⁡−1xyzf ( x , y , z ) = \tan ^ { - 1 } x y zf(x,y,z)=tan−1xyz

Find the Linearization of the Function at the Given Point $f ( x , y , z ) = \tan ^ { - 1 } x y z$