# Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 57

$$L(1,0,0) =y-3z \\ L(1,1,0) =x+y-z-1$$

#### Work Step by Step

$f_x (1,0,0) =y-3z =0-3(0)=0 \\f_y(1,0,0) =x+2z =1+0=1\\ f_y(1,0,0) =2y-3x =2(0)-(3)(1) \\= -3$ (a) $L(1,0,0) =0(x-1)+(1)(y-0) -3(z-0)\\ =y-3z$ and $f_x (1,1,0) =y-3z =1-0=1$ (b) $f_y (1,0,0) =x+2z =1+(2)(0)=1 \\ f_z(1,0,0) =2y-3x =(2)(0)-3(1) = -1$ and, $L(1,1,0) =1 (x-1)+(1)(y-1) -1 (z-0)\\ =x+y-z-1$

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