Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 27

a) $5+3x-4y$ b) $3x-4y+5$

a. As we know that $L(x,y)=f(x,y)+f_x(x-m)+f_y(y-n)$ Here, we have $f_x=3$; and $f_y=-4$ Thus, we get $L(0,0)=5+3(x-0)+(-4)(y-0) \\ \implies L(0,0)=5+3x-4y$ b. As we know that $L(x,y)=f(x,y)+f_x(x-m)+f_y(y-n)$ Here, we have $f_x=3$; $f_y=-4$ $L(1,2)=0+3(x-1)+(-4)(y-2)\\ \implies L(1,2)=3x-4y+5$