Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 30

a) $-x+2y+1$ b) $-e^3+2e^3y-2e^3$

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