$($a$)(12$ points$)$ Let $f(u,v)$ be a differentiable function such that $f(2,3)=1,f_1(2,3)=-4$ and $f_2(2,3)=7.$ Let us define $g(x,y)=f(xy,x+y).$ Find an equation of the tangent plane to the surface $z=g(x,y)$ at the point $(1,2).$

Solution: An equation of the tangent plane at $(1,2)$ is $z=g_x(1,2)(x-1)+g_y(1,2)(y-2)+g(1,2)(1),$ where $g(1,2)=f(2,3)=1(1),g_x(1,2)=f_1(xy,x+y)y+f_2(xy,x+y)*1{|}_x||=1,y=2=f_1(2,3)*2+f_2(2,3)*1=-1(4),$ and $g_y(1,2)=f_1(xy,x+y)x+f_2(xy,x+y)*1{|}_x||=1,y=2=f_1(2,3)*1+f_2(2,3)*1=3(4).$ Thus, $z=L(x,y)=-(x-1)+3(y-2)+1$ is an equation of the tangent plane $(2).$

$($b$)(6$ points$)$ Use a linear approximation to estimate $g(1\mathrm{.}1,2\mathrm{.}1).$

Solution: We have $g(1\mathrm{.}1,2\mathrm{.}1)$~~$L(1\mathrm{.}1,2\mathrm{.}1)$

$={?}^{(2)}-(1\mathrm{.}1-1)+3(2\mathrm{.}1-2)+1$

$(2)-0\mathrm{.}1+0\mathrm{.}3+1$

$(2)1\mathrm{.}2$