(3 points) The directional derivative of f(x, y) at (2, 1) in the direction going from (2, 1) toward the point (1, 3) is −2/ √ 5, and the directional derivative at (2, 1) in the direction going from (2, 1) toward the point (5, 5) is 1. Compute fx(2, 1) and fy(2, 1).

The vector pointing from (2, 1) to (1, 3) points in the same direction as the vector [tex]\vec u=(1,3)-(2,1)=(-1,2)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec u[/tex] is[tex]D_{\vec u}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec u}{\|\vec u\|}[/tex]We have[tex]\|\vec u\|=\sqrt{(-1)^2+2^2}=\sqrt5[/tex]Then[tex]D_{\vec u}f(2,1)=(f_x(2,1),f_y(2,1))\cdot\dfrac{(-1,2)}{\sqrt5}=\dfrac{-f_x(2,1)+2f_y(2,1)}{\sqrt5}=-\dfrac2{\sqrt5}[/tex][tex]\implies f_x(2,1)-2f_y(2,1)=2[/tex]The vector pointing from (2, 1) to (5, 5) has the same direction as the vector [tex]\vec v=(5,5)-(2,1)=(3,4)[/tex]. The derivative of [tex]f[/tex] at (2, 1) in the direction of [tex]\vec v[/tex] is[tex]D_{\vec v}f(2,1)=\nabla f(2,1)\cdot\dfrac{\vec v}{\|\vec v\|}[/tex][tex]\|\vec v\|=\sqrt{3^2+4^2}=5[/tex]so that[tex](f_x(2,1),f_y(2,1))\cdot\dfrac{(3,4)}5=1[/tex][tex]\implies3f_x(2,1)+4f_y(2,1)=5[/tex]Solving the remaining system gives [tex]f_x(2,1)=\dfrac95[/tex] and [tex]f_y(2,1)=-\dfrac1{10}[/tex].