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 partial derivatives are

fx =5/9

fy =(-13/18)

Step-by-step explanation:

defining the vector v (from (2,1) to (1,3))

v=(1,3)-(2,1) = (-1,2)

the unit vector will be

v'=(-1,2)/√5 = (-1/√5,2/√5)

the directional derivative is

fv(x,y) = fx*v'x + fy*v'y = fx*(-1/√5)+fy(2/√5) =-2/√5

then defining the vector u ( from (2, 1) toward the point (5, 5) )

u=(5,5)-(2,1) = (3,4)

the unit vector will be

u'=(3,4)/5 = (3/5,4/5)

the directional derivative is

fu(x,y) = fx*ux + fy*uy = fx*(3/5)+fy(4/5)=1

thus we have the set of linear equations

-fx/√5*+2*fy/√5 =(-2/√5) → -fx + 2*fy = -2

(3/5) fx+(4/5)*fy=1 → 3* fx+4*fy = 5

subtracting the first equation twice to the second

 3*fx+4*fy -(- 2fx)*-4*fy = 5 -2*(-2)

5*fx=9

fx=5/9

thus from the first equation

-fx + 2*fy = -2

fy= fx/2 -1 = 5/18 -1 = -13/18

thus we have

fx =5/9

fy =(-13/18)

it would be 6xy and -16xy.

step-by-step explanation: