Answer:
the partial derivatives are
fx =5/9
fy =(-13/18)
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)