Given the function:
#f(x,y) = x^2+xy+y^2#
the gradient of the function is the vector:
#nabla f = ( f_x, f_y) = (2x+y, 2y+x)#
So, in the point #(-1,1)# the value of the gradient is:
#[nabla f]_(-1,1) = (-2+1, 2-1) = (-1,1)#
while the unit vector originating from the point can be represented as:
#vec s = (costheta,sintheta) # with #theta in [0,2pi)#
#d/(ds) f(x,y) = vec s * nabla f =(-1,1)*(costheta,sintheta) = sintheta-costheta#
We can see that the derivative is zero when:
#sintheta = costheta#, that is for #theta= pi/4# and #theta=-(3pi)/4#
Now we maximize the value of #s(theta) = sintheta-costheta# with respect to #theta#.
#(ds)/(d theta) = costheta+sintheta= 0 => theta = -pi/4 # or #theta = 3/4pi#
#(d^2s)/(d theta^2) = -sintheta+ costheta #
so, for #theta = -pi/4# we have:
#[(d^2s)/(d theta^2)]_(-pi/4) = sqrt(2)/2+sqrt(2)/2 = sqrt2 > 0#
and this is a maximum. while for #theta = (3pi)/4#we have:
#[(d^2s)/(d theta^2)]_((3pi)/4) = -sqrt(2)/2-sqrt(2)/2 = -sqrt2 < 0#
So the maximum derivative is in the direction:
#theta= -pi/4#
and its value is:
#[(ds)/(d theta)]_(-pi/4) = sin(-pi/4)+cos(-pi/4) = sqrt2#
