SOLUTION: Let f(x,y)=(x^2+2y^2)/ (x+y) if (x,y)otherthan(0,0) 0 otherwise Then the directional derivative of f at (0, 0) along u = (1, 1) is

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Question 1106994: Let
f(x,y)=(x^2+2y^2)/
(x+y) if (x,y)otherthan(0,0)
0 otherwise

Then the directional derivative of f at (0, 0) along u = (1, 1) is
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Since the function is not defined at (0,0) then the directional derivative does not exist at (0,0).
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
If "s" is the parameter on the straight line x = y along the vector (1,1), then we have  


    s = , y = ,   or, equivalently,  x = ,  y = .


Therefore, the numerator is  +  =  +  =  +  = ,  


while the denominator is x + y =  +  = .


Then the ratio itself is

f(x,y) = f(s) =  = .


Thus the function f(s) is LINEAR on s along this direction, and is zero at  x= y= 0= s by the definition, which is consistent with the linear behavior.


So (and therefore), the derivative    DOES EXIST  and is equal to .

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To avoid misunderstanding, let me note (highlight/underline) that for the given function the derivative "along a direction"
DEPENDS on direction, so the function f(x,y) is NOT differentiate at (0,0) in the classic sense as a function of two variables.

It is ONLY differentiate "along a direction", and is a classic example of such a function.


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