Question

prove that f(x,y)=1+ sqrt (x^2+y^2) is not differentiable on (0,0)

Answer 1

\[f(x,y)=1+\sqrt{x^2+y^2}\]

Answer 2

i tried aproaching (0,0) by y=x and y=-x but i get the same results witch implies that the function is differentiable...

Answer 3

take the derivative... you will end up with:
f' = 0 + (1/2)(x^2+y^2)^(-1/2)(2x+2y) = (x+y)/sqrt(x^2+y^2)

Answer 4

now try plugging in x=0 and y=0, you will end up with sqrt 0 = 0 in the denominator u cant have that..

Answer 5

partial derivatives... i should remember that.. thanks!

Answer 6

lol np.. I didn't even notice this was like a multivariable problem... haha i just used that good old calc 1