Partial Derivatives

Question

Let g(x,y)=(xy)^(1/3).

e.Does the graph of g have a tangent plane at (0,0)? Explain briefly.

f.Is g differentiable at (0,0)? Explain briefly.

 

I asked this question yesterday but was still confused. I know that g is continuous at (0,0) and that the partial derivatives at (0,0) exist but aren't continuous. What I'm confused about is whether g has a tangent plane at (0,0) and if g is differentiable at (0,0). These two things probably go together, but when I evaluated 

lim(x,y)→(0,0) (f(x,y)−f(0,0)−∂f∂x∣∣∣(0,0)(x−0)−∂f∂y∣∣∣(0,0)(y−0))/‖(x,y)‖, I got 0, which according to my book means that g is differentiable at that point since the partial derivatives exist. But the answer I received yesterday says it's not differentiable, but I'm still not sure why. Thank you.

Answer 1

Greetings! I will be glad to clear this up for you !

My response to this is no, it is not differentiable because the slope is undefined at the point (x,y) = (0,0). The slope yields the partial derivatives at the given point (0,0) as

(∂g/∂x) = (y/3)(xy)^-2/3 = (y / 3(xy)^2/3)

(∂g/∂y) = (x/3)(xy)^-2/3 = (x / 3(xy)^2/3)

Both partial derivatives with respect to x and y give the same output of (0/0) at the point (0,0).

The verbal translation would be that the function is undefined and not differentiable at point (0,0).

I hope this helped!