Partial derivatives (multiple parts)

Question

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

a. Is g continuous at (0,0)?

I put down yes since it's not undefined there.

 

b.Calculate ∂g/∂x and ∂g/∂y when xy≠0.

This wasn't too bad and pretty straightforward. I just have ∂g/∂x=1/3(xy)^(-2/3)y and ∂g/∂y=1/3(xy)^(-2/3)x.

 

c.Show that gx(0,0) and gy(0,0) exist and find their values. 

I got really stuck here. I tried the limit definition but just got 0 over 0. The h's don't seem to cancel out.

 

d.Are ∂g/∂x  and ∂g/∂y continuous at (0,0)? Explain briefly.

I'm guessing here, I should just show that the limit is equal to the actual value. But, I couldn't do part c, which probably would help me answer part d.

 

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

Um.. please explain this to me. Not sure how to go about this.

 

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

I'm guessing this is the accumulation of all parts, but I can't do the majority of them, so yeah.

 

Please help, and thank you!

Answer 1

a. Yes. To see this, note that any polynomial function is continuous and so is f(x) = x^1/3. Now, use the fact that the composition of continuous functions is continuous to get that g(x,y) = (xy)^1/3 is everywhere continuous.

 

b. Your answer is right: Or we can write it as ∂g/∂x = y^1/3/(3x^2/3) and ∂g/∂x = x^1/3/(3y^2/3)

 

c. They are just directional derivatives in the x and y directions respectively, and so since g(x,0)=g(0,y)=0 identically, we have g_x(0,0)=g_y(0,0)=0.

 

d. No. They are only continuous in the restricted domains, x-axis for g_x(x,0) and y-axis for g_y(0,y).

 

e,f. No. along the line y=mx, we have g(x,mx)=m^1/3x^2/3 with infinite slope at x=0, so directional derivatives only exist on the positive and negative x and y directions. But the formula D_ug(x,y) = u·∇g(x,y) predicts them to exist with value 0 in all directions. Thus g(x,y) is not differentiable at (0,0) and there is no tangent plane at that point.