- #1
_Steve_
- 19
- 0
Hey guys, I'm just wondering if I got this question right:
Discuss where the following in [itex]R^{2}[/itex] is differentiable:
[itex]f(x,y)=(x^{3}+y^{3})^{2/3}[/itex]
So I take the partial derivative:
[itex]f_{x}(x,y)=\frac{2x^{2}}{(x^{3}+y^{3})^{1/3}}[/itex]
and see that f(x,y) might not be differentiable at (0,0), so I used the definition of a partial derivative to see if the partial derivatives exist at (0,0):
[itex]f_{x}(0,0)=lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=0[/itex]
I thought this was the end of the question, since it shows that the partial derivative at (0,0) does exist, and thus it's continuous for all (x,y) (showing that f(x,y) is differentiable)
But under the question there's a hint, it says:
Hint: Verify the handy inequality [itex]|x^{3}+y^{3}| \leq 2(x^{2}+y^{2})^{3/2}[/itex]
Did I miss something?
Discuss where the following in [itex]R^{2}[/itex] is differentiable:
[itex]f(x,y)=(x^{3}+y^{3})^{2/3}[/itex]
So I take the partial derivative:
[itex]f_{x}(x,y)=\frac{2x^{2}}{(x^{3}+y^{3})^{1/3}}[/itex]
and see that f(x,y) might not be differentiable at (0,0), so I used the definition of a partial derivative to see if the partial derivatives exist at (0,0):
[itex]f_{x}(0,0)=lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=0[/itex]
I thought this was the end of the question, since it shows that the partial derivative at (0,0) does exist, and thus it's continuous for all (x,y) (showing that f(x,y) is differentiable)
But under the question there's a hint, it says:
Hint: Verify the handy inequality [itex]|x^{3}+y^{3}| \leq 2(x^{2}+y^{2})^{3/2}[/itex]
Did I miss something?