For f(x,y)=(xy)1/3, show the existence of directional derivatives on (0,0)

[ezqmarquez]

Can you solve this?

If f(x,y)=(xy)^1/3. Show the existence of directional derivatives on (0,0)

Thanks

 

[HallsofIvy]

Consider the "direction" making angle \(\displaystyle \theta\) with the x-axis and let \(\displaystyle m= tan(\theta)\). A line through (0, 0) in that direction is of the form y= mx. On that line \(\displaystyle f(x,y)= (xy)^{1/3}= x^{2/3}m^{1/3}\). The derivative in that direction is \(\displaystyle f'= (2/3)x^{1/3}m^{1/3}\) which is 0 at (0, 0).

That works for all \(\displaystyle \theta\) except \(\displaystyle \theta= 90\) degrees where m does not exist. In that case the line is just x= 0 so \(\displaystyle f(0,y)= 0^{1/3}= 0\), a constant, so that the derivative is again 0.