Is f(x,y) Differentiable at (0,0)?

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SUMMARY

The function f(x,y) = x³/(x²+y²) for (x,y) ≠ (0,0) and f(0,0) = 0 is not differentiable at the point (0,0). The evaluation of the partial derivative with respect to y yields -2x³y/(x²+y²)², which indicates that the limit does not approach a single value as (x,y) approaches (0,0) from different directions. Specifically, substituting y = mx leads to a limit that depends on the slope m, confirming that the partial derivative does not exist at (0,0). Thus, f(x,y) is concluded to have no derivative at this point.

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Barioth
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Hi,

I'm having trouble évaluation the differentiability at (0,0) of the function

$$f(x,y)=\frac{x^3}{x^2+y^2}$$ for (x,y) not nul, and $$f(x,y)=0$$ if (x,y)=0

I know f is differentiable if (x,y) isn't nul since the partial derivative are continuous, but I don't know how to evaluate it at (0,0)

Thanks for passing by
 
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Barioth said:
Hi,

I'm having trouble évaluation the differentiability at (0,0) of the function

$$f(x,y)=\frac{x^3}{x^2+y^2}$$ for (x,y) not nul, and $$f(x,y)=0$$ if (x,y)=0

I know f is differentiable if (x,y) isn't nul since the partial derivative are continuous, but I don't know how to evaluate it at (0,0)

Thanks for passing by

Let's take the partial derivative with respect to y:
$$\frac{\partial}{\partial y}f(x,y) = - \frac{2x^3 y}{(x^2+y^2)^2}$$

Is it the same in all directions when approaching (0,0)?
You may want to switch to polar coordinates.
 
I like Serena said:
$$\frac{\partial}{\partial y}f(x,y) = - \frac{2x^3 y}{(x^2+y^2)^2}$$

tacking y=m*x we end up with

$$\frac{-2*m*x^4}{(m^2+1)^2*x^4}= \frac{-2*m}{(m²+1)²}$$

This telling me that the limite when (x,y)->(0,0) isn't contiunous.

There for the partial derivative doesn't exist at 0. I can conclude that f(x,y) has no derivative at (0,0).

Thank you for the help!
 

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