Does anyone have any idea on how can I prove if this partial derivative is continuous at the origin:

f(x,y)=xy(x2+2y2)(x2+y2)3/2f(x,y) = \frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}

Of course I know the definition: for every ϵ\epsilon there is a δ\delta if ||(x,y)|| ||(x,y)|| is less that δ\delta than |f(x,y)||f(x,y)| is less than ϵ\epsilon.77

I’m finding difficult do work with |f(x,y)||f(x,y)|

I already tried |f(x,y|=|xy(x2+2y2)(x2+y2)3/2|≤|x(x2+2y2)(x2+y2)1/2||f(x,y| = |\frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}| \leq |\frac{x(x^2 + 2y^2)}{(x^2+y^2)^{1/2}}| because |y(x2+y2)|≤1|\frac{y}{(x^2+y^2)}|\leq1

But I’m know kind of stucked in there. Can someone help me or give a hint on how to proceed? Thanks!

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1

How is ff defined at (0,0)(0,0)?

– Dr. MV

2 days ago

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1 Answer

1

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HINT:

Transforming to polar coordinates (ρ,ϕ)(\rho,\phi), we have

f(x,y)=xy(x2+2y2)(x2+y2)3/2=ρ4sin(ϕ)cos(ϕ)(cos2(ϕ)+2sin2(ϕ))ρ3\begin{align}

f(x,y)&=\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\\\\

&=\frac{\rho^4\sin(\phi)\cos(\phi)(\cos^2(\phi)+2\sin^2(\phi))}{\rho^3}

\end{align}

Then, the limit as ρ→0\rho\to0 is trivially 00. If f(0,0)=0f(0,0)=0, then ff is continuous at the origin.

If one wishes to proceed without a coordinate transformation, simply note that

|xy(x2+2y2)(x2+y2)3/2|≤|2xy(x2+y2)(x2+y2)3/2|=|2xy√x2+y2|≤x2+y2√x2+y2\begin{align}

\left|\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\right|&\le \left|\frac{2xy(x^2+y^2)}{(x^2+y^2)^{3/2}}\right|\\\\

&=\left|\frac{2xy}{\sqrt{x^2+y^2}}\right| \\\\

&\le \frac{x^2+y^2}{\sqrt{x^2+y^2}}

\end{align}