Given $$g(x,y)=\bigg\{\frac{x^2y^4}{(x^2+y^2)} \text{ if }(x,y)\neq(0,0)$$ $$g(0,0)=0$$

My attempt: $\frac{\partial g}{\partial x}=\frac{2xy^6}{(x^2+y^2)^2},\frac{\partial g}{\partial y}=\frac{x^2(4x^2y^3+2y^5)}{(x^2+y^2)^2}$ in the neighborhood around $(0,0)$ $$\therefore \frac{\partial g}{\partial x}(0,0)=\lim_{t\rightarrow 0} \frac{f(t,0)-f(0,0)}{t}=\frac{t^2(0)}{t^4}\cdot\frac{1}{t}=0$$ $$\therefore \frac{\partial g}{\partial y}(0,0)=\lim_{t\rightarrow 0} \frac{f(0,t)-f(0,0)}{t}=\frac{0(0+2t^5)}{t^4}\cdot\frac{1}{t}=0$$ Since $g_x,g_y$ is continuous.


You have to prove that $\frac {\partial g} {\partial x} (x,y) \to 0$ as $(x,y) \to 0$ and $\frac {\partial g} {\partial y} (x,y) \to 0$ as $(x,y) \to 0$. For this use the following: $2|x||y| \leq x^{2}+y^{2}$ and $y^{2} \leq x^{2}+y^{2}$; from these we get $|\frac {\partial g} {\partial x} (x,y)| \leq y^{4} \to 0$. Similarly, use $x^{2} \leq x^{2}+y^{2}$ and $y^{2} \leq x^{2}+y^{2}$ for the partial derivative w.r.t. $y$.

  • $\begingroup$ Does this only show continuously differentiable at $(0,0)$? $\endgroup$ – Dillain Smith Feb 27 at 2:19
  • 1
    $\begingroup$ @DillainSmith Continuity of the partial derivative at other points is obvious. $\endgroup$ – Kavi Rama Murthy Feb 27 at 5:03

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