# Showing $g:\mathbb{R}^2\rightarrow\mathbb{R}$ is continuously differentiable.

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. $$y$$.
• Does this only show continuously differentiable at $(0,0)$? – Dillain Smith Feb 27 at 2:19