Differentiability - Real Analysis I'm trying to show that $xy^2$ is differentiable everywhere.
I've tried doing it from the definition but the algebra gets a little too complicated for me. If anyone could help that would be great, thanks
 A: Let $(x,y)=(a+h, b+k)$ and $(x_0,y_0)=(a,b)$.
Then we must show that $\displaystyle\lim_{(h,k)\rightarrow(0,0)}\frac{f(a+h,b+k)-f(a,b)-f_{x}(a,b)h-f_{y}(a,b)k}{\sqrt{h^2+k^2}}=0$, or that
$\displaystyle\lim_{(h,k)\rightarrow(0,0)}\frac{(a+h)(b+k)^2-ab^2-b^2h-2abk}{\sqrt{h^2+k^2}}=\lim_{(h,k)\rightarrow(0,0)}\frac{ak^2+2bkh+hk^2}{\sqrt{h^2+k^2}}=0$.
Letting $h=r\cos\theta$ and $k=r\sin\theta$, we get that
$\displaystyle\lim_{(h,k)\rightarrow(0,0)}\frac{ak^2+2bkh+hk^2}{\sqrt{h^2+k^2}}=a\lim_{r\to0}r\sin^{2}\theta+2b\lim_{r\to0}r\sin\theta\cos\theta+\lim_{r\to0}r^2\cos\theta\sin^{2}\theta=0$.
A: Consider a 2-D function $f(x,y)$. $f(x,y)$ is differentiable at $(x_0,y_0)$ if there is a unique plane tangent $z$ to the surface at that point. In that case if the tangent plane exists its equation is, $$z=f(x_0,y_0)+(x-x_0)f_x(x_0,y_0)+(y-y_0)f_y(x_0,y_0)$$
$f_x,f_y$, partial derivative of $f$ with respect to $x$ and $y$.
Here, $f(x,y)= xy^2$, consider any point $(x_0,y_0)$ then,
$$z=x_0{y_0}^2+(x-x_0){y_0}^2+(y-y_0)2x_0y_0={y_0}^2x+2x_0y_0y-2x_0{y_0}^2=ax+by+c$$
where $a={y_0}^2, b=2x_0y_0, c=-2x_0{y_0}^2$.
Thus here for any $(x_0,y_0)$ a unique tangent plane $z$ exists to the surface of $f$ at $(x_0,y_0)$. Hence $f$ is differentiable everywhere.
