# Continuous partial derivatives

I have the following function and I want to show that it is differentiable. I am going to do this by showing that the partial derivatives are continuous and so I will show that they are continuous at (0,0). So, i am going to show that as the limit of (x,y) approaches (0,0) the derivative approaches 0.

$$f(x,y)=\left\{\begin{array}{l} \frac{x^2y^2}{\sqrt{x^2+y^2}},\:\text{if (x,y) \not= (0,0);}\\ 0,\:\text{if (x,y)=(0,0);} \end{array}\right.$$

$$\begin{cases} \dfrac {\partial f}{\partial x}(x,y)= \dfrac{x^3y^2+2xy^4}{({x^2+y^2})^\frac{3}{2}}\\ \dfrac {\partial f}{\partial x}(0,0)=0\end{cases}$$

I am having trouble however and was wondering if anyone could work me through this case.

• Where on Earth did you get the cube root in the denominator on the RHS in the formula for the derivative? Jun 10, 2014 at 11:09
• Yeah, that's meant to be to the power 3/2, ill fix it now, thanks!
– J-C
Jun 10, 2014 at 11:13

We have

$$\left|\dfrac{x^3y^2+2xy^4}{{(x^2+y^2)^{3/2}}}\right|\le \dfrac{|x|(x^2+y^2)^2+2|x|(x^2+y^2)^2}{(x^2+y^2)^{3/2}}=3|x|(x^2+y^2)^{1/2}\xrightarrow{(x,y)\to(0,0)}0$$ so $\dfrac {\partial f}{\partial x}(x,y)$ is continuous at $(0,0)$

• This term $$\dfrac{|x|^3(x^2+y^2)+2|x|y^2(x^2+y^2)}{\sqrt[3]{x^2+y^2}}(|x|^3+2|x|y^2)(x^2+y^2)^{2/3}$$ I understand how you can get the first fraction but am not to sure how you are able to multiply it by the following terms?
– J-C
Jun 10, 2014 at 11:09
• My apologies, i posted an incorrect equation
– J-C
Jun 10, 2014 at 11:23

We have $|x|\le\sqrt{x^2+y^2}$ and $|y|\le\sqrt{x^2+y^2}$. Thus, the absolute value of the numerator does not exceed $3(x^2+y^2)^{5/2}$ and the fraction does not exceed $3(x^2+y^2)$ in absolute value and hence tends to zero as $(x,y)\to(0,0)$

• Thankyou! Finally it makes sense!
– J-C
Jun 10, 2014 at 11:22