0
$\begingroup$

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.

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

2 Answers 2

Reset to default
1
$\begingroup$

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 $(0,0)$

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – J-C
    Jun 10, 2014 at 11:09
  • $\begingroup$ My apologies, i posted an incorrect equation $\endgroup$
    – J-C
    Jun 10, 2014 at 11:23
0
$\begingroup$

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 $(x,y)\to(0,0)$

$\endgroup$
1
  • $\begingroup$ Thankyou! Finally it makes sense! $\endgroup$
    – J-C
    Jun 10, 2014 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.