# Check differentiability in function

• Jan 18th 2010, 12:11 PM
Bop
Check differentiability in function
Hello, I can't find any way to prove if this funtion is or isn't differentiable in if $(x,y)=(0,0)$:

${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$ if $(x,y) \neq(0,0)$

$f(x,y)=0$ if $(x,y)=(0,0)$

Thank you!
• Jan 18th 2010, 12:12 PM
VonNemo19
Quote:

Originally Posted by Bop
Hello, I can't find any way to prove if this funtion is or isn't differentiable in if $(x,y)=(0,0)$:

${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$ if $(x,y) \neq(0,0)$

$f(x,y)=0$ if $(x,y)=(0,0)$

Thank you!

Start by taking the derivative and seeing whether or not it is defined at $(0,0)$.
• Jan 18th 2010, 12:14 PM
General
Quote:

Originally Posted by Bop
Hello, I can't find any way to prove if this funtion is or isn't differentiable in if $(x,y)=(0,0)$:

${f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}$ if $(x,y) \neq(0,0)$

$f(x,y)=0$ if $(x,y)=(0,0)$

Thank you!

Use the following formulas to find $f_x(0,0)$ and $f_y(0,0)$:
$f_x(a,b)=\lim_{h\to0} \frac{f(a+h,b)-f(a,b)}{h}$
$f_y(a,b)=\lim_{h\to0} \frac{f(a,b+h)-f(a,b)}{h}$
• Jan 18th 2010, 12:19 PM
Bop
I have calculated derivative respect x: $\dfrac{\partial f}{\partial x}=\displaystyle\frac{x^{4}+3x^{4}y^{2}}{{(x^{2}+y ^{2})}^{2}}$, respect y: $\dfrac{\partial f}{\partial y}=\displaystyle\frac{-2yx^{3}}{{(x^{2}+y^{2})}^{2}}$ but then?
I have calculate that limit of $\dfrac{\partial f}{\partial x}$ doesn't exist.. How can I continue?