# Partial derivative question

• Oct 10th 2011, 08:05 PM
Kuma
Partial derivative question
Hi there. I have been given this problem:

I want to find the partial of y at the point (0,0) of the following.

f(x,y) = [2x^2 +3y^2]/[x^2+y^2] when (x,y) =/= (0,0)

f(x,y) = 0 when (x,y) = (0,0)

There is discontinuity at the point I am supposed to find the partial of y to, so can I say the function is not differentiable at that point?
• Oct 10th 2011, 09:27 PM
Prove It
Re: Partial derivative question
Quote:

Originally Posted by Kuma
Hi there. I have been given this problem:

I want to find the partial of y at the point (0,0) of the following.

f(x,y) = [2x^2 +3y^2]/[x^2+y^2] when (x,y) =/= (0,0)

f(x,y) = 0 when (x,y) = (0,0)

There is discontinuity at the point I am supposed to find the partial of y to, so can I say the function is not differentiable at that point?

\displaystyle f(x, y) = \begin{cases} \begin{align*}\frac{2x^2 + 3y^2}{x^2 + y^2} \textrm{ if }(x, y) &\neq (0, 0) \\ 0 \textrm{ if }(x, y) &= (0, 0)\end{align*}\end{cases}

A function is differentiable at a point if it is continuous at that point, and if its partial derivatives are continuous at that point.

So to check continuity, we need to check that the limit exists and is equal to 0 at the point (x, y) = (0, 0). The easiest way is to convert to polars.

\displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)} \frac{2x^2 + 3y^2}{x^2 + y^2} &= \lim_{r \to 0}\frac{2(r\cos{\theta})^2 + 3(r\sin{\theta})^2}{r^2} \\ &= \lim_{r \to 0}\frac{2r^2\cos^2{\theta} + 3r^2\sin^2{\theta}}{r^2} \\ &= \lim_{r \to 0}(2\cos^2{\theta} + 3\sin^2{\theta}) \\ &= \lim_{r \to 0}(2 + \sin^2{\theta}) \end{align*}

Clearly, this value will change depending on the value of $\displaystyle \theta$, so the limit does not exist at that point.

Since the function is not continuous at that point, the function is not differentiable at that point.
• Oct 10th 2011, 09:32 PM
Kuma
Re: Partial derivative question

But I haven't learnt polars to check continuity yet. Can I take a different approach? For example if i approach along the x axis the limit is is = 2, and along the y axis, the limit is = 3. Thus you can conclude the limit does not exist?
• Oct 10th 2011, 09:34 PM
Prove It
Re: Partial derivative question
Quote:

Originally Posted by Kuma

But I haven't learnt polars to check continuity yet. Can I take a different approach? For example if i approach along the x axis the limit is is = 2, and along the y axis, the limit is = 3. Thus you can conclude the limit does not exist?

Yes that's fine as well :)
• Oct 11th 2011, 07:24 AM
HallsofIvy
Re: Partial derivative question
Quote:

Originally Posted by Kuma
Hi there. I have been given this problem:

I want to find the partial of y at the point (0,0) of the following.

f(x,y) = [2x^2 +3y^2]/[x^2+y^2] when (x,y) =/= (0,0)

f(x,y) = 0 when (x,y) = (0,0)

There is discontinuity at the point I am supposed to find the partial of y to, so can I say the function is not differentiable at that point?

Yes, if a function is not continuous at a point, it is not differentiable there BUT that does not mean its partial derivatives do not exist! For example, the simple function f(x,y)= 0 if xy= 0, 0 otherwise is not continuous anywhere on the axes and in particular at the origin. But the partial derivatives at (0, 0) both exist (and are both 0). For a function of two or more variables, being "differentiable" is different from having partial derivatives.

To find the partial derivative with respect to y at (0, 0), just use the definition:
$\lim_{h\to 0}\frac{f(0,h)- f(0,0)}{h}= \iim_{h\to 0}\frac{2(0^2)+ 3h^2}{(0^2+ h^2)h}= \lim_{h\to 0}\frac{3}{h}$
which does not exist.

So the partial derivative with respect to y does not exist at (0, 0) but NOT because the function was not continuous there.