@Mann great !!

Got it and thank you :)

I am using this defintion of differentiability, $ \lim_{ \Delta \rho \to 0} \cfrac{ \Delta z - dx }{\Delta \rho} = 0 $

Just for the sake of reference here is the function, $$f (x,y) \begin{cases} (x^2 + y^2) \arctan(y/x) & \text{if x $\ne$ 0} \\ y^2 \cfrac{\pi}{2} & \text{if x = 0} \end{cases} $$

dx is what?

$dz = z_x \Delta x + z_y \Delta y$

Sorry, that's $dz$

The problem comes in finding $z_x$.

The definition is to check differentiability at (0,0) only right?

Let me formulate the problem in the way I know

Okay, we can discuss about the applicabiility later, because problem comes in finding the partial derivative with respect to x

Why is that?

$$z_x = \lim_{h \to 0} \cfrac{ f(h,y) - f(0,y)}{h} = \lim_{h \to 0} \cfrac{ (h^2 + y^2) \arctan(y/h) - y^2 \cfrac{\pi}{2} }{h} $$

depending on the sign of $y$ , $\arctan(y/h)$ can be $\pm \pi /2$

Even assuming that h > 0

but in answer it is stated that the function is differentiable.

But does the limit exists for different cases of y>0 , y<0 and y=0?

It seems like there will be an issue

issue regarding what?

h can approach from both positive and negative side of 0 for a given y

Ahh but

We have to check differentiabilty at only (0,0)

Yeah, there wil be two possible values for all different sign combination.

@Mann :(

What?

We don't care whether the partial derivative exist or not I think

Didn't saw that!

Ahhh xD

So its okay now?

Thanks!

Try changing to polar coordinates

@Mann yeah, boils down to epsilon-delta stuffs, fine!

It may be better

Yeah, did that earlier.

Okayy

$2 \Delta T * sin \frac { \delta \theta }{2} = \Delta T * \delta \theta$

Can someone justify this? (It is actually approximately equal to, Not exactly equal to)

( $ \delta \theta $ is obviously, small). Also note that $ \Delta T$ is some constant.

@McSuperbX1 limit x tends to 0 sinx/x= 1

correct..

Hello!

Every convergent sequence is bounded right?

Then what about $\dfrac{1}{1-n}$