- #1

RiotRick

- 42

- 0

## Homework Statement

Examine if the function is differentiable in (0,0)##\in \mathbb{R}^2##? If yes, calculate the differential Df(0,0).

##f(x,y) = x + y## if x > 0 and ##f(x,y) =x+e^{-x^2}*y## if ##x \leq 0 ## (it's one function)

## Homework Equations

##lim_{h \rightarrow 0} \frac{||f(x_0+h)-f(x_0)-J(h)||}{||h||}=0##

## The Attempt at a Solution

I'm not familiar with two subfunctions in one function.

So my plan is: I show the definition for both subfunctions if x> and <0 and then i show that both are equal at 0.

I calculated the partial derivative for ##x \leq 0##:

##\frac{f(0+h,0)-f(0,0)}{h} = \frac{h+e^{-h^2}*0}{h}= 1##

##\frac{f(0,0+h)-f(0,0)}{h} = \frac{0+e^{0}*h}{h}= 1##

so J = (1,1)

##\frac{f(0+x,0+y)-f(0,0)-(1,1)*(x,y)^T}{\sqrt{x^2+y^2}}=\frac{x+e^{-x^2}*y-0-(x+y)}{\sqrt{x^2+y^2}}##

now if x is less 0 we can simplify to:

##\frac{y*e^{x^2}-y)}{\sqrt{x^2+y^2}}##

##\frac{r*sin(\theta)*e^{(r*cos(\theta))^2}-r*sin(\theta))}{r}##

##\frac{sin(\theta)*e^{(r*cos(\theta))^2}-sin(\theta))}{1}##

if r approaches 0 then

##\frac{sin(\theta)*e^{0}-sin(\theta))}{1}=\frac{sin(\theta)*1-sin(\theta))}{1}=\frac{0}{1}=0##

Does this work so far? Is my idea okay and can I proceed this way for x>0?