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 November 14th, 2011, 11:40 AM #1 Newbie   Joined: Oct 2011 Posts: 14 Thanks: 0 differentiable function? Let $f$ be a function on $A=\{ (x,y) \in \mathbb{R}^n: xy>-1 \}=$ $f(x,y)=\frac{\sqrt{1+xy}-1}{y}$ and $f(x,y)=\frac{x}{2}$ when $y=0$. Is $f$ differentiable at $(0,0)$? Is the differential of function is continuous at this point? Is this function differentiable at $(1,0)$? Can anyone give me a hand?
November 14th, 2011, 12:37 PM   #2
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Re: differentiable function?

Quote:
 Originally Posted by Camille91 Let $f$ be a function on $A=\{ (x,y) \in \mathbb{R}^n: xy>-1 \}=$ $f(x,y)=\frac{\sqrt{1+xy}-1}{y}$ and $f(x,y)=\frac{x}{2}$ when $y=0$. Is $f$ differentiable at $(0,0)$? Is the differential of function is continuous at this point? Is this function differentiable at $(1,0)$? Can anyone give me a hand?
You should clarify what you mean by differentiable - y only or both x and y.

For x = 1 and y near 0, f(x,y) ? 1/(2?y), which is not differentiable.
For x and y near 0, f(x,y) ? {?(x/y)}/2

 November 17th, 2011, 08:22 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: differentiable function? Mathman, "differentiable" is defined for functions of two variables, separately from the existance of the partial derivatives. It can be shown, however, that if a function is differentiable at a point, its partial derivatives must be continuous in a neighborhood of that point.

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