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Suppose $f:R^2\to R$ is differentiable and $F(x)=f(x,-x)$. I have tried to compute the derivative through 2 methods and had a sign problem. Could someone please point out where I messed up? The derivative is supposed to be unique!

Method 1: Chain rule:

${dg\over dx}={\partial f\over\partial x}{d x\over d x}+{\partial f\over\partial y}{d y\over d x}={\partial f\over\partial x}-{\partial f\over\partial y}$

Method 2: Definition:

$f(x+t,y+t)-f(x,y)-tL(x,y)$ $=f(x+t,y+t)-f(x+t,y)+f(x+t,y)-f(x,y)-tL(x,y)$ $=t{\partial f\over\partial y}+t{\partial f\over\partial x}+O(t^2)-tL(x,y)$

So it seems like I should take $L(x,y)={\partial f\over\partial y}+ {\partial f\over\partial x}$

Why is there a sign error?

Thanks

2 Answers 2

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Method two is incorrect. When you perturb $x$ by a $+t$ you perturb $y$ by a $-t$ in effect. I.e. you should have looked at the expression

$f(x+t,y(x+t)).$

Observe $y(x+t)=-(x+t)=(-x)-t=y(x)-t$.

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    $A$ha! Very well-spotted, thanks!2011-11-23
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Method $1$ is good. In method 2, just notice that F'(x) = \lim_{t \to 0} \frac{F(x+t) - F(x)}t = \lim_{t \to 0} \frac{f(x+t,-x-t) - f(x,-x)}t and the same steps should apply.

Hope that helps,

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    @lucy : No problem =) And I see that you're new to MSE, so welcome! Have fun and feel free to ask anything you want here. ^_^2011-11-23