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I'm trying to understand this paragraph in a real analysis book:

Let $f:U\to \mathbb R$ be defined in an open set $U\subset \mathbb R^n$. Since $U$ is open, there exists $\epsilon \gt 0$ such that $-\epsilon \lt t\lt \epsilon\implies \lambda(t)=a+te_i\in U$ . The partial derivative of $f$ at a point $a$ is the derivative, at the point $t=0$, of the function $f\circ\lambda:(-\epsilon,\epsilon)\to\mathbb R$, i.e., $\frac{\partial f}{\partial x_i}(a)=(f\circ\lambda)'(0)$.

I didn't understand why $-\epsilon \lt t\lt \epsilon\implies \lambda(t)=a+te_i\in U$ and why $\frac{\partial f}{\partial x_i}(a)=(f\circ\lambda)'(0)$.

Thanks

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  • $\begingroup$ Because the set $U$ is open, so there is an ball of radius $\epsilon$ centered at $a$ contained in $U$. $\endgroup$ – Gregory Grant May 6 '15 at 22:18
  • $\begingroup$ What he's trying to do is start at a point $a\in U$ and move in a straight line in one of the standard directions, but he wants to make sure he stays inside $U$ so that the function is defined. $\endgroup$ – Gregory Grant May 6 '15 at 22:20
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    $\begingroup$ He's ultimately trying to say that a partial derivative is the same as just moving in a line in one of the standard directions and taking the one-dimensional derivative. He's trying to give you intuition about the meaning of the partial derivative. $\endgroup$ – Gregory Grant May 6 '15 at 22:23
  • $\begingroup$ @WillJagy He didn't really use the chain rule, he's just equating the partial derivative with the derivative of a composition of functions. The chain rule would say $(f\circ\lambda)'(0)=f'(\lambda(0))\lambda'(0)$. $\endgroup$ – Gregory Grant May 6 '15 at 22:25
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There exists an open ball $B(a,\epsilon)\subset U$. Then,

$$-\epsilon<t<\epsilon\implies \|a-(a+te_i)\|=|t|\|e_i\|<\epsilon$$

and

$$\frac{\partial f}{\partial x_i}(a)=\lim_{t\to 0}\frac{f(a+te_i)-f(a)}t=\lim_{t\to0}\frac{f(\lambda(t))-f(\lambda(0))}t=(f\circ\lambda)'(0)$$

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