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$\bf 9.18\ \ $ Example $\ \ $ Let $\gamma$ be a differentiable mapping of the segment $(a,b)\subset {\Bbb R}^1$ into an open set $E\subset {\Bbb R}^n$, in other words, $\gamma$ is a differentiable curve in $E$. Let $f$ be a real-valued differentiable function with domain $E$. Thus $f$ is a differentiable mapping of $E$ into ${\Bbb R}^1$. Define $$\tag{31} g(t)=f(\gamma(t))\quad(a<t<b).$$ $\qquad$ The chain rule asserts then that $$\tag{32} g'(t)=f'(\gamma(t))\gamma'(t)\quad(a<t<b).$$ Since $\gamma'(t)\in L(\mathbb R^1,{\Bbb R}^n)$ and $\color{red}{\underline{\color{black}{f'(\gamma(t))\in L(\mathbb{R}^n,{\Bbb R}^1)}}}$, $(32)$ defines $g'(t)$ as a linear operator on ${\bf R}^1$. This agrees with the fact that $g$ maps $(a,b)$ into ${\Bbb R}^1$.

We know that $\gamma: (a,b)\to E\subset \mathbb{R}^n$ and $f:E\to \mathbb{R}^1$. Hence $f'(\gamma(t))\in L(\mathbb{R}^1, \mathbb{R}^1)$ since to any point $t\in(a,b)$ it corresponds some real number.

Why Rudin regarded it as $f'(\gamma(t))\in L(\mathbb{R}^n, \mathbb{R}^1)$?

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$t$ is fixed there, so set $\gamma(t)=b$. Thus he's referring to $f'(b)$, which is a linear map from $\Bbb R^n$ to $\Bbb R$.

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  • $\begingroup$ I guees that $f'(\gamma(t))$ is function from $(a,b)$ to $\mathbb{R}^1$. But I can't understand why I am wrong $\endgroup$ – ZFR Jan 24 '16 at 7:42
  • $\begingroup$ No, for each $t$ you have a linear map from $\Bbb R^n$ to $\Bbb R$. You need to review basic composition of functions. $f'(\gamma(t))$ is NOT $(f\circ\gamma)’(t)$. $\endgroup$ – Ted Shifrin Jan 24 '16 at 7:50
  • $\begingroup$ But $f'(\gamma(t))$ is composition of $f'$ and $\gamma$. Right? $\endgroup$ – ZFR Jan 24 '16 at 7:55
  • $\begingroup$ Yes, so you get exactly what we've been telling you. Reread my answer! $\endgroup$ – Ted Shifrin Jan 24 '16 at 7:57
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    $\begingroup$ You're welcome. Sometimes mental blocks are annoying. :) $\endgroup$ – Ted Shifrin Jan 24 '16 at 8:09
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Since $f : E \to \mathbb R$ and $E$ is open in $\mathbb R^n$, we have for each $y\in E$, $f'(y) \in L(\mathbb R^n, \mathbb R)$. In particular, at $y = \gamma(t)$ we have $f'(\gamma(t)) \in L(\mathbb R^n, \mathbb R)$.

You might be thinking of $(f\circ \gamma)' (t)$ instead.

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  • $\begingroup$ Dear John! I understand your post but what's wrong in my reasoning? $\endgroup$ – ZFR Jan 24 '16 at 7:35
  • $\begingroup$ $f'(\gamma(t))$ is function from $\mathbb{R}$ to $\mathbb{R}$. $\endgroup$ – ZFR Jan 24 '16 at 7:36
  • $\begingroup$ "to any point $t\in(a,b)$ it corresponds some real number". That makes it a function from $(a, b) \to \mathbb R$. That function cannot be $f$. ($t\mapsto f'(\gamma(t))$ is not a function from $\mathbb R$ to $\mathbb R$, it's image is not $\mathbb R$)@RFZ $\endgroup$ – user99914 Jan 24 '16 at 7:38
  • $\begingroup$ Sorry for stupid question! But why it's image is not $\mathbb{R}$? $\endgroup$ – ZFR Jan 24 '16 at 7:40
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    $\begingroup$ $f'(\gamma(t))$ it's composition of two functions, namely $f'\circ \gamma$ where $\gamma:(a,b)\to \mathbb{R}^n$ and $f'\in L(\mathbb{R}^n,\mathbb{R})$. Hence $f'\circ \gamma$ it's from $\mathbb{R}$ to $\mathbb{R}$. I can't understand my mistake :( $\endgroup$ – ZFR Jan 24 '16 at 7:47
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The derivative of $f$ at the point $\gamma(t)$ is a linear map from $\mathbb{R}^n$ to $\mathbb{R}$. This is a linear approximation of $f$ near that point, which you can visualize as a hyperplane in $\mathbb{R}^{n+1}$ that is tangent to the graph of $f$. In multivariable calculus, the gradient of $f$ is essentially the transpose of this linear map.

Read about the Fréchet derivative for a more abstract version of this idea.

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