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As I was going through differentiable functions in my notes on multivariable calculus while preparing for exam in a few days, it states that a function $f:A \rightarrow Y$ is said to be differentiable at $a \in A$ if there is a linear map $T \in L(X,Y) ,$ such that lim$_{r \to 0}\frac{\|f(a+r)-f(a)-T\text{r}\|}{\|\text{r}\|}=0$.

It is called the derivative of $f$ at $a \in A$.

There is a note behind the definition which states that $f'(a)$ is a linear operator.

It is still not clear to me what does $f'$ maps each small value around $a$ to?

coool
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1 Answers1

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I'm not sure you understand the whole thing. Here $f'(a)$ is just the name of the unique continuous linear operator such that etc. You could just write $L_a$. In general, $f'(a)$ is not a function of $a$, in the sense that you believe. For a generic $f$, you may have differentiability at $a$ but not at any point near $a$ itself.

On the other hand, if $f$ is differentiable in a whole neighborhood $U$ of $a$, then $f'$ maps points of this neighborhood to elements of $L(X,Y)$. Hence $$f' \colon U \to L(X,Y).$$

Siminore
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  • Your answer has almost clarity .But 1 thing I want to know is how $f′$ maps points of $U$ neighborhood to elements of $L(X,Y)$.Is there some proof for this.Please help me with this... – coool Sep 25 '14 at 09:35
  • The derivative is a map from $U$ to $L(X,Y)$ by definition! – Siminore Sep 25 '14 at 09:36