doubt regarding Definition of differentiability . The definition of a differentiable function is as follows:

A function $f:A\to Y$ is said to be differentiable at a $\in A$ if there is a linear map $T\in L(X,Y), $ such that : $$\text{lim}_{~r\to 0}\frac{\|f(a+r)-f(a)-Tr\|}{\|r\|}=0$$
If such a $T:X\to Y$ exists ,then it is unique.It is denoted as derivative of $f$ at $a\in A$ and is denoted as $f'(a)$.Note that $f'(a):X\to Y$ is a linear operator and its value at $x\in X$ is written as $f'(a)(x)\in Y$

I can't understand if we map all values from  $X\to Y$ using $f'(a)$ what do we get... I know it is $f'(a)(x)\in Y$ at a particular $x\in X$..but what are we doing with this mapping..
Kindly help me..
 A: Look at the case in dimension one, so $X = Y = \mathbb{R}$. Further take $A$ the positive reals and $f : A \to Y$, $x \mapsto \log x$. 
I assume that you know what the derivative (in the old sense) is namely $f'(a)= 1/a$. This seems like something quite different as it is a real number not a linear map from $X$ to $Y$, so $\mathbb{R}$ to $\mathbb{R}$. But recall what it takes to define a linear map from $\mathbb{R}$ to $\mathbb{R}$, just one real number or $1$ by $1$ matirx if you prefer. 
So the linear map $T$ for the point $a$ is the map $x \mapsto \frac{1}{a}x$. So, $f'(a)(x)$ is $\frac{1}{a} x$.  
What does this mapping represent. It is a line through the origin parallel to the tangent of $f$ at $a$, and $x \mapsto f(a) + f'(a)(x)$ is just the equation of the tangent of $f$ at $a$. 
So the definition means in $a$ there is a tangent line (or in general space) that approximates $f$ "well."
The space $f'(a)(x)$ for $x \in X$ is the linear subspace 'parallel' to the 'tangent space.'
What you need this for, well, it depends. But what you actually might want to know is the reason why the definition is written as it is. This is simply due to the fact that  it "only" works in this way. (Of course the only is not literal, there are always different ways to do things, but what I mean is it is actually something familiar in disguise.)
Recall the onedimensional case again. 
You have 
$$
\lim_{r\to 0} \frac{f(a+r)-f(a)}{r}= f'(a)
$$
This is the same as
$$
\lim_{r\to 0} \frac{f(a+r)-f(a) - f'(a)r}{r}= 0
$$
So, you could say $f'(a)$ is the $T$, if it exists, such that 
$$
\lim_{r\to 0} \frac{f(a+r)-f(a) - Tr}{r}= 0
$$
and of course you cal also add absolute values
$$
\lim_{r\to 0} \frac{|f(a+r)-f(a) - Tr|}{|r|}= 0
$$
to get what you have now. 
However, in you general situation you cannot go "back" from this last equation since you cannot divide by $r$ only by $|r|$.
To sum this up, the definition is on the one hand a rewrite of the usual defintiion that also work in more generaility, and on the other hand it is "nothing but" the original motivation for considering differentiability; can we find a line tangent to the graph. 
