Intuition behind Modern Idea of Differential Calculus of several variables I have read that The modern idea of differential calculus of several variables is it tries to approximate increment in dependent variable by a linear map acting on independent variable.
What is the Intuition behind approximating something by a linear map?Also why the linear map has to act on increment in independent variable?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$Though calculus may appear to be a study of "variables", the modern mathematical viewpoint is really that calculus studies functions or mappings, i.e., causal relationships between variables. (Causal here means that in a functional relationship $\Vec{y} = f(\Vec{x})$, each value of $\Vec{x}$ uniquely determines a value of $\Vec{y}$. Relationship between variables refers to the fact that the derivative is a property of $f$, not of $\Vec{x}$ or $\Vec{y}$. This is not the explicit stance in all quantitative fields, and may not even be widely recognized by students of mathematics.)
In mathematics, a functional relationship $\Vec{y} = f(\Vec{x})$ is given by a formula or other deterministic specification in the input variable(s) $\Vec{x}$. In the sciences, "laws of nature" are assumed to have the character of functional relationships (e.g., the current state $\Vec{x}$ of a system uniquely determines the future state $\Vec{y}$), but all we are able to do is prepare isolated values of $\Vec{x}$, perform experiments, and measure outcomes $\Vec{y}$.
To "explore" a law of nature, a good experimenter will prepare a system depending on a finite number of numerical quantities, say $\Vec{x} = (x_{1}, x_{2}, \dots, x_{n})$, and then make a number of numerical measurements of $\Vec{y} = (y_{1}, y_{2}, \dots, y_{m})$ by (i) changing the variables one at a time (or controlling the variables) (ii) by small amounts.
The simplest systems are those for which increments have independent effects on the output. These are precisely linear (or really, affine) mappings. In more detail, suppose our "initial" system is in a state represented by the values $\Vec{x}_{0}$, and write $\Vec{y}_{0} = f(\Vec{x}_{0})$. Suppose $\Delta\Vec{x}_{1}$ and $\Delta\Vec{x}_{2}$ are "changes in the input variable", and write
\begin{align*}
  \Delta\Vec{y}_{1} &= f(\Vec{x}_{0} + \Delta\Vec{x}_{1}) - f(\Vec{x}_{0}), \\
  \Delta\Vec{y}_{2} &= f(\Vec{x}_{0} + \Delta\Vec{x}_{2}) - f(\Vec{x}_{0}).
\end{align*}
To say increments have independent effects is to say that for all real parameters $t_{1}$ and $t_{2}$,
\begin{multline*}
  f(\Vec{x}_{0} + t_{1}\, \Delta\Vec{x}_{1} + t_{2}\, \Delta\Vec{x}_{2}) - f(\Vec{x}_{0}) \\
  \begin{aligned}
  &= \underbrace{\bigl[f(\Vec{x}_{0} + t_{1}\, \Delta\Vec{x}_{1} + t_{2}\, \Delta\Vec{x}_{2}) - f(\Vec{x}_{0} + t_{2}\, \Delta\Vec{x}_{2})\bigr]}_{t_{1}\, \Delta\Vec{y}_{1}}
  + \bigl[f(\Vec{x}_{0} + t_{2}\, \Delta\Vec{x}_{2}) - f(\Vec{x}_{0})\bigr] \\
  &= t_{1}\, \Delta\Vec{y}_{1} + t_{2}\, \Delta\Vec{y}_{2}.
  \end{aligned}
\end{multline*}
(The underbraced quantity manifests independence, i.e., the increment in the output is the same regardless of where that increment is applied.)
By induction on the number of increments, the same is true for any finite number of increments. Formally, $f(\Vec{x}_{0} +\;\cdot\;) - f(\Vec{x}_{0})$ distributes over linear combinations of increments in the input.
In reality, laws of nature tend not to be so simple. Nonetheless, it is an empirical fact that many phenomena are approximately linear for small changes in the input. And that leads us to define differentiable mappings to be approximately linear.
Precisely, if $f$ is an $\Reals^{m}$-valued mapping defined in a neighborhood of $\Vec{x}_{0}$ in $\Reals^{n}$, we say $f$ is differentiable at $\Vec{x}_{0}$ if there exists a linear transformation $Df(\Vec{x}_{0}):\Reals^{n} \to \Reals^{m}$ and an $\Reals^{m}$-valued mapping $E$ such that
$$
f(\Vec{x}_{0} + \Delta\Vec{x}) = f(\Vec{x}_{0}) + Df(\Vec{x}_{0})\, \Delta\Vec{x} + E(\Delta\Vec{x}),\qquad
\lim_{\Delta\Vec{x} \to \Vec{0}} \frac{\|E(\Delta\Vec{x})\|}{\|\Delta\Vec{x}\|} = 0.
$$
