Why is $\sqrt{1 + x^2}$ approximately equal to $1 + \frac{x^2}{2}$? I saw this in Shankar’s Physics book and couldn’t make out the reasoning behind it. I would assume the dx and derivative have nothing to do with it.
https://i.imgur.com/iyTxPwW.jpg
 A: Notice that if $f(x) = \sqrt{1+x}$ , then $f'(x) = \dfrac{1}{2 \sqrt{1+x}}$. Moreover, the equation of the tangent line of $f(x)$ at $x=0$ is given by 
$$ y- f(0) = f'(0)(x-0) \implies y = 1 + \frac{1}{2}x $$
Therefore,  for values of $x \approx 0$, we have 
$$ \sqrt{1+x} \approx 1 +\frac{1}{2} x $$
in particular, if we replace $x$ with $x^2$, one has 
$$ \boxed{ \sqrt{1+x^2} \approx 1 +\frac{1}{2} x^2  }$$
A: $\sqrt{1+x} \approx 1+x/2$ is the linear approximation of $\sqrt{1+x}$ near $x=0$, which is based on the fact that $\left. \frac{d}{dx} \sqrt{1+x} \right |_{x=0}=1/2$. 
A: Source: How to find square roots
Here $x$ is close to $0$.
Imagine you have a square of area $1+x^2$. The first estimate of the side length is about $1$, sounds reasonable?
But then that's an underestimation of that side length, because the estimated $1\times1$ square $\square$ is missing an L-shaped region $\blacksquare$, whose area is $x^2$!
$$\left.\begin{array}{ccccccc}
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare\\
\end{array}\quad\right\}\sqrt{1+x^2}=1+\ldots?$$
Then the next estimate is to pretend that the L-shaped region is just two $1\times \frac{x^2}2$ rectangles. This give an over-estimated side length of the square
$$l = \sqrt{1+x^2} \approx 1+\frac{x^2}2$$
