Q.Find the approximate value of $\sqrt{48.96}$.

Approximation Using Differentials

A small change in a function is estimated from its derivative. If $y=f(x)$ and $x$

changes by a small amount $\Delta x$, then

$\Delta y\approx dy=f'(x)\,\Delta x,$ so

$f(x+\Delta x)\approx f(x)+f'(x)\,\Delta x.$

This is the linear (first-order) approximation of $f$ near $x$.

To approximate a value such as $\sqrt[3]{345}$, choose a nearby point where $f$ is

known exactly: take $f(x)=x^{1/3}$, $x=343$ (since $7^3=343$), $\Delta x=2$. Then

$f'(x)=\dfrac1{3x^{2/3}}=\dfrac1{3\cdot49}$, giving

$\sqrt[3]{345}\approx7+\dfrac{2}{147}\approx7.013$. The same idea estimates

$\sqrt{\,},\ \sin,\ \log$ and error propagation. Accuracy is best when $\Delta x$ is

small relative to $x$. …