Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 28

Estimate $\sqrt{99.8}$ by using linear approximation or differentials
Using Linear Approximation
$L(x) = f(a) + f'(a)(x-a)$
Lt $f(x) = \sqrt{x}$, $x = a = 100$



$
\begin{equation}
\begin{aligned}
f'(a) = f'(100) &= \frac{d}{dx}(x)^{\frac{1}{2}}\\
\\
f'(100) &= \frac{1}{2} (x)^{\frac{-1}{2}}\\
\\
f'(100) &= \frac{1}{2(x)^{\frac{1}{2}}}\\
\\
f'(100) &= \frac{1}{2\sqrt{100}}\\
\\
f'(100) &= \frac{1}{2(10)}\\
\\
f'(100) &= \frac{1}{20}
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
L(x) &= 10 + \frac{1}{20} ( x - 100 )\\
\\
L(x) &= 10 + \frac{x-100}{20}\\
\\
L(x) &= \frac{200+x-100}{20}\\
\\
L(x) &= \frac{x+100}{20}\\
\\
L(x) &= \frac{99.8+100}{20}\\
\\
L(x) &= \frac{199.8}{20}\\
\\
L(x) &= 9.99
\end{aligned}
\end{equation}
$

So,
$\sqrt{99.8} \approx 9.99$