We can use differentials (which is the same as linear
approximation) to estimate some complicated functions.
Example 1:
Estimate $\sqrt{4.036}$ without a calculator.
Solution: We are given $x$ and $f$ and we choose
$a$. We need a value of $a$ for
which we can compute $f$ and $f'$, and which is close to our
given $x$. We are looking at the function
$f(x)=\sqrt{x}$, $x=4.036$ and we choose $a=4$.
Notice that $f'(x)=\displaystyle{\frac{1}{2\sqrt{x}}}$.
Since $f(a)=2$ and $f'(a) = \frac{1}{4}$, we can
estimate
\begin{eqnarray*}f(x)\approx L(x)&=&f(a) + f'(a) (x-a) \cr
&=& 2 + \frac{.0036}{4} \cr & = & 2.009.
\end{eqnarray*} This is an extremely accurate approximation.
DO: check this
approximation against the approximation your calculator gives.
Example 2: Estimate $e^{.03}$ without a calculator.