Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 30

Prove in terms of Linear Approximations or differentials, why the approximation of $(1.0)1^6 \approx 1.06 $ is reasonable.
Assume that $f(x) = \sec x$, by using linear approximation
$L(x) = f(a) + f'(a)(x-a)$, for $x$ close enough to $a$
Since $x = 1.01$, we let $a = 1$

So we have,

$
\begin{equation}
\begin{aligned}
f(a) = f(1) &= (1)6\\
\\
f(1) &= 1\\
\\
f'(a) = f'(1) &= \frac{d}{dx}(x^6)\\
\\
f'(1) &= 6x^5\\
\\
f'(1) &= 6(1)^5\\
\\
f'(1) &= 6
\end{aligned}
\end{equation}
$

Thus,


$
\begin{equation}
\begin{aligned}
L(x) &= 1+ 6 (x-1)\\
\\
L(x) &= 1 + 6x - 6\\
\\
L(x) &= 6x - 5\\
\\
L(1.01) &= 6(1.01) - 5\\
\\
L(1.01) &= 6.06 -5 \\
\\
L(1.01) &= 1.06
\end{aligned}
\end{equation}
$


Therefore,

$(1.01)^6 \approx 1.06$