Bài tập 52 trang 221 SGK Toán 11 NC

Ta có:

\(\begin{array}{l}
df(x) = \frac{{ - 2(1 + \tan x)\frac{1}{{{{\cos }^2}x}}}}{{{{(1 + \tan x)}^4}}}.\Delta x\\
 = \frac{{ - 2\Delta x}}{{{{\cos }^2}x{{(1 + \tan x)}^3}}}
\end{array}\)

\(\begin{array}{l}
 \Rightarrow df(\pi 6) = \frac{{ - 2.\frac{\pi }{{360}}}}{{{{\cos }^2}\frac{\pi }{6}{{\left( {1 + \tan \frac{\pi }{6}} \right)}^3}}}\\
 = \frac{{ - \pi }}{{180.\frac{3}{4}{{\left( {1 + \frac{1}{{\sqrt 3 }}} \right)}^3}}} \approx  - 0,0059
\end{array}\)

-- Mod Toán 11 HỌC247