Bài tập 5.8 trang 199 SBT Toán 11

Ta có:

\(\begin{array}{l}
{\rm{\Delta }}y = \sin 2\left( {x + {\rm{\Delta }}x} \right) - \sin 2x\\
 = \sin \left( {2x + 2{\rm{\Delta }}x} \right) - \sin 2x\\
 = 2\cos \left( {2x + {\rm{\Delta }}x} \right)\sin {\rm{\Delta }}x
\end{array}\)

Suy ra \(\frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}} = \frac{{2\cos \left( {2x + {\rm{\Delta }}x} \right)\sin {\rm{\Delta }}x}}{{{\rm{\Delta }}x}}\)

Vậy \(\frac{{{\rm{\Delta }}y}}{{{\rm{\Delta }}x}}\) tại \(x = \frac{\pi }{4}\) là \(\frac{{2\cos \left( {{\rm{\Delta }}x + \frac{\pi }{2}} \right)\sin {\rm{\Delta }}x}}{{{\rm{\Delta }}x}} =  - \frac{{2\cos \left( {\frac{\pi }{2} - {\rm{\Delta }}x} \right)\sin {\rm{\Delta }}x}}{{{\rm{\Delta }}x}} = \frac{{ - 2{{\sin }^2}{\rm{\Delta }}x}}{{{\rm{\Delta }}x}}\)

Chọn A.

-- Mod Toán 11 HỌC247