Đáp án:
\[y' = \frac{{ - 4}}{{{{\sin }^2}\left( {1 - 2x} \right)}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {\tan x} \right)' = \frac{1}{{{{\cos }^2}x}}\\
y = \frac{{ - 2}}{{\tan \left( {1 - 2x} \right)}}\\
\Rightarrow y' = \frac{{\left( { - 2} \right)'.tan\left( {1 - 2x} \right) - \left( {\tan \left( {1 - 2x} \right)} \right)'.\left( { - 2} \right)}}{{{{\tan }^2}\left( {1 - 2x} \right)}}\\
= \frac{{2\left( {\tan \left( {1 - 2x} \right)} \right)'}}{{{{\tan }^2}\left( {1 - 2x} \right)}}\\
= \dfrac{{2.\frac{{\left( {1 - 2x} \right)'}}{{{{\cos }^2}\left( {1 - 2x} \right)}}}}{{\frac{{{{\sin }^2}\left( {1 - 2x} \right)}}{{{{\cos }^2}\left( {1 - 2x} \right)}}}}\\
= \dfrac{{\frac{{ - 4}}{{{{\cos }^2}\left( {1 - 2x} \right)}}}}{{\frac{{{{\sin }^2}\left( {1 - 2x} \right)}}{{{{\cos }^2}\left( {1 - 2x} \right)}}}}\\
= \frac{{ - 4}}{{{{\sin }^2}\left( {1 - 2x} \right)}}
\end{array}\)
