Đáp án:
i. \(\dfrac{{3 - 2x}}{{1 - x}}\)
Giải thích các bước giải:
\(\begin{array}{l}
i.y = \dfrac{{x + 1}}{{\sqrt {1 - x} }}\\
y' = \dfrac{{\sqrt {1 - x} - \dfrac{{ - 1}}{{2\sqrt {1 - x} }}}}{{1 - x}}\\
= \dfrac{{2\left( {1 - x} \right) + 1}}{{1 - x}}\\
= \dfrac{{2 - 2x + 1}}{{1 - x}}\\
= \dfrac{{3 - 2x}}{{1 - x}}\\
n.y' = \dfrac{{\sin x\left( {1 + \tan x} \right) + x.\cos x.\left( {1 + \tan x} \right) - \dfrac{1}{{{{\cos }^2}x}}.\left( {x.\cos x} \right)}}{{1 + \tan x}}\\
= \dfrac{{\sin x.\dfrac{{\sin x}}{{\cos x}} + x.\cos x + x.\cos x.\dfrac{{\sin x}}{{\cos x}} - \dfrac{{x.\cos x}}{{{{\cos }^2}x}}}}{{1 + \dfrac{{\sin x}}{{\cos x}}}}\\
= \dfrac{{\dfrac{{{{\sin }^2}x}}{{\cos x}} + x.\cos x + x.\sin x - \dfrac{x}{{\cos x}}}}{{\dfrac{{\sin x + \cos x}}{{\cos x}}}}\\
= \dfrac{{{{\sin }^2}x + x.{{\cos }^2}x + x.\sin x.\cos x - x.\cos x}}{{\cos x}}.\dfrac{{\cos x}}{{\sin x + \cos x}}\\
= \dfrac{{{{\sin }^2}x + x.{{\cos }^2}x + x.\sin x.\cos x - x.\cos x}}{{\sin x + \cos x}}
\end{array}\)
